1 00:00:00,120 --> 00:00:02,450 The following content is provided under a Creative 2 00:00:02,450 --> 00:00:03,830 Commons license. 3 00:00:03,830 --> 00:00:06,060 Your support will help MIT OpenCourseWare 4 00:00:06,060 --> 00:00:10,170 continue to offer high quality educational resources for free. 5 00:00:10,170 --> 00:00:12,710 To make a donation or to view additional materials 6 00:00:12,710 --> 00:00:16,620 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,620 --> 00:00:17,325 at ocw.mit.edu. 8 00:00:22,630 --> 00:00:23,800 PROFESSOR: Right. 9 00:00:23,800 --> 00:00:24,960 And we've done some review. 10 00:00:36,090 --> 00:00:37,130 Something else. 11 00:00:37,130 --> 00:00:38,046 AUDIENCE: [INAUDIBLE]. 12 00:00:42,880 --> 00:00:50,500 PROFESSOR: Yeah, symmetry, review principal axes ideas. 13 00:00:55,910 --> 00:00:56,420 What else? 14 00:01:07,310 --> 00:01:09,140 I think we spent quite a bit of time 15 00:01:09,140 --> 00:01:13,825 on generalized forces and ways for computing them. 16 00:01:39,270 --> 00:01:41,920 All right. 17 00:01:41,920 --> 00:01:46,069 And I'm trying to begin to summarize what we've 18 00:01:46,069 --> 00:01:47,610 done up to this point because we have 19 00:01:47,610 --> 00:01:49,730 a quiz coming next Tuesday. 20 00:01:49,730 --> 00:01:52,830 So some of yesterday's lecture was 21 00:01:52,830 --> 00:01:54,690 intended to kind of start pulling together, 22 00:01:54,690 --> 00:01:58,060 comparing direct versus Lagrange, advantages, 23 00:01:58,060 --> 00:01:59,900 disadvantages. 24 00:01:59,900 --> 00:02:03,360 And on Tuesday next, we'll do more examples. 25 00:02:03,360 --> 00:02:05,550 OK, so the problem you're going to work on today-- 26 00:02:05,550 --> 00:02:07,090 and you really are going to work in groups. 27 00:02:07,090 --> 00:02:08,298 And how many do we have here? 28 00:02:08,298 --> 00:02:13,380 We've got four-- about enough for three groups, let's say. 29 00:02:13,380 --> 00:02:17,170 We have a double pendulum. 30 00:02:17,170 --> 00:02:20,440 It's made out of two-- now I have two pieces of chalk. 31 00:02:24,440 --> 00:02:28,110 It's two slender rods, but an approximation-- this 32 00:02:28,110 --> 00:02:29,110 is a double pendulum. 33 00:02:29,110 --> 00:02:33,100 They have many-- that one and this one have a lot in common. 34 00:02:33,100 --> 00:02:36,755 Takes how many coordinates to completely describe the motion? 35 00:02:39,670 --> 00:02:41,450 One for each rod, right? 36 00:02:41,450 --> 00:02:44,360 And the coordinates up there that are drawn in that diagram 37 00:02:44,360 --> 00:02:47,640 is the angle of the first one with the vertical 38 00:02:47,640 --> 00:02:49,640 and the angle of the second one with a vertical. 39 00:02:49,640 --> 00:02:51,570 And with that, you can completely 40 00:02:51,570 --> 00:02:54,190 describe any allowable motion of the system. 41 00:02:54,190 --> 00:02:56,755 So a double pendulum has a property 42 00:02:56,755 --> 00:03:01,100 that it's got two natural frequencies and two mode 43 00:03:01,100 --> 00:03:01,600 shapes. 44 00:03:01,600 --> 00:03:04,160 That's the shape of the first mode. 45 00:03:04,160 --> 00:03:07,260 Both masses go in the same direction. 46 00:03:07,260 --> 00:03:10,990 And the second mode looks like that. 47 00:03:10,990 --> 00:03:14,080 The two masses go in opposite directions. 48 00:03:14,080 --> 00:03:17,532 Not with equal amplitudes, but opposite directions. 49 00:03:17,532 --> 00:03:18,990 And it's a different frequency too. 50 00:03:18,990 --> 00:03:20,030 It's higher. 51 00:03:20,030 --> 00:03:22,670 So it has two natural frequencies, two mode shapes. 52 00:03:22,670 --> 00:03:25,400 And if you had-- in this case it shows 53 00:03:25,400 --> 00:03:29,610 a force acting on the system, pushing this thing 54 00:03:29,610 --> 00:03:30,960 back and forth. 55 00:03:30,960 --> 00:03:37,800 And we're interested in the generalized forces, 56 00:03:37,800 --> 00:03:40,140 or thinking in terms of the work done 57 00:03:40,140 --> 00:03:41,880 by the non-conservative forces. 58 00:03:41,880 --> 00:03:49,750 So the exercise to do in groups-- like four or five 59 00:03:49,750 --> 00:03:50,970 of you are a group. 60 00:03:50,970 --> 00:03:52,270 You five are a group. 61 00:03:52,270 --> 00:03:57,840 And you two, four, five are a group there. 62 00:03:57,840 --> 00:03:59,080 Work this out. 63 00:03:59,080 --> 00:04:00,650 Find this vector. 64 00:04:00,650 --> 00:04:03,400 This is the position vector to the point 65 00:04:03,400 --> 00:04:06,370 of application of the force. 66 00:04:06,370 --> 00:04:08,520 And we want you to do that. 67 00:04:08,520 --> 00:04:10,170 Actually, I did something last time, 68 00:04:10,170 --> 00:04:12,780 and I'm almost forgetting to do it. 69 00:04:12,780 --> 00:04:14,880 I want to remind you of something. 70 00:04:14,880 --> 00:04:17,600 Professor Gossard, each week for the recitations-- 71 00:04:17,600 --> 00:04:21,430 he teaches Thursday ones-- does a write-up of the recitation. 72 00:04:21,430 --> 00:04:22,660 And they're posted. 73 00:04:22,660 --> 00:04:25,990 So this week the solution to this problem is posted. 74 00:04:25,990 --> 00:04:29,510 But also, a little quick review of the important stuff. 75 00:04:29,510 --> 00:04:30,980 So this would be posted. 76 00:04:30,980 --> 00:04:34,850 And basically I've redrawn it up here on the board. 77 00:04:34,850 --> 00:04:37,820 So this is this kinematic method for obtaining 78 00:04:37,820 --> 00:04:39,430 generalized forces. 79 00:04:39,430 --> 00:04:45,120 So a body with N forces on it. 80 00:04:45,120 --> 00:04:48,480 Here is the ith one. 81 00:04:48,480 --> 00:04:52,590 At each point of application of the force, 82 00:04:52,590 --> 00:04:56,850 there is a total virtual displacement 83 00:04:56,850 --> 00:04:58,820 that basically comes from the sum 84 00:04:58,820 --> 00:05:01,240 of all the individual generalized 85 00:05:01,240 --> 00:05:03,480 coordinate virtual motions. 86 00:05:03,480 --> 00:05:06,100 You sum them up, you get the total. 87 00:05:06,100 --> 00:05:11,960 And so therefore, at the point of application of each force, 88 00:05:11,960 --> 00:05:16,270 there is a total amount of virtual work that's done. 89 00:05:16,270 --> 00:05:18,780 So the total non-conservative work 90 00:05:18,780 --> 00:05:23,870 is the sum of the virtual work done 91 00:05:23,870 --> 00:05:27,710 at the point of application of every force summed over 92 00:05:27,710 --> 00:05:29,880 all the forces. 93 00:05:29,880 --> 00:05:32,820 And remember, inside of here, the total displacement 94 00:05:32,820 --> 00:05:36,450 at every location is a summation over of all the generalized 95 00:05:36,450 --> 00:05:37,300 coordinates. 96 00:05:37,300 --> 00:05:41,030 So it really ends up as a double sum, this dot product, 97 00:05:41,030 --> 00:05:43,960 the forces times delta ri. 98 00:05:43,960 --> 00:05:48,440 It's a double summation over the forces 99 00:05:48,440 --> 00:05:53,410 that are applied and over the generalized coordinates. 100 00:05:53,410 --> 00:05:56,850 But at the end of the day, what you end up doing, 101 00:05:56,850 --> 00:05:58,450 the calculation you want to do, is 102 00:05:58,450 --> 00:06:03,630 you need to know the generalized force-- the force associated 103 00:06:03,630 --> 00:06:05,970 with each generalized coordinate. 104 00:06:05,970 --> 00:06:10,240 And that's the summation of the contribution 105 00:06:10,240 --> 00:06:18,080 of the virtual work caused by that coordinate's motion dotted 106 00:06:18,080 --> 00:06:21,540 into the force at every location. 107 00:06:21,540 --> 00:06:24,730 So at every location, there's a little contribution 108 00:06:24,730 --> 00:06:27,500 to the total work done. 109 00:06:27,500 --> 00:06:32,190 And therefore, the total generalized force is this sum. 110 00:06:35,115 --> 00:06:40,570 So today's problem is easy in the sense that i is 1. 111 00:06:40,570 --> 00:06:43,140 You only have to deal with one force. 112 00:06:43,140 --> 00:06:44,980 But this is a little messier problem 113 00:06:44,980 --> 00:06:47,660 in that in order to do this, you need 114 00:06:47,660 --> 00:06:49,840 to be able-- this is the r sub i in this case. 115 00:06:49,840 --> 00:06:55,010 Here's point P. I want you to first, just in groups, 116 00:06:55,010 --> 00:07:00,180 work out what this vector is, rP, in the system as drawn. 117 00:07:00,180 --> 00:07:03,530 Here's the inertial system xy, generalized 118 00:07:03,530 --> 00:07:05,610 coordinates theta 1, theta 2. 119 00:07:05,610 --> 00:07:10,840 And this bar is L1 long, and this bar is L2 long. 120 00:07:10,840 --> 00:07:14,280 And come up with an answer in a form like this, something 121 00:07:14,280 --> 00:07:18,181 in the i hat direction plus something in the j hat 122 00:07:18,181 --> 00:07:18,680 direction. 123 00:07:18,680 --> 00:07:20,810 And then as your group finishes-- this is group 124 00:07:20,810 --> 00:07:22,830 one here-- come put your answer up. 125 00:07:22,830 --> 00:07:24,390 Second group, put your answer up. 126 00:07:24,390 --> 00:07:25,730 Third group, put your answer up. 127 00:07:25,730 --> 00:07:26,820 And we'll move on. 128 00:07:26,820 --> 00:07:29,628 This will go pretty fast. 129 00:07:29,628 --> 00:07:31,083 All right, looks like everybody's 130 00:07:31,083 --> 00:07:31,550 pretty much in agreement. 131 00:07:31,550 --> 00:07:32,508 Not much to talk about. 132 00:07:32,508 --> 00:07:33,900 So let's do the next piece. 133 00:07:33,900 --> 00:07:39,310 So the next part is to compute your variations, 134 00:07:39,310 --> 00:07:43,270 this part of the calculation, for each of the coordinates. 135 00:07:43,270 --> 00:07:53,500 So do rP with respect to theta 1 and rP with respect to theta 2 136 00:07:53,500 --> 00:07:56,280 and write down your answers. 137 00:08:00,770 --> 00:08:06,375 So the last step here is get q theta 1 q theta 2. 138 00:08:12,810 --> 00:08:15,530 So move on to getting the two generalized forces now. 139 00:08:20,590 --> 00:08:21,506 AUDIENCE: [INAUDIBLE]. 140 00:08:25,720 --> 00:08:26,595 PROFESSOR: You could. 141 00:08:26,595 --> 00:08:28,840 And then if you did, then this would 142 00:08:28,840 --> 00:08:32,419 have to be a summation over two components. 143 00:08:32,419 --> 00:08:33,780 Remember, this is a sum. 144 00:08:33,780 --> 00:08:35,690 It just happens in this example, we only 145 00:08:35,690 --> 00:08:38,171 have i goes from 1 to 1. 146 00:08:38,171 --> 00:08:42,400 But if there were two forces, then you would do this twice. 147 00:08:42,400 --> 00:08:47,480 You'd have an r1 with respect to theta 1 and an r2 with respect 148 00:08:47,480 --> 00:08:49,460 to theta 1. 149 00:08:49,460 --> 00:08:53,520 And you'd do this product twice, add the two pieces together, 150 00:08:53,520 --> 00:08:57,220 to get the total-- the total generalized force is 151 00:08:57,220 --> 00:09:00,950 the sum of the bits that come from all 152 00:09:00,950 --> 00:09:03,368 of the individual forces. 153 00:09:03,368 --> 00:09:05,813 AUDIENCE: So if [INAUDIBLE] just doing theta 1, 154 00:09:05,813 --> 00:09:08,747 but [INAUDIBLE] two forces, you just do one force, 155 00:09:08,747 --> 00:09:10,710 and then you do [INAUDIBLE] force-- 156 00:09:10,710 --> 00:09:12,450 PROFESSOR: Yeah, so now here let's 157 00:09:12,450 --> 00:09:17,550 have a B. We'll call it just B force. 158 00:09:17,550 --> 00:09:22,760 And it has to Bx components and By components. 159 00:09:22,760 --> 00:09:28,172 You would now also have to compute what would be rB. 160 00:09:28,172 --> 00:09:29,350 You would find it. 161 00:09:29,350 --> 00:09:32,980 Then you'd have rB would be this vector, which 162 00:09:32,980 --> 00:09:37,910 is just that piece plus this piece, right? 163 00:09:37,910 --> 00:09:42,250 So you'd have just those-- rB would be this plus this, 164 00:09:42,250 --> 00:09:44,370 and then you would do the derivative of rB 165 00:09:44,370 --> 00:09:45,830 with respect to theta 1. 166 00:09:45,830 --> 00:09:46,930 And you'd get something. 167 00:09:46,930 --> 00:09:50,152 And derivative of rB with respect to theta 2. 168 00:09:50,152 --> 00:09:54,232 AUDIENCE: So then you'd just do the rB equals [INAUDIBLE] 169 00:09:54,232 --> 00:09:58,224 for one force, and then you do the rPe [INAUDIBLE]? 170 00:09:58,224 --> 00:09:58,890 PROFESSOR: Yeah. 171 00:09:58,890 --> 00:10:01,650 So you have two-- so this thing would end up 172 00:10:01,650 --> 00:10:03,870 being the summation of two contributions, right? 173 00:10:03,870 --> 00:10:16,710 This would look like an FB dot rB with respect to-- 174 00:10:16,710 --> 00:10:19,400 and this is just for just one of them. 175 00:10:19,400 --> 00:10:21,250 So we'll do the theta 1. 176 00:10:21,250 --> 00:10:27,165 With respect to theta 1 plus F-- I'm 177 00:10:27,165 --> 00:10:35,350 just calling this one F-- dot derivative rP with respect 178 00:10:35,350 --> 00:10:37,780 to theta 1. 179 00:10:37,780 --> 00:10:42,920 So you get two contributions from the two forces 180 00:10:42,920 --> 00:10:45,110 give you the total q. 181 00:10:45,110 --> 00:10:47,070 This would be q theta 1. 182 00:10:47,070 --> 00:10:49,920 And then you'd do it again for the theta 2. 183 00:10:49,920 --> 00:10:51,715 And you'd have two possible pieces. 184 00:10:58,882 --> 00:11:02,900 AUDIENCE: So you try to find r's for every force [INAUDIBLE]. 185 00:11:02,900 --> 00:11:05,420 PROFESSOR: At every point of application of a force, 186 00:11:05,420 --> 00:11:07,820 you define an r. 187 00:11:07,820 --> 00:11:12,875 You eventually need to do this summation for every force. 188 00:11:17,320 --> 00:11:17,820 All right. 189 00:11:17,820 --> 00:11:20,230 Everybody's got the same answers. 190 00:11:20,230 --> 00:11:21,790 So now I'm going to try to-- this 191 00:11:21,790 --> 00:11:24,670 is an indirect way of answering your question. 192 00:11:24,670 --> 00:11:28,910 I'm now going to try to convince you that you're wrong. 193 00:11:28,910 --> 00:11:33,172 And I want you to tell me why my logic is flawed. 194 00:11:33,172 --> 00:11:36,120 AUDIENCE: No. 195 00:11:36,120 --> 00:11:37,750 PROFESSOR: OK? 196 00:11:37,750 --> 00:11:44,520 So I look at these answers here, and so this 197 00:11:44,520 --> 00:11:49,940 says that if-- this line right here, if I caused 198 00:11:49,940 --> 00:11:57,190 a little delta theta 1 here, that swings this arm 199 00:11:57,190 --> 00:11:59,360 through that amount. 200 00:11:59,360 --> 00:12:06,540 And it moves a distance in the x direction, L1 cosine theta 201 00:12:06,540 --> 00:12:07,470 delta theta 1. 202 00:12:07,470 --> 00:12:10,110 And it moves over that little bit right here, right? 203 00:12:10,110 --> 00:12:10,730 OK. 204 00:12:10,730 --> 00:12:14,560 And you're telling me that this point over here 205 00:12:14,560 --> 00:12:16,900 moves that amount. 206 00:12:16,900 --> 00:12:21,030 The point of application of this force 207 00:12:21,030 --> 00:12:27,980 due to delta x, delta theta 1, is this delta theta 1. 208 00:12:27,980 --> 00:12:30,770 There's no L2 involved. 209 00:12:30,770 --> 00:12:37,600 So to me, it seems like here's this pendulum. 210 00:12:37,600 --> 00:12:44,150 And if I move this first piece by a little bit, delta theta 1, 211 00:12:44,150 --> 00:12:46,280 this piece also moves. 212 00:12:46,280 --> 00:12:48,370 And so if you just think of a straight piece, 213 00:12:48,370 --> 00:12:51,200 delta theta 1 times L1 moves a little bit. 214 00:12:51,200 --> 00:12:55,050 But down here it moves like twice as much. 215 00:12:55,050 --> 00:12:59,390 So I think that when you move this little delta theta 216 00:12:59,390 --> 00:13:03,880 one, that you ought to get even more motion down here. 217 00:13:03,880 --> 00:13:08,330 And yet your answers say that that's not true. 218 00:13:08,330 --> 00:13:13,096 Tell me why-- what's flawed in my reasoning? 219 00:13:13,096 --> 00:13:15,576 AUDIENCE: So when you-- I'll try. 220 00:13:15,576 --> 00:13:19,874 When you [INAUDIBLE] theta 2, [INAUDIBLE]. 221 00:13:19,874 --> 00:13:21,770 And now when you move by theta 1, 222 00:13:21,770 --> 00:13:25,090 [INAUDIBLE] all points move the same theta. 223 00:13:25,090 --> 00:13:26,232 So even though [INAUDIBLE]. 224 00:13:31,050 --> 00:13:31,790 PROFESSOR: OK. 225 00:13:31,790 --> 00:13:35,220 So you're saying that we argue that when 226 00:13:35,220 --> 00:13:39,890 you have one virtual displacement, or in this case 227 00:13:39,890 --> 00:13:46,630 a rotation delta theta 1, we freeze all other generalized 228 00:13:46,630 --> 00:13:47,530 coordinates. 229 00:13:47,530 --> 00:13:51,970 That means we freeze delta theta 2 and don't let it change. 230 00:13:51,970 --> 00:13:55,420 Delta theta 2 is measured with respect to the vertical, right? 231 00:13:55,420 --> 00:13:59,090 So if that moves over delta theta 1, 232 00:13:59,090 --> 00:14:01,500 this whole thing would-- this angle here 233 00:14:01,500 --> 00:14:04,980 we go from delta theta 2 to delta theta 2 plus delta theta 234 00:14:04,980 --> 00:14:08,630 1 if my argument's correct. 235 00:14:08,630 --> 00:14:12,160 But for my argument to be correct, 236 00:14:12,160 --> 00:14:14,730 that would have to swing by delta theta 1 over here 237 00:14:14,730 --> 00:14:15,530 as well. 238 00:14:15,530 --> 00:14:19,990 But the fact that we freeze theta 2, in fact, 239 00:14:19,990 --> 00:14:24,720 the angle between these two things has to change. 240 00:14:24,720 --> 00:14:26,340 This has to open up a little bit. 241 00:14:26,340 --> 00:14:27,740 So when this is moving over, this 242 00:14:27,740 --> 00:14:30,670 is dropping down so that this point, this whole body 243 00:14:30,670 --> 00:14:31,620 only translates. 244 00:14:31,620 --> 00:14:34,890 So that whoever described it as just pure translation 245 00:14:34,890 --> 00:14:38,310 of the second body, that's another way of saying it. 246 00:14:38,310 --> 00:14:41,570 So you've got to remember what it means, 247 00:14:41,570 --> 00:14:43,940 freezing all the other generalized coordinates 248 00:14:43,940 --> 00:14:46,920 and only allowing the one you picked to move. 249 00:14:46,920 --> 00:14:48,000 Good. 250 00:14:48,000 --> 00:14:48,660 All right. 251 00:14:48,660 --> 00:14:50,520 Nice work. 252 00:14:50,520 --> 00:14:52,190 We've got some time left. 253 00:14:52,190 --> 00:14:55,370 So do you have any questions from problem 254 00:14:55,370 --> 00:14:57,950 sets, lectures, just-- 255 00:14:57,950 --> 00:14:59,450 AUDIENCE: [INAUDIBLE]. 256 00:14:59,450 --> 00:15:01,575 PROFESSOR: Yeah. 257 00:15:01,575 --> 00:15:02,950 AUDIENCE: When you're calculating 258 00:15:02,950 --> 00:15:04,450 kinetic energy of a rotating body, 259 00:15:04,450 --> 00:15:05,890 there's an omega [INAUDIBLE]. 260 00:15:09,610 --> 00:15:11,750 In the homework, the last homework, 261 00:15:11,750 --> 00:15:13,690 I didn't know which omega could take, 262 00:15:13,690 --> 00:15:16,200 whether to take it from the wheel or from the axis. 263 00:15:16,200 --> 00:15:18,417 PROFESSOR: Oh, this is the rotating thing with the-- 264 00:15:18,417 --> 00:15:19,000 AUDIENCE: Yes. 265 00:15:19,000 --> 00:15:19,700 PROFESSOR: Ah, OK. 266 00:15:19,700 --> 00:15:21,158 AUDIENCE: There's two omegas there, 267 00:15:21,158 --> 00:15:23,870 and I didn't know which omega [INAUDIBLE]. 268 00:15:23,870 --> 00:15:27,890 PROFESSOR: So should we just talk about that problem? 269 00:15:27,890 --> 00:15:32,830 Are other people-- this is the one where you had basically 270 00:15:32,830 --> 00:15:36,120 the rod, and on the end of the rod 271 00:15:36,120 --> 00:15:39,350 was the wheel, which was rotating. 272 00:15:39,350 --> 00:15:44,940 So this was going around at some omega 1 in the z, right? 273 00:15:44,940 --> 00:15:47,750 And this was rotating, and I think 274 00:15:47,750 --> 00:15:55,550 this was like z, maybe x and y. 275 00:15:55,550 --> 00:15:56,800 So it was rotating about that. 276 00:15:56,800 --> 00:15:58,632 AUDIENCE: [INAUDIBLE] it had a negative. 277 00:15:58,632 --> 00:16:00,010 It was a negative [INAUDIBLE]. 278 00:16:00,010 --> 00:16:01,301 PROFESSOR: Oh, it was a-- yeah. 279 00:16:01,301 --> 00:16:06,350 Actually, I think I have it here with its correct definition. 280 00:16:06,350 --> 00:16:08,440 So rather than try to-- 281 00:16:08,440 --> 00:16:11,821 AUDIENCE: Yeah, because there was the positive [INAUDIBLE], 282 00:16:11,821 --> 00:16:13,742 and then there's the [INAUDIBLE]. 283 00:16:13,742 --> 00:16:14,658 PROFESSOR: Here it is. 284 00:16:14,658 --> 00:16:16,860 Let's get the picture right. 285 00:16:29,940 --> 00:16:37,930 And there was an x, a y, and a z. 286 00:16:37,930 --> 00:16:41,205 And this was cap omega. 287 00:16:44,170 --> 00:16:47,670 And this was omega 1. 288 00:16:47,670 --> 00:16:51,210 And that's minus, in minus direction, 289 00:16:51,210 --> 00:16:53,000 for this system, right? 290 00:16:53,000 --> 00:16:55,560 OK. 291 00:16:55,560 --> 00:17:05,670 So then we're trying to compute T for this system. 292 00:17:09,210 --> 00:17:12,829 1/2, and what'd we call for the masses here? 293 00:17:12,829 --> 00:17:17,710 This was m1, and this was m2, I guess. 294 00:17:17,710 --> 00:17:19,140 Yeah. 295 00:17:19,140 --> 00:17:25,280 So this T for 2-- T1 you don't have a problem with, right? 296 00:17:25,280 --> 00:17:28,820 T1 is just a shaft, just a rod. 297 00:17:28,820 --> 00:17:34,100 It's pivoting about its end 1/2i about-- if this 298 00:17:34,100 --> 00:17:36,930 is-- I call it o. 299 00:17:36,930 --> 00:17:40,510 1/2i, i with respect to o, mL squared 300 00:17:40,510 --> 00:17:44,760 over 3 times that squared. 301 00:17:44,760 --> 00:17:45,370 You're done. 302 00:17:45,370 --> 00:17:47,200 That's the energy of this part. 303 00:17:47,200 --> 00:17:49,310 This is the problematic one, right? 304 00:17:49,310 --> 00:17:50,320 The piece of it. 305 00:17:50,320 --> 00:17:55,830 So 1/2, I would say mass 2. 306 00:17:55,830 --> 00:17:57,150 And this does have a point. 307 00:17:57,150 --> 00:18:00,110 They called this point B here. 308 00:18:00,110 --> 00:18:03,410 Velocity of B dot velocity of B. 309 00:18:03,410 --> 00:18:06,600 So that's the 1/2mv squared piece of it. 310 00:18:06,600 --> 00:18:21,240 Plus 1/2 omega dot h with respect to G 311 00:18:21,240 --> 00:18:26,270 for-- this is mass 2. 312 00:18:26,270 --> 00:18:28,460 So you have a contribution to the kinetic energy 313 00:18:28,460 --> 00:18:33,700 of this that comes from its center of gravity translating, 314 00:18:33,700 --> 00:18:38,320 and a contribution from the rotor rotating with respect 315 00:18:38,320 --> 00:18:42,260 to its center of mass here, G here. 316 00:18:45,610 --> 00:18:47,170 And that'd be the expression you use. 317 00:18:47,170 --> 00:18:56,420 And then the problem is, what omega is the trick here, right? 318 00:18:56,420 --> 00:18:57,050 OK. 319 00:18:57,050 --> 00:19:00,000 This is the one that had the concept question I talked 320 00:19:00,000 --> 00:19:01,730 about in class because it came up, 321 00:19:01,730 --> 00:19:05,630 and I hadn't thought about this, about the body 322 00:19:05,630 --> 00:19:10,660 coordinates attached to this are actually rotating with this. 323 00:19:10,660 --> 00:19:15,326 And that gets messy trying to figure out 324 00:19:15,326 --> 00:19:16,700 what are the-- how do you break-- 325 00:19:16,700 --> 00:19:19,810 you're supposed to express the angular 326 00:19:19,810 --> 00:19:21,800 rotation in the body coordinates. 327 00:19:27,500 --> 00:19:30,960 But this angular rotation, this one 328 00:19:30,960 --> 00:19:36,960 actually is lined up with this axis. 329 00:19:36,960 --> 00:19:39,250 And you could have body coordinates on here. 330 00:19:39,250 --> 00:19:45,420 You could have a-- if this is a y, 331 00:19:45,420 --> 00:19:55,310 you could have a body coordinate x and z that rotate with it. 332 00:19:55,310 --> 00:20:02,450 But this Component would just be still along this axis, right? 333 00:20:02,450 --> 00:20:10,170 So the concept question [INAUDIBLE] 334 00:20:10,170 --> 00:20:15,700 was basically, are these body coordinates, x, y, z, 335 00:20:15,700 --> 00:20:18,770 attached to the rod? 336 00:20:18,770 --> 00:20:23,040 Principal axes for this body. 337 00:20:23,040 --> 00:20:28,650 And the complication here is that this body is rotating. 338 00:20:28,650 --> 00:20:34,570 And so this x and this z don't rotate with that body. 339 00:20:34,570 --> 00:20:39,620 And so they're not body fixed coordinates for that body. 340 00:20:39,620 --> 00:20:42,980 But in fact, they're still principal coordinates 341 00:20:42,980 --> 00:20:44,070 for that body. 342 00:20:44,070 --> 00:20:45,350 Why is that true? 343 00:20:48,284 --> 00:20:51,710 AUDIENCE: Because you have [INAUDIBLE]. 344 00:20:51,710 --> 00:20:55,601 And you can take whatever you want for the [INAUDIBLE]. 345 00:20:55,601 --> 00:20:56,600 PROFESSOR: You're close. 346 00:20:56,600 --> 00:21:01,930 This body is axially symmetric. 347 00:21:01,930 --> 00:21:05,330 And so at any instant in time, you 348 00:21:05,330 --> 00:21:07,900 could say that-- you could define 349 00:21:07,900 --> 00:21:10,640 a principal coordinate on this body that 350 00:21:10,640 --> 00:21:12,240 is lined up with these. 351 00:21:16,520 --> 00:21:19,484 So for axially symmetric bodies, you could do that. 352 00:21:19,484 --> 00:21:20,150 You can do that. 353 00:21:20,150 --> 00:21:21,858 You can just say, at an incident in time, 354 00:21:21,858 --> 00:21:24,620 let's compute the kinetic energy. 355 00:21:24,620 --> 00:21:30,140 And we have the rotation of that body 356 00:21:30,140 --> 00:21:34,970 defined in the instantaneous principal coordinates 357 00:21:34,970 --> 00:21:36,290 of the body. 358 00:21:36,290 --> 00:21:43,540 So now omega for the second body is 359 00:21:43,540 --> 00:21:58,050 this in the k minus omega 1 in the j of the x, y, z system 360 00:21:58,050 --> 00:22:01,680 rotating with the rod. 361 00:22:01,680 --> 00:22:06,870 And then to complete this, what's H? 362 00:22:14,130 --> 00:22:30,650 So H with respect to G is Ixx Iyy Izz times-- 363 00:22:30,650 --> 00:22:38,570 and now this one's rotation is 0 minus omega 1 and cap omega. 364 00:22:42,570 --> 00:22:46,610 And I multiply that times this with the definition 365 00:22:46,610 --> 00:22:49,640 that the first one gives me the I, second one J. 366 00:22:49,640 --> 00:22:52,990 And so I end up with two pieces. 367 00:22:52,990 --> 00:22:54,730 The first one is zero. 368 00:22:54,730 --> 00:22:57,150 I get an Iyy. 369 00:22:57,150 --> 00:22:59,240 And remember, these are defined with respect 370 00:22:59,240 --> 00:23:03,130 to center of mass of the disk. 371 00:23:03,130 --> 00:23:17,200 Iyy minus Iyy omega 1 j plus Izz cap omega k. 372 00:23:17,200 --> 00:23:20,340 And that'd be the angular momentum for the disk. 373 00:23:23,120 --> 00:23:23,890 Right? 374 00:23:23,890 --> 00:23:34,400 And then the kinetic energy, the 1/2 omega dot H, 375 00:23:34,400 --> 00:23:41,110 you now just have another omega 1/2, and you have a vector, 376 00:23:41,110 --> 00:23:49,530 0 minus omega 1 J cap omega k dotted 377 00:23:49,530 --> 00:24:06,080 with the H vector, which is your 0 Iyy omega 1 378 00:24:06,080 --> 00:24:10,540 j and Izz cap omega k. 379 00:24:10,540 --> 00:24:13,359 AUDIENCE: I have a question. [INAUDIBLE]. 380 00:24:13,359 --> 00:24:14,025 PROFESSOR: Yeah. 381 00:24:14,025 --> 00:24:18,610 And if that one's negative, then it'll 382 00:24:18,610 --> 00:24:20,826 happily fix this one because it ought 383 00:24:20,826 --> 00:24:23,300 to come out looking like omega squareds, right? 384 00:24:23,300 --> 00:24:32,500 And so out of this, you get Iyy omega 1 squared plus Izz cap 385 00:24:32,500 --> 00:24:33,425 omega squared. 386 00:24:48,370 --> 00:24:54,440 And that's the second piece, and you add it to this piece. 387 00:24:54,440 --> 00:24:57,860 AUDIENCE: So is there ever a time where the omega's inside 388 00:24:57,860 --> 00:24:58,360 [INAUDIBLE]. 389 00:25:19,200 --> 00:25:37,820 PROFESSOR: So let me-- we have a total-- so let's just 390 00:25:37,820 --> 00:25:48,846 talk-- we have H with respect to G. 391 00:25:48,846 --> 00:25:50,990 Or a slightly simpler example. 392 00:26:28,770 --> 00:26:32,490 So H for this system with respect to the center 393 00:26:32,490 --> 00:26:39,720 is H with respect to G plus rG with respect 394 00:26:39,720 --> 00:26:44,740 to the center cross PG, right? 395 00:26:44,740 --> 00:26:47,720 I can write the angular momentum that way. 396 00:26:47,720 --> 00:26:52,950 And first piece we know. 397 00:26:52,950 --> 00:27:01,060 The second piece, rG, is in a r hat direction, if you will. 398 00:27:01,060 --> 00:27:06,300 The P is going around the circle. 399 00:27:06,300 --> 00:27:09,630 And so the cross product of those two 400 00:27:09,630 --> 00:27:13,280 gives me a component, what's the direction of the result. 401 00:27:13,280 --> 00:27:21,110 So this is r cross theta hat. 402 00:27:21,110 --> 00:27:22,850 You get a k. 403 00:27:22,850 --> 00:27:26,150 This ends up being-- this gives you a k component. 404 00:27:26,150 --> 00:27:31,590 And this one we write out. 405 00:27:31,590 --> 00:27:48,350 So this is minus Iyy omega 1 j plus Izz cap omega k 406 00:27:48,350 --> 00:27:54,080 plus some stuff that's in the k. 407 00:27:54,080 --> 00:27:56,400 Agreed? 408 00:27:56,400 --> 00:28:02,680 And I want to know the torques required to make 409 00:28:02,680 --> 00:28:05,750 this system do what it's doing. 410 00:28:05,750 --> 00:28:08,770 How do I get the torques required 411 00:28:08,770 --> 00:28:11,620 to make the system go around? 412 00:28:15,412 --> 00:28:16,644 AUDIENCE: [INAUDIBLE]. 413 00:28:16,644 --> 00:28:17,310 PROFESSOR: Yeah. 414 00:28:17,310 --> 00:28:24,760 Sum of the external torques equals dH. 415 00:28:24,760 --> 00:28:28,540 Now, with respect-- remember, we would ordinarily usually write 416 00:28:28,540 --> 00:28:33,130 that as an A. We'll call it H with respect to O dt 417 00:28:33,130 --> 00:28:36,910 plus velocity-- usually we call it A-- with respect to O, 418 00:28:36,910 --> 00:28:39,800 cross P. Right? 419 00:28:39,800 --> 00:28:41,590 In this system, what's the velocity 420 00:28:41,590 --> 00:28:43,140 of this piece, this term? 421 00:28:43,140 --> 00:28:43,640 Zero. 422 00:28:43,640 --> 00:28:45,690 So we don't have to worry about it. 423 00:28:45,690 --> 00:28:47,740 So the sum of the external torques 424 00:28:47,740 --> 00:28:52,320 should just be the time derivative of this angular 425 00:28:52,320 --> 00:28:55,500 momentum vector. 426 00:28:55,500 --> 00:28:57,400 Right? 427 00:28:57,400 --> 00:29:12,810 And dH then O dt is equal to the partial derivative with respect 428 00:29:12,810 --> 00:29:16,705 to t of H. And I'll write it like this. 429 00:29:16,705 --> 00:29:18,365 This is in the rotating frame. 430 00:29:22,700 --> 00:29:27,040 So this is the piece from inside the frame plus, 431 00:29:27,040 --> 00:29:29,900 this is just the derivative of a rotating vector. 432 00:29:29,900 --> 00:29:40,060 Omega cross H. And the issue here is, what's the omega? 433 00:29:40,060 --> 00:29:42,040 That's really what this question comes down to. 434 00:29:53,340 --> 00:29:55,450 When we did this problem, we found 435 00:29:55,450 --> 00:30:01,250 that H came from taking the omega, the vector, 436 00:30:01,250 --> 00:30:04,950 multiplying it by this, and we got these two pieces. 437 00:30:04,950 --> 00:30:08,930 And so it started off with components in the j and k 438 00:30:08,930 --> 00:30:11,920 direction, and it came out with components in the j and k 439 00:30:11,920 --> 00:30:15,510 because this was diagonal. 440 00:30:15,510 --> 00:30:17,260 So we have principal axes. 441 00:30:17,260 --> 00:30:19,860 We came out with these two pieces. 442 00:30:19,860 --> 00:30:22,360 So the angular momentum has a j component and a k component. 443 00:30:25,190 --> 00:30:27,350 What is actually-- now, this vector, 444 00:30:27,350 --> 00:30:29,740 this angular momentum vector, the reason 445 00:30:29,740 --> 00:30:34,480 we have this second term is because of the change 446 00:30:34,480 --> 00:30:36,610 of direction of a vector. 447 00:30:36,610 --> 00:30:40,270 This piece comes from taking the time derivatives of the unit 448 00:30:40,270 --> 00:30:41,880 vectors in the problem. 449 00:30:41,880 --> 00:30:44,190 The time derivatives of the other stuff in the problem 450 00:30:44,190 --> 00:30:46,110 has been taken care of by this. 451 00:30:46,110 --> 00:30:48,880 So this only deals with the time derivatives of the unit 452 00:30:48,880 --> 00:30:51,150 vectors. 453 00:30:51,150 --> 00:31:00,190 So what is the actual rotation rate 454 00:31:00,190 --> 00:31:05,070 in this problem, the rate at which things 455 00:31:05,070 --> 00:31:06,650 are changing direction? 456 00:31:06,650 --> 00:31:10,240 Which one's changing direction? 457 00:31:10,240 --> 00:31:11,406 AUDIENCE: [INAUDIBLE]. 458 00:31:11,406 --> 00:31:13,155 PROFESSOR: Is this one changing direction? 459 00:31:13,155 --> 00:31:13,580 No. 460 00:31:13,580 --> 00:31:15,450 So actually, it doesn't give you any derivative, right? 461 00:31:15,450 --> 00:31:16,810 Is this one changing direction? 462 00:31:16,810 --> 00:31:17,310 Yeah. 463 00:31:21,980 --> 00:31:33,410 And at what rotation rate is it changing direction? 464 00:31:33,410 --> 00:31:35,549 AUDIENCE: [INAUDIBLE]. 465 00:31:35,549 --> 00:31:36,590 PROFESSOR: Capital omega. 466 00:31:36,590 --> 00:31:44,810 But why not-- but it's not changing little omega, is it? 467 00:31:44,810 --> 00:31:49,180 The math, the vectors works that all out for you. 468 00:31:49,180 --> 00:31:53,920 Because this is in the j direction, and this 469 00:31:53,920 --> 00:31:55,030 is the same. 470 00:31:55,030 --> 00:31:59,540 You just use the same total rotation vector here. 471 00:31:59,540 --> 00:32:02,440 And this will have in it the zero, 472 00:32:02,440 --> 00:32:09,430 the omega minus omega 1 j, the plus cap omega k. 473 00:32:09,430 --> 00:32:13,450 And the parts that you don't care-- this 474 00:32:13,450 --> 00:32:16,290 crossed with itself goes to zero. 475 00:32:16,290 --> 00:32:21,390 So even though you're just going ahead and leaving this in here, 476 00:32:21,390 --> 00:32:26,400 it doesn't result in anything because the cross product 477 00:32:26,400 --> 00:32:28,940 would-- cross product with itself gives you nothing. 478 00:32:28,940 --> 00:32:32,100 So the only piece that actually contributes nonzero 479 00:32:32,100 --> 00:32:35,510 contribution to this answer is this crossed 480 00:32:35,510 --> 00:32:39,490 with the pieces in here. 481 00:32:39,490 --> 00:32:43,940 So when you're taking derivatives 482 00:32:43,940 --> 00:32:47,640 of rotating vectors, remember the original formula. 483 00:32:47,640 --> 00:32:50,290 And the original formula is the derivative 484 00:32:50,290 --> 00:32:53,330 of the vector in the rotating frame-- 485 00:32:53,330 --> 00:32:57,090 is this magnitude getting more or less-- 486 00:32:57,090 --> 00:33:00,869 plus just the rotation rate crossed 487 00:33:00,869 --> 00:33:01,910 with the original vector. 488 00:33:04,640 --> 00:33:07,670 And the pieces that are common just fall out. 489 00:33:07,670 --> 00:33:12,600 The piece of this that is due to its own rotation 490 00:33:12,600 --> 00:33:13,780 doesn't enter into it. 491 00:33:13,780 --> 00:33:17,840 This rotation doesn't contribute to that d 492 00:33:17,840 --> 00:33:19,205 by dt of the rotation. 493 00:33:21,792 --> 00:33:22,750 Just works out for you. 494 00:33:22,750 --> 00:33:23,290 Yeah. 495 00:33:23,290 --> 00:33:26,194 AUDIENCE: What would happen if that just [INAUDIBLE] 496 00:33:26,194 --> 00:33:29,098 rotates like this, but also rotates on its own axis 497 00:33:29,098 --> 00:33:31,520 as it does it? 498 00:33:31,520 --> 00:33:34,092 PROFESSOR: You mean if this one is going in the k? 499 00:33:34,092 --> 00:33:35,940 AUDIENCE: Yeah. 500 00:33:35,940 --> 00:33:37,682 PROFESSOR: Then-- 501 00:33:37,682 --> 00:33:42,056 AUDIENCE: [INAUDIBLE] you didn't have the big omega term 502 00:33:42,056 --> 00:33:45,570 because the actual thing isn't-- the big omega would be taken 503 00:33:45,570 --> 00:33:50,380 into account when looking at the first term of kinetic energy. 504 00:33:50,380 --> 00:33:55,720 PROFESSOR: If this one also had some k rotation in addition 505 00:33:55,720 --> 00:34:02,500 relative to here, we'll call it omega 2 in the k direction, 506 00:34:02,500 --> 00:34:05,840 then what's the total rotation rate in the k 507 00:34:05,840 --> 00:34:07,422 direction for the disk? 508 00:34:07,422 --> 00:34:09,030 AUDIENCE: [INAUDIBLE]. 509 00:34:09,030 --> 00:34:12,440 PROFESSOR: You add cap omega, 2 omega 2, 510 00:34:12,440 --> 00:34:17,489 and you'd have its total k directed rotation, right? 511 00:34:17,489 --> 00:34:20,350 And if you did that, you'd end up in here 512 00:34:20,350 --> 00:34:26,330 with an omega plus omega-- cap omega plus omega 2 k. 513 00:34:29,480 --> 00:34:36,550 But when you came over to do the time derivative, 514 00:34:36,550 --> 00:34:38,917 you just leave all those things in there. 515 00:34:38,917 --> 00:34:40,500 And the ones that you don't care about 516 00:34:40,500 --> 00:34:42,110 will just cancel out because they're 517 00:34:42,110 --> 00:34:45,498 cross products with themselves. 518 00:34:45,498 --> 00:34:49,929 AUDIENCE: [INAUDIBLE] omega also appears [INAUDIBLE]. 519 00:34:49,929 --> 00:34:51,900 PROFESSOR: Oh yeah. 520 00:34:51,900 --> 00:34:54,800 Well, the cap omega appears in this piece. 521 00:34:58,040 --> 00:35:00,060 But this goes to k. 522 00:35:00,060 --> 00:35:01,784 We did a particular problem. 523 00:35:01,784 --> 00:35:03,450 I'm not trying to prove this in general, 524 00:35:03,450 --> 00:35:05,241 although there's probably a way to do that. 525 00:35:05,241 --> 00:35:10,110 This problem, this term is in the k direction. 526 00:35:10,110 --> 00:35:15,270 And therefore, k hat does not change direction with time. 527 00:35:15,270 --> 00:35:19,980 Therefore, the dH dt of this piece goes to zero. 528 00:35:19,980 --> 00:35:22,370 So we only had to deal with this part 529 00:35:22,370 --> 00:35:25,080 in taking the time derivative that dealt 530 00:35:25,080 --> 00:35:28,300 with changes in direction. 531 00:35:28,300 --> 00:35:29,830 It's only the changes in direction 532 00:35:29,830 --> 00:35:32,140 that the omega cross something matters, right? 533 00:35:36,790 --> 00:35:37,700 Good question. 534 00:35:37,700 --> 00:35:40,630 Something else? 535 00:35:40,630 --> 00:35:41,435 Recent problems. 536 00:35:48,440 --> 00:35:51,466 Somebody in the last class asked about, 537 00:35:51,466 --> 00:35:55,275 had a question about this thing. 538 00:35:55,275 --> 00:35:57,830 That bring anything to memory? 539 00:35:57,830 --> 00:35:59,690 AUDIENCE: [INAUDIBLE]. 540 00:35:59,690 --> 00:36:01,780 PROFESSOR: I see some grimaces. 541 00:36:01,780 --> 00:36:05,230 What troubled you about that problem? 542 00:36:05,230 --> 00:36:08,460 What troubled the last group about that problem was the 543 00:36:08,460 --> 00:36:09,210 posted solution. 544 00:36:11,890 --> 00:36:14,300 They didn't understand the posted solution. 545 00:36:14,300 --> 00:36:18,390 So I ran through-- I showed why the posted 546 00:36:18,390 --> 00:36:20,390 solution works the way it does. 547 00:36:20,390 --> 00:36:21,890 AUDIENCE: Well, one confusion that I 548 00:36:21,890 --> 00:36:26,043 had had doing this problem was that the spring for the wheel 549 00:36:26,043 --> 00:36:31,610 mass, when that exerts displacements, Newton's 550 00:36:31,610 --> 00:36:34,120 third being what it is, should it also 551 00:36:34,120 --> 00:36:36,165 exert some force on the large base mass 552 00:36:36,165 --> 00:36:38,680 as well, so you should take that into account 553 00:36:38,680 --> 00:36:40,910 for the total displacemenet caused by springs 554 00:36:40,910 --> 00:36:44,452 for big mass, the big-- 555 00:36:44,452 --> 00:36:46,660 PROFESSOR: Are you talking about not computing-- just 556 00:36:46,660 --> 00:36:49,215 giving equations of motion, or computing generalized forces, 557 00:36:49,215 --> 00:36:50,470 or what's the context? 558 00:36:50,470 --> 00:36:51,605 AUDIENCE: Yes, when you're computing 559 00:36:51,605 --> 00:36:52,670 your equations of motion, you have 560 00:36:52,670 --> 00:36:54,390 to take into account from the larger 561 00:36:54,390 --> 00:36:56,200 mass of the displacements in the spring, 562 00:36:56,200 --> 00:36:58,040 it's attached to the wall, [INAUDIBLE]. 563 00:36:58,040 --> 00:37:00,286 PROFESSOR: As well as the-- yeah. 564 00:37:00,286 --> 00:37:04,040 So if you're doing this by the direct method, 565 00:37:04,040 --> 00:37:06,820 you have to figure that out. 566 00:37:06,820 --> 00:37:08,370 So here's a way of doing it. 567 00:37:08,370 --> 00:37:11,050 This is a way that I think the-- the way the student 568 00:37:11,050 --> 00:37:12,830 just read off his paper. 569 00:37:12,830 --> 00:37:17,440 So I said, let's talk about the forces in the x direction 570 00:37:17,440 --> 00:37:19,690 on the main mass. 571 00:37:19,690 --> 00:37:23,410 Well, that's mass 1 times its acceleration 572 00:37:23,410 --> 00:37:25,940 must be equal to all the external forces. 573 00:37:25,940 --> 00:37:28,663 This is minus kx minus bx dot. 574 00:37:28,663 --> 00:37:34,590 This is minus the x component of the spring force 575 00:37:34,590 --> 00:37:37,840 minus the x component of the friction 576 00:37:37,840 --> 00:37:43,500 force from the wheel minus the x component of any normal force 577 00:37:43,500 --> 00:37:45,620 from the wheel. 578 00:37:45,620 --> 00:37:47,895 And I do the same thing for the second mass, M2. 579 00:37:50,570 --> 00:37:53,490 The total forces on it must be equal to its mass 580 00:37:53,490 --> 00:37:54,640 times its acceleration. 581 00:37:54,640 --> 00:37:56,390 Well, its acceleration I'm doing this just 582 00:37:56,390 --> 00:37:57,990 to get in the x direction. 583 00:37:57,990 --> 00:38:02,480 Its acceleration in the x is the main mass, x double dot, 584 00:38:02,480 --> 00:38:04,750 which it's sitting on. 585 00:38:04,750 --> 00:38:12,420 And the x component of the relative coordinate 586 00:38:12,420 --> 00:38:13,605 x1 double dot. 587 00:38:13,605 --> 00:38:16,320 This is the total acceleration in the x direction 588 00:38:16,320 --> 00:38:18,030 of the second mass. 589 00:38:18,030 --> 00:38:20,440 And the total forces on that second mass 590 00:38:20,440 --> 00:38:23,450 are-- signs are changed. 591 00:38:23,450 --> 00:38:26,460 Internal force, internal force, internal force, 592 00:38:26,460 --> 00:38:29,820 plus the force that was applied. 593 00:38:29,820 --> 00:38:31,520 Actually, here it is written out. 594 00:38:31,520 --> 00:38:34,910 Here is that external force that was applied to that wheel. 595 00:38:34,910 --> 00:38:39,690 If you take these two equations, you just add them together. 596 00:38:39,690 --> 00:38:43,920 All of the internal forces drop out. 597 00:38:43,920 --> 00:38:46,480 Plus fx minus fx. 598 00:38:46,480 --> 00:38:47,610 All of those drop out. 599 00:38:47,610 --> 00:38:50,750 And you end up with an expression that says, 600 00:38:50,750 --> 00:38:53,620 this is a system. 601 00:38:53,620 --> 00:38:58,330 The total external forces on the system in the x direction 602 00:38:58,330 --> 00:39:03,710 must be equal to the total mass of the system 603 00:39:03,710 --> 00:39:07,620 times the acceleration of its center of mass. 604 00:39:07,620 --> 00:39:08,120 Right? 605 00:39:08,120 --> 00:39:10,050 That's what Newton said. 606 00:39:10,050 --> 00:39:14,290 Well, the acceleration of the center of mass 607 00:39:14,290 --> 00:39:18,030 is the sum of the individual pieces 608 00:39:18,030 --> 00:39:21,040 times their individual accelerations. 609 00:39:21,040 --> 00:39:26,790 The sum of the total mass times the acceleration 610 00:39:26,790 --> 00:39:28,960 of the center of mass is equal to the sum 611 00:39:28,960 --> 00:39:32,430 of the individual masses times their individual accelerations. 612 00:39:32,430 --> 00:39:34,700 Mass one, its acceleration. 613 00:39:34,700 --> 00:39:39,090 Mass two, its acceleration due to its local coordinate 614 00:39:39,090 --> 00:39:43,540 plus its acceleration due to the main mass acceleration. 615 00:39:43,540 --> 00:39:46,060 And all of that's got to be equal to the external forces, 616 00:39:46,060 --> 00:39:53,100 which are now only minus kx, minus bx dot, and plus F. 617 00:39:53,100 --> 00:39:55,820 And this term here, this stuff, all that, 618 00:39:55,820 --> 00:40:00,260 including this, these three M X double dot terms, 619 00:40:00,260 --> 00:40:01,820 all added together as the same thing 620 00:40:01,820 --> 00:40:05,020 is the total mass times the acceleration of the center 621 00:40:05,020 --> 00:40:06,250 of mass, which is somewhere. 622 00:40:08,900 --> 00:40:13,510 But that's a completely legitimate equation 623 00:40:13,510 --> 00:40:14,750 for this system. 624 00:40:14,750 --> 00:40:17,470 And it involves both masses, both coordinates. 625 00:40:20,590 --> 00:40:23,670 So this one here would just be some x1 double dot times 626 00:40:23,670 --> 00:40:25,754 a cosine phi or a sine phi or something like that. 627 00:40:25,754 --> 00:40:27,378 AUDIENCE: So for this problem, is there 628 00:40:27,378 --> 00:40:28,652 only one equation of motion? 629 00:40:28,652 --> 00:40:29,360 PROFESSOR: Oh no. 630 00:40:29,360 --> 00:40:30,670 There still has to be two. 631 00:40:30,670 --> 00:40:35,900 But I've just shown you a way of getting at one of them 632 00:40:35,900 --> 00:40:40,730 without ever solving for the internal forces. 633 00:40:40,730 --> 00:40:44,370 By just realizing that if I separate them, 634 00:40:44,370 --> 00:40:47,190 that Newton's third law, all the internal ones 635 00:40:47,190 --> 00:40:50,007 will cancel when I add the two pieces together. 636 00:40:50,007 --> 00:40:51,590 I still have to get a second equation. 637 00:40:51,590 --> 00:40:52,548 Where would you get it? 638 00:41:02,080 --> 00:41:05,496 AUDIENCE: Sum of the forces in x1? 639 00:41:05,496 --> 00:41:07,470 AUDIENCE: They're rotating [INAUDIBLE]. 640 00:41:07,470 --> 00:41:08,390 PROFESSOR: Yeah, you want to do something 641 00:41:08,390 --> 00:41:10,306 where you don't have to solve-- you don't want 642 00:41:10,306 --> 00:41:13,170 to have to solve for any of these internal forces 643 00:41:13,170 --> 00:41:15,060 that you don't know, right? 644 00:41:15,060 --> 00:41:19,040 So the spring force you can know because it 645 00:41:19,040 --> 00:41:22,420 is a minus k times the coordinate that you 646 00:41:22,420 --> 00:41:25,630 have to use, minus kx1. 647 00:41:25,630 --> 00:41:27,330 So I would go in here, and I'd say 648 00:41:27,330 --> 00:41:32,280 maybe the-- what's the sum of the torques about that point, 649 00:41:32,280 --> 00:41:33,325 perhaps. 650 00:41:33,325 --> 00:41:34,700 AUDIENCE: There's no slip, right? 651 00:41:34,700 --> 00:41:35,658 PROFESSOR: And no slip. 652 00:41:35,658 --> 00:41:37,340 So these things-- that avoids having 653 00:41:37,340 --> 00:41:39,490 to deal with any of these forces. 654 00:41:39,490 --> 00:41:41,550 The real force acting on it, real force, 655 00:41:41,550 --> 00:41:42,720 you have to deal with that. 656 00:41:42,720 --> 00:41:46,760 And the minus kx1 force you need. 657 00:41:46,760 --> 00:41:50,810 It comes into your equation of motion. 658 00:41:50,810 --> 00:41:53,370 If you do that particular approach, 659 00:41:53,370 --> 00:41:57,640 it has one problematic-- it's not terrible 660 00:41:57,640 --> 00:41:59,890 but you have to remember to deal with it-- 661 00:41:59,890 --> 00:42:06,250 is that the sum of the x torques about that point we'll call A 662 00:42:06,250 --> 00:42:19,100 is d HA dt plus vAo cross P, right? 663 00:42:19,100 --> 00:42:21,370 Or I gave you an equation yesterday. 664 00:42:21,370 --> 00:42:23,500 This is easier to do usually if you 665 00:42:23,500 --> 00:42:39,940 do d HG dt plus rG with respect to A cross the mass 666 00:42:39,940 --> 00:42:45,270 times the acceleration of g with respect to o. 667 00:42:45,270 --> 00:42:49,810 This would be the second mass, M2. 668 00:42:49,810 --> 00:42:51,540 You can derive. 669 00:42:51,540 --> 00:42:53,200 It's about three lines. 670 00:42:53,200 --> 00:42:57,390 You can prove that this statement's the same as that. 671 00:42:57,390 --> 00:43:02,280 This one, turns out this one's less work. 672 00:43:02,280 --> 00:43:05,850 And the reason that this one's more work is this term always 673 00:43:05,850 --> 00:43:07,520 cancels out. 674 00:43:07,520 --> 00:43:10,310 Cancels out with a piece of it is generated 675 00:43:10,310 --> 00:43:11,815 when you do this calculation. 676 00:43:11,815 --> 00:43:13,440 But you know how messy this stuff gets. 677 00:43:13,440 --> 00:43:15,000 You've got all these unit vectors running around 678 00:43:15,000 --> 00:43:16,590 and cosine thetas and sine thetas. 679 00:43:16,590 --> 00:43:19,050 It's a lot of work to get this piece. 680 00:43:19,050 --> 00:43:21,340 And it's twice the work because you 681 00:43:21,340 --> 00:43:24,794 have to go in and find the piece that cancels it in here. 682 00:43:24,794 --> 00:43:27,210 So if you actually-- I sat down about a week ago and said, 683 00:43:27,210 --> 00:43:29,970 there's got to be a way to get rid of this thing. 684 00:43:29,970 --> 00:43:32,160 And there is, and it comes out like this. 685 00:43:32,160 --> 00:43:33,940 Now there's no pieces that cancel. 686 00:43:33,940 --> 00:43:38,030 You cancelled out in the proof, and get this piece to go away. 687 00:43:38,030 --> 00:43:39,650 And you're left with this. 688 00:43:39,650 --> 00:43:44,750 So this is the distance from your point 689 00:43:44,750 --> 00:43:46,160 to the center of mass. 690 00:43:46,160 --> 00:43:48,794 In this case, it's just r. 691 00:43:48,794 --> 00:43:50,460 And the only thing you have to calculate 692 00:43:50,460 --> 00:43:52,080 is the acceleration of the center 693 00:43:52,080 --> 00:43:57,750 of mass, which is capital X double dot plus X1 double dot. 694 00:43:57,750 --> 00:43:59,480 You know that. 695 00:43:59,480 --> 00:44:02,342 So this is vastly easier to do. 696 00:44:02,342 --> 00:44:04,772 AUDIENCE: So I understand how to use that if I'm 697 00:44:04,772 --> 00:44:06,230 using different [INAUDIBLE]. 698 00:44:10,604 --> 00:44:13,034 PROFESSOR: Center of mass, I didn't get the last bit. 699 00:44:13,034 --> 00:44:13,950 AUDIENCE: [INAUDIBLE]. 700 00:44:18,224 --> 00:44:19,390 PROFESSOR: That's the other. 701 00:44:19,390 --> 00:44:20,889 Because you're doing it around point 702 00:44:20,889 --> 00:44:23,260 A is what gets you into this mess. 703 00:44:23,260 --> 00:44:25,830 If you're doing it-- as soon as you work with respect-- 704 00:44:25,830 --> 00:44:28,520 if you're working with G, the center of mass, 705 00:44:28,520 --> 00:44:34,200 then this is vG cross vG, which is always zero. 706 00:44:34,200 --> 00:44:36,800 This has a vG in it. 707 00:44:36,800 --> 00:44:40,320 So that's the reason we like to work around center of the mass, 708 00:44:40,320 --> 00:44:42,260 but we don't do centers of mass when 709 00:44:42,260 --> 00:44:45,490 it causes us to have a whole bunch of external things 710 00:44:45,490 --> 00:44:46,280 we don't know. 711 00:44:50,740 --> 00:44:52,840 Thanks. 712 00:44:52,840 --> 00:44:53,440 Yes. 713 00:44:53,440 --> 00:44:54,802 AUDIENCE: [INAUDIBLE].