1 00:00:00,070 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,800 Commons license. 3 00:00:03,800 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,580 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,580 --> 00:00:17,260 at ocw.mit.edu. 8 00:00:37,410 --> 00:00:40,110 PROFESSOR: OK, let's get started. 9 00:00:40,110 --> 00:00:42,260 Can we get them up? 10 00:00:42,260 --> 00:00:43,690 So this is our thing spinning. 11 00:00:43,690 --> 00:00:46,180 What are the units of the generalized force 12 00:00:46,180 --> 00:00:48,060 in this problem, which is going to be related 13 00:00:48,060 --> 00:00:50,130 to the torque at the bottom. 14 00:00:50,130 --> 00:00:54,000 So most people said Newton meters, 15 00:00:54,000 --> 00:00:56,340 which is the units of torque. 16 00:00:56,340 --> 00:01:00,550 And that would be correct. 17 00:01:00,550 --> 00:01:03,290 So you're going to get some i theta double dot 18 00:01:03,290 --> 00:01:05,940 kind of equation of motion with this, 19 00:01:05,940 --> 00:01:08,380 as units of moment or torque. 20 00:01:08,380 --> 00:01:12,440 And any external non-conservative force on it 21 00:01:12,440 --> 00:01:14,660 would in this case have units of torque. 22 00:01:14,660 --> 00:01:17,440 OK, next. 23 00:01:17,440 --> 00:01:19,920 So we have a pendulum, kind of an odd shape. 24 00:01:19,920 --> 00:01:21,590 That is, does it have symmetries? 25 00:01:24,290 --> 00:01:27,665 Name a symmetry that this thing has. 26 00:01:27,665 --> 00:01:29,640 AUDIENCE: [INAUDIBLE]. 27 00:01:29,640 --> 00:01:32,390 PROFESSOR: You mean axial then, or a plane, or what? 28 00:01:32,390 --> 00:01:35,030 AUDIENCE: [INAUDIBLE]. 29 00:01:35,030 --> 00:01:36,850 PROFESSOR: Axis or a plane, OK. 30 00:01:36,850 --> 00:01:38,880 So this thing has symmetries. 31 00:01:38,880 --> 00:01:40,930 You could convince yourself pretty quickly 32 00:01:40,930 --> 00:01:42,820 that the principal axes with one big 33 00:01:42,820 --> 00:01:44,590 break down the center of it. 34 00:01:44,590 --> 00:01:47,650 And the other two would be perpendicular to that. 35 00:01:47,650 --> 00:01:52,800 So is it appropriate to use the parallel axis theorem to find 36 00:01:52,800 --> 00:01:57,150 an equation of motion of this thing for when it's pinned 37 00:01:57,150 --> 00:01:58,470 at the top? 38 00:01:58,470 --> 00:02:01,790 And most people said yes if you said no. 39 00:02:01,790 --> 00:02:03,380 I think parallel axis theorem works 40 00:02:03,380 --> 00:02:05,440 just great in this problem. 41 00:02:05,440 --> 00:02:11,670 It's planar motion, and the, for example, kinetic energy 42 00:02:11,670 --> 00:02:20,245 you can write as 1/2i about o up there, the hinge point. 43 00:02:20,245 --> 00:02:24,160 i with respect to that point theta double dot. 44 00:02:24,160 --> 00:02:26,220 And to do i about that point, you'd 45 00:02:26,220 --> 00:02:27,740 use the parallel axis theorem. 46 00:02:27,740 --> 00:02:34,970 So i about g plus the distance l squared times m. 47 00:02:34,970 --> 00:02:35,820 This one. 48 00:02:35,820 --> 00:02:39,300 Do you expect a centripetal acceleration term 49 00:02:39,300 --> 00:02:42,730 to show up in your equations of motion? 50 00:02:42,730 --> 00:02:45,175 And some said yes, some said no. 51 00:02:48,010 --> 00:02:50,010 So the equation of motion for this, 52 00:02:50,010 --> 00:02:54,580 I would probably write in terms of some rotation 53 00:02:54,580 --> 00:02:58,890 theta of the big disk. 54 00:02:58,890 --> 00:03:02,055 And that will cause the mass on the string to move up and down. 55 00:03:05,280 --> 00:03:12,770 But for the rotating parts of the system, 56 00:03:12,770 --> 00:03:20,810 are there-- I'll think of a clear way to word this. 57 00:03:20,810 --> 00:03:25,630 Let's ignore the little mass going up and down for a moment. 58 00:03:25,630 --> 00:03:29,370 The axis of rotation of that double disk and hub 59 00:03:29,370 --> 00:03:32,270 and the bigger disk, does the axis of rotation 60 00:03:32,270 --> 00:03:35,636 go through the center of mass of that disk? 61 00:03:35,636 --> 00:03:36,552 AUDIENCE: [INAUDIBLE]. 62 00:03:36,552 --> 00:03:37,135 PROFESSOR: OK. 63 00:03:37,135 --> 00:03:39,713 Is it statically balanced? 64 00:03:47,220 --> 00:03:50,150 We haven't talked about balancing in a while. 65 00:03:50,150 --> 00:03:52,940 Good review question for thinking about a quiz 66 00:03:52,940 --> 00:03:55,260 next Tuesday. 67 00:03:55,260 --> 00:04:00,840 So what does it take for something 68 00:04:00,840 --> 00:04:07,400 which is rotating to be considered statically balanced? 69 00:04:07,400 --> 00:04:10,694 The axis of rotation must what? 70 00:04:10,694 --> 00:04:11,630 AUDIENCE: [INAUDIBLE]. 71 00:04:11,630 --> 00:04:13,754 PROFESSOR: He says pass through the center of mass. 72 00:04:13,754 --> 00:04:16,910 Anybody else have a suggestion, different suggestion? 73 00:04:16,910 --> 00:04:21,370 If the axis of rotation passes through the center of mass, 74 00:04:21,370 --> 00:04:24,105 is it statically balanced? 75 00:04:24,105 --> 00:04:27,210 AUDIENCE: It has to be a principal axis? 76 00:04:27,210 --> 00:04:29,210 PROFESSOR: Does it have to be a principal axis? 77 00:04:29,210 --> 00:04:30,850 What do you think? 78 00:04:30,850 --> 00:04:33,600 Is that axis passing through g have to be a principal 79 00:04:33,600 --> 00:04:36,945 axis in order for it to be statically balanced? 80 00:04:42,200 --> 00:04:42,865 Let's see. 81 00:04:49,270 --> 00:04:53,680 I don't have a little-- no, it's too big. 82 00:04:59,720 --> 00:05:01,780 This is just a wheel. 83 00:05:01,780 --> 00:05:05,270 And this is my axle going through the center of mass. 84 00:05:05,270 --> 00:05:07,400 And it goes through the center of mass, 85 00:05:07,400 --> 00:05:15,650 and it'll have no tendency to swing down to a low point 86 00:05:15,650 --> 00:05:19,710 because the weight, its weight, is acting right 87 00:05:19,710 --> 00:05:20,820 on its center of mass. 88 00:05:20,820 --> 00:05:23,885 And with respect to this axle, there's no moment arm. 89 00:05:23,885 --> 00:05:25,740 So there's no torque caused by gravity 90 00:05:25,740 --> 00:05:31,320 that could cause this to put its center of mass below the axle. 91 00:05:31,320 --> 00:05:33,380 So it doesn't matter, even if I had 92 00:05:33,380 --> 00:05:35,830 the axis going through here. 93 00:05:35,830 --> 00:05:39,850 As long as that axis passes through g, 94 00:05:39,850 --> 00:05:42,750 there is no tendency for this thing to swing to a low side 95 00:05:42,750 --> 00:05:47,630 because the mass mg is acting right on the axle. 96 00:05:47,630 --> 00:05:48,690 No moment arm. 97 00:05:48,690 --> 00:05:53,660 So the only condition for static balance 98 00:05:53,660 --> 00:05:56,820 for a rotor of any kind, something rotating, 99 00:05:56,820 --> 00:06:00,320 to be statically balanced is for the axis of rotation 100 00:06:00,320 --> 00:06:01,680 to pass through the mass center. 101 00:06:05,360 --> 00:06:11,060 So the rotating part of this system, the axis of rotation 102 00:06:11,060 --> 00:06:15,770 passes through the mass center, that big disk. 103 00:06:15,770 --> 00:06:18,875 Does the little mass, m, rotate? 104 00:06:22,420 --> 00:06:28,450 So it can't be statically or dynamically in balance. 105 00:06:28,450 --> 00:06:30,390 It's not a rotating system. 106 00:06:30,390 --> 00:06:32,570 It's just a little translating mass 107 00:06:32,570 --> 00:06:34,540 that happens to be pulled up and down 108 00:06:34,540 --> 00:06:36,770 by the action of this thing. 109 00:06:36,770 --> 00:06:39,610 So the question was, do you expect 110 00:06:39,610 --> 00:06:42,040 a centripetal acceleration term to show up? 111 00:06:42,040 --> 00:06:43,720 The answer's no. 112 00:06:43,720 --> 00:06:44,840 So let's get back to that. 113 00:06:44,840 --> 00:06:48,730 If you have a static imbalance-- let's say my axis of rotation 114 00:06:48,730 --> 00:06:51,570 is here. 115 00:06:51,570 --> 00:06:54,190 This is definitely statically imbalanced now. 116 00:06:54,190 --> 00:06:57,640 Its mass center is down here. 117 00:06:57,640 --> 00:07:02,080 The force of gravity pulls it down until it hangs straight. 118 00:07:02,080 --> 00:07:04,520 That's how you can just test to see if you're passing 119 00:07:04,520 --> 00:07:06,280 through the mass center. 120 00:07:06,280 --> 00:07:12,980 So if you have an axis that does not pass through g, 121 00:07:12,980 --> 00:07:14,760 then you're statically imbalanced. 122 00:07:14,760 --> 00:07:18,700 And if you rotate about that axis, 123 00:07:18,700 --> 00:07:25,250 is there a net centripetal force required as it goes around? 124 00:07:25,250 --> 00:07:26,510 Why? 125 00:07:26,510 --> 00:07:28,810 AUDIENCE: Because the center of mass is on the outside. 126 00:07:28,810 --> 00:07:31,620 PROFESSOR: So the center of mass is some distance away. 127 00:07:31,620 --> 00:07:33,820 And you're swinging it around and around 128 00:07:33,820 --> 00:07:37,150 and make that center of mass go in a circle. 129 00:07:37,150 --> 00:07:39,809 You're applying a centripetal acceleration 130 00:07:39,809 --> 00:07:41,350 to the center of mass, and that takes 131 00:07:41,350 --> 00:07:45,330 a force, which you sometimes think of as a centrifugal force 132 00:07:45,330 --> 00:07:45,887 pulling out. 133 00:07:45,887 --> 00:07:47,720 That's what you're going to have to pull in, 134 00:07:47,720 --> 00:07:51,800 some m r omega squared. 135 00:07:51,800 --> 00:07:55,710 So no centripetal term would be expected in this problem. 136 00:07:55,710 --> 00:07:57,420 Let's go to the next one. 137 00:07:57,420 --> 00:07:59,500 This one, you had two different conditions. 138 00:07:59,500 --> 00:08:03,740 Either rolling without slip or rolling with slip. 139 00:08:03,740 --> 00:08:08,030 And the question is, for which conditions 140 00:08:08,030 --> 00:08:10,290 are the generalized forces associated 141 00:08:10,290 --> 00:08:13,730 with a virtual displacement of delta theta, 142 00:08:13,730 --> 00:08:18,560 the rotation of the wheel, equal to zero? 143 00:08:18,560 --> 00:08:20,200 So remember, generalized forces are 144 00:08:20,200 --> 00:08:23,820 the forces that account for the non-conservative forces 145 00:08:23,820 --> 00:08:25,350 in the problem. 146 00:08:25,350 --> 00:08:30,070 So if it's rolling down the hill without slip, 147 00:08:30,070 --> 00:08:35,900 are there any non-conservative forces acting on it? 148 00:08:35,900 --> 00:08:38,710 Is there a friction force acting on it? 149 00:08:38,710 --> 00:08:40,360 Yeah, but does it do any work? 150 00:08:40,360 --> 00:08:44,019 No, because there's no delta r at the point of application 151 00:08:44,019 --> 00:08:44,560 of the force. 152 00:08:44,560 --> 00:08:46,300 There's no motion. 153 00:08:46,300 --> 00:08:50,650 So for which condition does the generalized forces 154 00:08:50,650 --> 00:08:51,300 equal to zero? 155 00:08:51,300 --> 00:08:53,960 Certainly for the condition when it does not slip. 156 00:08:53,960 --> 00:08:57,740 What about if it does slip? 157 00:08:57,740 --> 00:09:00,480 Would you expect a non-conservative force 158 00:09:00,480 --> 00:09:01,590 to do work on it? 159 00:09:01,590 --> 00:09:04,660 What force would that be? 160 00:09:04,660 --> 00:09:05,800 The friction force. 161 00:09:05,800 --> 00:09:11,630 So now, as the wheel turns, that point of application where it's 162 00:09:11,630 --> 00:09:17,020 sliding, you're actually getting a little delta omega, delta 163 00:09:17,020 --> 00:09:18,260 theta rather. 164 00:09:18,260 --> 00:09:22,540 You move a little distance r delta theta, 165 00:09:22,540 --> 00:09:24,140 dotted with the force. 166 00:09:24,140 --> 00:09:27,620 You get a certain little bit of virtual work 167 00:09:27,620 --> 00:09:31,560 done, r delta theta times f. 168 00:09:31,560 --> 00:09:36,370 So the only case a is where you get zero. 169 00:09:36,370 --> 00:09:37,215 Oh, this one. 170 00:09:41,070 --> 00:09:44,970 What's the generalized force associated with f? 171 00:09:44,970 --> 00:09:46,500 That's this applied force. 172 00:09:46,500 --> 00:09:50,320 It's applied to the rolling thing going down the hill. 173 00:09:50,320 --> 00:09:53,110 The force is horizontal. 174 00:09:53,110 --> 00:09:56,730 And you're asked what's the generalized force associated 175 00:09:56,730 --> 00:10:01,760 with f due to the virtual displacement delta x? 176 00:10:01,760 --> 00:10:07,980 And x is the motion of the main cart. 177 00:10:07,980 --> 00:10:11,130 So when you're doing generalized forces, 178 00:10:11,130 --> 00:10:14,230 you think in terms of virtual work. 179 00:10:14,230 --> 00:10:17,650 Can you imagine that that cart's moving a little distance 180 00:10:17,650 --> 00:10:23,100 delta x, the main x in the xyo system? 181 00:10:23,100 --> 00:10:26,560 It shifts a little distance delta x. 182 00:10:26,560 --> 00:10:29,360 That force, does it move-- does the point 183 00:10:29,360 --> 00:10:33,910 of application of that force move with that delta x? 184 00:10:33,910 --> 00:10:36,092 That's really what the question comes down to. 185 00:10:44,290 --> 00:10:44,895 This one. 186 00:11:01,380 --> 00:11:04,310 So this is your inertial system, and it's 187 00:11:04,310 --> 00:11:06,700 going to account for the motion of this object. 188 00:11:09,900 --> 00:11:15,300 And then up here, got your wheel. 189 00:11:15,300 --> 00:11:23,680 And we have some coordinate system attached to this object, 190 00:11:23,680 --> 00:11:27,470 noting the position of the wheel as it 191 00:11:27,470 --> 00:11:29,860 goes up and down the hill. 192 00:11:29,860 --> 00:11:34,380 So this coordinate's relative to the moving cart. 193 00:11:34,380 --> 00:11:37,240 This coordinate's inertial. 194 00:11:37,240 --> 00:11:43,580 The question's only asking then, what is the virtual work done 195 00:11:43,580 --> 00:11:49,440 that's associated width delta x here, some motion of the cart. 196 00:11:49,440 --> 00:11:55,960 And it's going to be what we're looking for, the Qx delta x. 197 00:11:55,960 --> 00:11:59,427 And when we say, OK. 198 00:11:59,427 --> 00:12:01,135 At the point of application of the force, 199 00:12:01,135 --> 00:12:04,170 there's literally F here. 200 00:12:04,170 --> 00:12:06,490 And it's in the horizontal direction, which is exactly 201 00:12:06,490 --> 00:12:15,060 the same direction as capital X. So this is equal to the force. 202 00:12:15,060 --> 00:12:19,280 And in the problem it's called-- this is just 203 00:12:19,280 --> 00:12:22,105 called F. It's a vector. 204 00:12:24,850 --> 00:12:28,310 And we're trying to deduce the deflection 205 00:12:28,310 --> 00:12:30,140 at the point of application. 206 00:12:30,140 --> 00:12:34,460 I'm going to call it some-- it can be many forces. 207 00:12:34,460 --> 00:12:36,350 I'm going to call it F sub i so that we 208 00:12:36,350 --> 00:12:39,580 think of it as one of many. 209 00:12:39,580 --> 00:12:57,260 Times the displacement delta ri due to delta x. 210 00:12:57,260 --> 00:12:58,740 So this is a vector. 211 00:12:58,740 --> 00:13:00,100 This is a displacement. 212 00:13:00,100 --> 00:13:01,580 This is a force. 213 00:13:01,580 --> 00:13:04,390 The amount of work done as this force moves 214 00:13:04,390 --> 00:13:09,040 through that displacement is some F 215 00:13:09,040 --> 00:13:14,300 dot delta r, the movement at this point i where it's 216 00:13:14,300 --> 00:13:19,490 applied, due to this motion. 217 00:13:19,490 --> 00:13:24,450 So what is the delta r at this point due to that motion? 218 00:13:29,250 --> 00:13:31,720 So another concept here, when you're doing this, 219 00:13:31,720 --> 00:13:34,700 you think of these mentally one at a time. 220 00:13:34,700 --> 00:13:37,120 How many degrees of freedom do we have in this problem? 221 00:13:40,890 --> 00:13:43,190 Takes two to completely describe the motion. 222 00:13:43,190 --> 00:13:46,230 x and this one attached to the cart. 223 00:13:49,050 --> 00:13:51,750 And you think of these one at a time. 224 00:13:51,750 --> 00:13:53,370 So right now, we're asking what's 225 00:13:53,370 --> 00:13:59,060 the virtual work done by that force due to movement of this 226 00:13:59,060 --> 00:14:02,870 coordinate only, assuming this one is frozen? 227 00:14:02,870 --> 00:14:05,680 So this one is not allowed to move. 228 00:14:05,680 --> 00:14:07,180 So if this is not allowed to move, 229 00:14:07,180 --> 00:14:12,020 does the position of this change relative to this cart as you 230 00:14:12,020 --> 00:14:12,907 move the cart? 231 00:14:15,530 --> 00:14:16,360 No. 232 00:14:16,360 --> 00:14:17,910 That one is held fixed. 233 00:14:17,910 --> 00:14:19,950 It moves with the cart. 234 00:14:19,950 --> 00:14:23,760 So if you move the cart a little distance here, 235 00:14:23,760 --> 00:14:29,060 delta x in this frame, then this wheel 236 00:14:29,060 --> 00:14:31,810 moves with the cart that distance. 237 00:14:31,810 --> 00:14:35,500 So how much work gets done? 238 00:14:35,500 --> 00:14:44,430 So this delta ri in this case is equal to delta x. 239 00:14:44,430 --> 00:14:46,860 And the amount of work that gets done 240 00:14:46,860 --> 00:15:00,610 is delta w at point i due to delta x here is F dot delta x, 241 00:15:00,610 --> 00:15:04,170 but F is in the capital I direction. 242 00:15:04,170 --> 00:15:06,770 This is in the capital I direction. 243 00:15:06,770 --> 00:15:10,580 So it's just F delta x. 244 00:15:10,580 --> 00:15:11,930 is the work done. 245 00:15:11,930 --> 00:15:14,760 And that's Qx delta x. 246 00:15:14,760 --> 00:15:23,580 So the generalized force is just F. 247 00:15:23,580 --> 00:15:24,500 All right. 248 00:15:24,500 --> 00:15:25,310 What else we got? 249 00:15:28,040 --> 00:15:29,106 OK, wait a minute. 250 00:15:32,509 --> 00:15:33,800 Oh, this is due to the dashpot. 251 00:15:36,420 --> 00:15:38,010 What is generalized force associated 252 00:15:38,010 --> 00:15:42,670 with the dashpot due to a virtual deflection x1? 253 00:15:42,670 --> 00:15:45,640 Now x1's the coordinate, that thing 254 00:15:45,640 --> 00:15:46,850 rolling up and down the hill. 255 00:15:49,490 --> 00:15:52,725 So now the same thing, the dashpot force. 256 00:16:02,460 --> 00:16:05,142 Here's your cart. 257 00:16:05,142 --> 00:16:05,975 Here's your dashpot. 258 00:16:16,960 --> 00:16:20,190 And a free body diagram of the cart 259 00:16:20,190 --> 00:16:25,740 would show a dashpot force here, bx dot in that direction. 260 00:16:29,170 --> 00:16:37,890 And if we move the cart, again, a little bit delta 261 00:16:37,890 --> 00:16:40,655 x, how much work gets done? 262 00:16:40,655 --> 00:16:42,930 AUDIENCE: [INAUDIBLE]. 263 00:16:42,930 --> 00:16:45,790 PROFESSOR: This is not the question asked up there. 264 00:16:45,790 --> 00:16:47,530 Just first this. 265 00:16:47,530 --> 00:16:52,370 How much virtual work gets done by that in a deflection delta 266 00:16:52,370 --> 00:16:54,180 x? 267 00:16:54,180 --> 00:17:01,800 Well, it's some Fi dot delta r due to delta x, which 268 00:17:01,800 --> 00:17:03,760 is just delta x. 269 00:17:03,760 --> 00:17:09,349 So this is some minus bx dot in the I direction times delta 270 00:17:09,349 --> 00:17:10,900 x in the I again. 271 00:17:10,900 --> 00:17:13,609 It's just in this case it's minus. 272 00:17:13,609 --> 00:17:28,669 So this virtual work done this time due to just the dashpot. 273 00:17:28,669 --> 00:17:29,960 I don't know how to write that. 274 00:17:29,960 --> 00:17:31,250 I don't have a subscript. 275 00:17:31,250 --> 00:17:35,590 So this dashpot only here, bx dot, 276 00:17:35,590 --> 00:17:40,020 is bx dot minus in the I direction-- 277 00:17:40,020 --> 00:17:46,270 that's the force-- dotted with the dr that it moves. 278 00:17:48,910 --> 00:17:51,870 And the dr that it moves is just delta x. 279 00:17:51,870 --> 00:18:00,120 So this would be minus bx I dot delta x I is 280 00:18:00,120 --> 00:18:02,280 your virtual work that's done. 281 00:18:02,280 --> 00:18:08,750 And this is equal to Qx delta x. 282 00:18:08,750 --> 00:18:14,270 But this is a part of Qx due only to the dashpot. 283 00:18:14,270 --> 00:18:20,530 And you get minus bx dot delta x. 284 00:18:20,530 --> 00:18:26,990 So now we've gotten both parts of the total generalized force 285 00:18:26,990 --> 00:18:32,375 associated with the motion delta x is F minus bx dot. 286 00:18:39,550 --> 00:18:44,960 In general, this expression Qx delta x 287 00:18:44,960 --> 00:18:49,740 is the summation over all of the applied forces 288 00:18:49,740 --> 00:19:02,640 i dot delta r at i due to, in this case, 289 00:19:02,640 --> 00:19:07,050 the motion in the x direction. 290 00:19:07,050 --> 00:19:10,150 And we have two contributions here. 291 00:19:10,150 --> 00:19:14,190 They come from the applied force F and this. 292 00:19:14,190 --> 00:19:16,990 So Qx in this problem is going to turn out 293 00:19:16,990 --> 00:19:22,554 to be F minus bx dot. 294 00:19:22,554 --> 00:19:23,970 Now, what about the other-- what's 295 00:19:23,970 --> 00:19:25,720 really asked in this problem is how much-- 296 00:19:25,720 --> 00:19:29,070 what's the generalized force associated with the motion 297 00:19:29,070 --> 00:19:32,330 of the wheel down the hill? 298 00:19:32,330 --> 00:19:34,270 This is in the x1 direction. 299 00:19:34,270 --> 00:19:37,770 So now you have a little virtual deflection delta x1. 300 00:19:41,830 --> 00:19:46,580 How much virtual work gets done by the dashpot? 301 00:19:46,580 --> 00:19:47,480 AUDIENCE: Zero. 302 00:19:47,480 --> 00:19:48,438 PROFESSOR: I hear zero. 303 00:19:50,780 --> 00:19:54,800 So a little virtual displacement of delta x1, 304 00:19:54,800 --> 00:19:58,190 does it move the main cart? 305 00:19:58,190 --> 00:20:00,750 That's really what's going on here. 306 00:20:00,750 --> 00:20:05,360 Does the main cart move because the wheel and the main cart 307 00:20:05,360 --> 00:20:07,615 change relative position a little bit? 308 00:20:11,054 --> 00:20:11,597 AUDIENCE: No. 309 00:20:11,597 --> 00:20:12,180 PROFESSOR: No. 310 00:20:12,180 --> 00:20:13,180 I hear no here. 311 00:20:13,180 --> 00:20:16,000 I mean, that's the-- the coordinate x1 312 00:20:16,000 --> 00:20:20,620 is the relative motion between the cart and the wheel. 313 00:20:20,620 --> 00:20:23,840 And it's independent of the motion of the cart with respect 314 00:20:23,840 --> 00:20:27,130 to the inertial frame. 315 00:20:27,130 --> 00:20:29,520 So any motion of the little wheel 316 00:20:29,520 --> 00:20:31,405 does not affect the main cart. 317 00:20:35,030 --> 00:20:38,200 So there's no virtual work done by that dashpot 318 00:20:38,200 --> 00:20:40,901 because the wheel moves up and down the hill. 319 00:20:40,901 --> 00:20:42,650 And it makes sense, physical sense, right? 320 00:20:42,650 --> 00:20:44,650 The wheel can sit and roll up and down the hill all day long. 321 00:20:44,650 --> 00:20:46,150 It's not going to move that dashpot. 322 00:20:49,310 --> 00:20:51,180 Next. 323 00:20:51,180 --> 00:20:53,110 This problem. 324 00:20:53,110 --> 00:20:56,471 I just realized this problem's harder than I thought it was. 325 00:20:56,471 --> 00:20:59,484 It's one of those things that you look at it, 326 00:20:59,484 --> 00:21:00,775 oh, that looks straightforward. 327 00:21:00,775 --> 00:21:02,630 Then I looked at Audrey's solution 328 00:21:02,630 --> 00:21:05,030 and said, oh, she did it right. 329 00:21:05,030 --> 00:21:08,140 And this is a little trickier than you might think. 330 00:21:08,140 --> 00:21:12,330 So are the Axyz axes, which rotate with the hub-- 331 00:21:12,330 --> 00:21:15,500 so there's a rotating there-- definitely 332 00:21:15,500 --> 00:21:19,010 principal coordinates of that rod? 333 00:21:19,010 --> 00:21:19,970 That's not a problem. 334 00:21:19,970 --> 00:21:21,730 But are they principal coordinates 335 00:21:21,730 --> 00:21:24,040 for that disk out on the end? 336 00:21:26,986 --> 00:21:28,361 AUDIENCE: [INAUDIBLE]. 337 00:21:28,361 --> 00:21:29,110 PROFESSOR: Pardon? 338 00:21:29,110 --> 00:21:31,090 AUDIENCE: Depends if it's slipping or not. 339 00:21:31,090 --> 00:21:32,256 PROFESSOR: Depends if it's-- 340 00:21:32,256 --> 00:21:34,260 AUDIENCE: Slipping or not. 341 00:21:34,260 --> 00:21:38,275 PROFESSOR: Actually, I don't think so. 342 00:21:42,370 --> 00:21:44,370 Let's just talk about-- principal coordinates 343 00:21:44,370 --> 00:21:47,000 need to be attached to the body, right? 344 00:21:47,000 --> 00:21:51,890 So problems like this do-- let's say we're using Lagrange, 345 00:21:51,890 --> 00:21:54,930 and we want to calculate the total kinetic-- you 346 00:21:54,930 --> 00:21:58,030 need to calculate the total kinetic energy of the system. 347 00:21:58,030 --> 00:22:02,190 So how would you go about breaking this thing down 348 00:22:02,190 --> 00:22:04,610 to compute the total kinetic energy? 349 00:22:04,610 --> 00:22:09,420 Would you break it into more than one part? 350 00:22:09,420 --> 00:22:12,259 What would be the natural things to break it into? 351 00:22:12,259 --> 00:22:13,550 AUDIENCE: The disk and the rod. 352 00:22:13,550 --> 00:22:15,174 PROFESSOR: The disk and the rod, right? 353 00:22:15,174 --> 00:22:19,090 So the kinetic energy of the rod, that's 354 00:22:19,090 --> 00:22:20,090 pretty straightforward. 355 00:22:20,090 --> 00:22:28,040 It's 1/2i with respect to the center, theta dot squared, 356 00:22:28,040 --> 00:22:29,380 omega squared. 357 00:22:29,380 --> 00:22:32,830 But the kinetic energy of that disk out 358 00:22:32,830 --> 00:22:41,920 there, you need to account for its rotation and its movement 359 00:22:41,920 --> 00:22:45,760 of its center of mass in the circle. 360 00:22:45,760 --> 00:22:53,160 So let's-- I'm winging it now, so bear with me if I make any 361 00:22:53,160 --> 00:22:53,720 mistakes. 362 00:22:53,720 --> 00:22:54,640 You can help me out. 363 00:22:57,780 --> 00:23:03,740 T rod would be-- and our coordinate system 364 00:23:03,740 --> 00:23:06,810 is an A at the center. 365 00:23:09,570 --> 00:23:15,330 There actually isn't a-- oh yeah, there's the A. 366 00:23:15,330 --> 00:23:17,310 So T of the rod, I would argue, is 367 00:23:17,310 --> 00:23:36,040 1/2M of the rod omega squared ML squared over 3. 368 00:23:36,040 --> 00:23:38,160 Because it's rotating about one end. 369 00:23:38,160 --> 00:23:40,490 So apply the parallel axis theorem. 370 00:23:40,490 --> 00:23:44,630 The kinetic energy of the rod is 1/2M i 371 00:23:44,630 --> 00:23:49,220 with respect to that central axis, omega squared. 372 00:23:49,220 --> 00:23:52,880 And so that gets you from ML squared over 12 373 00:23:52,880 --> 00:23:56,830 to ML squared over 3 because you're swinging around one end. 374 00:23:56,830 --> 00:23:59,240 But now we need T of the disk. 375 00:24:01,950 --> 00:24:04,160 And to do T of the disk, I would do it 376 00:24:04,160 --> 00:24:12,870 by saying 1/2M of the disk velocity of this center of mass 377 00:24:12,870 --> 00:24:19,220 of the disk in o dot VGo-- so that 378 00:24:19,220 --> 00:24:24,830 takes care of its kinetic energy due to motion 379 00:24:24,830 --> 00:24:43,170 of its center of mass-- plus 1/2 omega dot H with respect to G. 380 00:24:43,170 --> 00:24:45,990 With respect to G, can you express 381 00:24:45,990 --> 00:24:50,870 H for the disk in terms of a mass moment of inertia 382 00:24:50,870 --> 00:24:53,487 and some rotations, rotation rates? 383 00:24:53,487 --> 00:24:54,195 Is it legitimate? 384 00:25:04,380 --> 00:25:08,440 Remember, I started-- a few days ago, I started the top 385 00:25:08,440 --> 00:25:12,530 and said, here's the general expression for kinetic energy. 386 00:25:12,530 --> 00:25:14,170 Full 3D, right? 387 00:25:14,170 --> 00:25:20,550 And it basically was that expression. 388 00:25:20,550 --> 00:25:23,380 That works full 3D. 389 00:25:23,380 --> 00:25:28,220 And then when you fix a point about which something 390 00:25:28,220 --> 00:25:33,760 is rotating, a rigid body, then you can simplify it. 391 00:25:33,760 --> 00:25:38,070 But in this case, H, you could represent the angular momentum 392 00:25:38,070 --> 00:25:42,600 around G as some I omega. 393 00:25:42,600 --> 00:25:46,920 And then you have to multiply it by-- so this inside of here 394 00:25:46,920 --> 00:25:51,830 will be-- you can represent this in here 395 00:25:51,830 --> 00:25:57,650 as some I omega dotted with omega, 396 00:25:57,650 --> 00:26:00,030 and you will get the kinetic energy 397 00:26:00,030 --> 00:26:02,350 due to the rotation of the disk. 398 00:26:02,350 --> 00:26:05,160 What makes this problem a little trickier than I thought-- 399 00:26:05,160 --> 00:26:06,740 I wasn't thinking really clearly-- 400 00:26:06,740 --> 00:26:12,520 is that omega is in-- what frame do you express omega 401 00:26:12,520 --> 00:26:17,000 in when you're computing H in terms 402 00:26:17,000 --> 00:26:18,320 of mass moments of inertia? 403 00:26:20,930 --> 00:26:25,580 In order to compute I, you have to use what coordinate system? 404 00:26:25,580 --> 00:26:26,550 AUDIENCE: [INAUDIBLE]. 405 00:26:26,550 --> 00:26:27,980 PROFESSOR: Body fixed. 406 00:26:27,980 --> 00:26:31,610 And when you compute I omega, what's the omega? 407 00:26:31,610 --> 00:26:35,394 What unit vectors is the omega expressed in? 408 00:26:35,394 --> 00:26:37,814 AUDIENCE: [INAUDIBLE]. 409 00:26:37,814 --> 00:26:40,240 It's rotating, but it's [INAUDIBLE]. 410 00:26:40,240 --> 00:26:42,560 PROFESSOR: But in what-- so the unit vector's 411 00:26:42,560 --> 00:26:45,330 associated with what coordinate system are the ones that you 412 00:26:45,330 --> 00:26:47,458 use to define omega? 413 00:26:47,458 --> 00:26:48,880 AUDIENCE: [INAUDIBLE]. 414 00:26:48,880 --> 00:26:52,110 PROFESSOR: The one attached to the body or not? 415 00:26:52,110 --> 00:26:54,940 Generally in terms of the one attached to the body. 416 00:26:54,940 --> 00:26:58,035 And the disk gets-- that means they're rotating. 417 00:26:58,035 --> 00:26:59,400 It gets a little messy. 418 00:26:59,400 --> 00:27:00,860 So I'm not going to do it because I 419 00:27:00,860 --> 00:27:02,930 can't do it in my head. 420 00:27:02,930 --> 00:27:07,780 So this one you need to do the kinetic energy of that. 421 00:27:07,780 --> 00:27:09,950 You need to use this approach. 422 00:27:09,950 --> 00:27:15,020 And you have to be careful with those. 423 00:27:15,020 --> 00:27:17,520 Food for thought at office hours and in recitation. 424 00:27:20,910 --> 00:27:21,445 All right. 425 00:27:31,160 --> 00:27:35,480 I had meant to start today-- we're well in, 426 00:27:35,480 --> 00:27:40,297 and I haven't even started where I was going to really go today. 427 00:27:40,297 --> 00:27:41,630 So I have kind of a broken play. 428 00:27:41,630 --> 00:27:43,030 I have to decide what to drop. 429 00:27:43,030 --> 00:27:43,755 Give me a moment. 430 00:28:30,520 --> 00:28:33,790 So the principal thing I wanted to talk about today 431 00:28:33,790 --> 00:28:38,210 was to dig a little deeper into finding generalized forces. 432 00:28:38,210 --> 00:28:40,830 And the way we've been doing it is 433 00:28:40,830 --> 00:28:42,840 what I would call kind of an intuitive approach. 434 00:28:47,910 --> 00:28:52,270 So let's imagine we've got a rigid body. 435 00:28:59,570 --> 00:29:03,356 And I have some forces acting on it, F1. 436 00:29:18,180 --> 00:29:21,780 And at the point of application of each of these forces, 437 00:29:21,780 --> 00:29:24,050 I can have a position vector that goes there. 438 00:29:24,050 --> 00:29:31,580 So there's R1, R2, R3. 439 00:29:34,910 --> 00:29:37,990 And they're all with respect to my o frame here, 440 00:29:37,990 --> 00:29:40,700 but I'm going to not write in all the o's. 441 00:29:51,960 --> 00:29:56,890 Let's say that this is a body in planar motion. 442 00:29:56,890 --> 00:29:58,300 So it's confined to a plane. 443 00:29:58,300 --> 00:29:59,280 It's a rigid body. 444 00:29:59,280 --> 00:30:01,380 It's like my disk there. 445 00:30:01,380 --> 00:30:03,940 So how many degrees of freedom at most 446 00:30:03,940 --> 00:30:07,700 would it have if there's no constraints? 447 00:30:07,700 --> 00:30:08,570 Three, right? 448 00:30:08,570 --> 00:30:12,480 It could move in the x, in the y, and it can rotate in the z. 449 00:30:12,480 --> 00:30:14,570 So potentially three degrees of freedom. 450 00:30:14,570 --> 00:30:20,770 So it'll take, let's say, G is maybe here. 451 00:30:20,770 --> 00:30:30,910 So we need three generalized coordinates. 452 00:30:30,910 --> 00:30:36,260 And they would be, say, x, y, and some rotation of theta. 453 00:30:40,860 --> 00:30:48,360 And with them, we have virtual displacements delta 454 00:30:48,360 --> 00:30:53,530 x, delta y, and delta theta. 455 00:30:53,530 --> 00:30:56,150 And we use those to figure out the amount 456 00:30:56,150 --> 00:31:00,450 of non-conservative work that happens 457 00:31:00,450 --> 00:31:03,255 when you make those little small motions occur. 458 00:31:05,980 --> 00:31:16,230 And we're trying to find-- we need to find Qx, Qy, and Q 459 00:31:16,230 --> 00:31:19,910 theta, the generalized forces associated 460 00:31:19,910 --> 00:31:24,631 with these small variations, virtual reflections in R3 461 00:31:24,631 --> 00:31:25,130 coordinates. 462 00:31:33,310 --> 00:31:34,810 So the jth one. 463 00:31:34,810 --> 00:31:36,430 I have a bunch of forces maybe up 464 00:31:36,430 --> 00:31:40,760 to here, someone here that's some Fj. 465 00:31:40,760 --> 00:31:47,430 So the generalized force-- I don't mean that. 466 00:31:47,430 --> 00:31:50,470 These are Fi's. 467 00:31:50,470 --> 00:31:55,420 The j's referred to are generalized coordinates. 468 00:31:55,420 --> 00:32:05,505 So generalized force Qj delta qj. 469 00:32:12,760 --> 00:32:19,170 This is the little bit of work done in the virtual deflection 470 00:32:19,170 --> 00:32:21,500 delta qj. 471 00:32:21,500 --> 00:32:27,630 And there's work done by all of these applied forces 472 00:32:27,630 --> 00:32:29,630 in the system, possibly. 473 00:32:29,630 --> 00:32:33,960 Every one, if I cause there a little delta x of this body, 474 00:32:33,960 --> 00:32:39,240 it moves over an amount delta x. 475 00:32:39,240 --> 00:32:41,840 All of those points of application of those forces 476 00:32:41,840 --> 00:32:44,320 move a little bit. 477 00:32:44,320 --> 00:32:49,290 And therefore, at every location a little bit of work gets done. 478 00:32:49,290 --> 00:32:52,360 So in order to account for all of the work, 479 00:32:52,360 --> 00:33:05,050 I have to do a summation of the Fi dot delta r at i. 480 00:33:05,050 --> 00:33:14,020 And now this is associated with the virtual displacement 481 00:33:14,020 --> 00:33:18,030 of generalized coordinate j. 482 00:33:18,030 --> 00:33:24,290 So the total virtual work done by a displacement 483 00:33:24,290 --> 00:33:30,090 of one of these is a summation of all the little bits of work 484 00:33:30,090 --> 00:33:34,660 done at all the points of application of forces 485 00:33:34,660 --> 00:33:37,190 dotted with the amount that the point 486 00:33:37,190 --> 00:33:42,740 of application of that force moves caused by delta qj. 487 00:33:42,740 --> 00:33:48,320 So in general, that's what the total qj would be. 488 00:33:48,320 --> 00:33:53,070 Now, we've done problems like that more or less intuitively. 489 00:33:53,070 --> 00:33:55,356 We figured it out and said, OK, if it moves this much, 490 00:33:55,356 --> 00:33:56,980 then it's going to move over that much. 491 00:33:56,980 --> 00:34:00,050 And if there is an angle between them, we take the component, 492 00:34:00,050 --> 00:34:01,350 and we just figure it out. 493 00:34:01,350 --> 00:34:06,164 Is there a more mathematical way of doing this? 494 00:34:06,164 --> 00:34:07,580 And so I'm going to show you that. 495 00:34:12,070 --> 00:34:19,880 So we've been doing this kind of the intuitive approach here, 496 00:34:19,880 --> 00:34:23,690 the reasoning out each of the deflections and calculating 497 00:34:23,690 --> 00:34:26,010 the result. That there's a kinematic-- there's 498 00:34:26,010 --> 00:34:27,874 an explicit kinematic way of doing this. 499 00:34:27,874 --> 00:34:29,040 AUDIENCE: I have a question. 500 00:34:29,040 --> 00:34:29,706 PROFESSOR: Yeah. 501 00:34:29,706 --> 00:34:32,388 AUDIENCE: What is it at the right corner? [INAUDIBLE]? 502 00:34:32,388 --> 00:34:33,679 PROFESSOR: Yeah, OK, I'm sorry. 503 00:34:33,679 --> 00:34:37,730 It says-- this is a delta ri. 504 00:34:37,730 --> 00:34:39,530 It has double subscripts here. 505 00:34:39,530 --> 00:34:45,360 This is the displacement at the point of application of force 506 00:34:45,360 --> 00:34:49,366 i due to the virtual displacement 507 00:34:49,366 --> 00:34:54,120 of generalized coordinate j. 508 00:34:54,120 --> 00:35:01,350 So that an example might be, if that's delta x over here, 509 00:35:01,350 --> 00:35:07,440 the summation is over i, is over F1 F2 to Fi. 510 00:35:07,440 --> 00:35:13,090 And the deflection is the deflection 511 00:35:13,090 --> 00:35:17,930 at the point of application i due to-- in this case 512 00:35:17,930 --> 00:35:22,830 I'm talking about delta x-- due to the virtual displacements 513 00:35:22,830 --> 00:35:25,010 delta x. 514 00:35:25,010 --> 00:35:27,930 And so that's-- we'd work each one of these out and add them 515 00:35:27,930 --> 00:35:29,210 up, and we'd have the answer. 516 00:35:29,210 --> 00:35:33,370 But there is a more explicit mathematical way 517 00:35:33,370 --> 00:35:34,350 of saying this. 518 00:35:34,350 --> 00:35:47,720 And that is to say that Qj delta qj is the summation over i 519 00:35:47,720 --> 00:35:52,760 equals 1 to N, however many there are, 520 00:35:52,760 --> 00:36:10,450 of Fi dot the derivative of ri with respect to qj delta qj. 521 00:36:13,240 --> 00:36:14,880 What's that mean? 522 00:36:14,880 --> 00:36:17,500 So let's look at one of these. 523 00:36:17,500 --> 00:36:21,100 So force one, there's a position vector R1. 524 00:36:24,760 --> 00:36:32,210 If I move in the x direction a little delta x, 525 00:36:32,210 --> 00:36:36,910 this point moves over delta x in that direction, 526 00:36:36,910 --> 00:36:40,810 in the x direction horizontally. 527 00:36:40,810 --> 00:36:50,190 Our position vector R1 has potentially components in the y 528 00:36:50,190 --> 00:36:51,860 as well as components in the x. 529 00:36:51,860 --> 00:36:55,520 But I'm only changing it in the x direction. 530 00:36:55,520 --> 00:36:58,590 So that portion of the possible movement 531 00:36:58,590 --> 00:37:06,110 of R1 due to changes in just one of the coordinates-- 532 00:37:06,110 --> 00:37:10,010 in this case, I was doing qx-- the derivative of R1 533 00:37:10,010 --> 00:37:12,850 with respect to qx, so only a part 534 00:37:12,850 --> 00:37:16,780 of its total possible movement is due to x. 535 00:37:16,780 --> 00:37:19,480 This gives us that portion. 536 00:37:19,480 --> 00:37:24,510 Times delta qx is the total movement 537 00:37:24,510 --> 00:37:29,890 in the direction of qx dotted with the force. 538 00:37:29,890 --> 00:37:31,610 You get the work done. 539 00:37:31,610 --> 00:37:37,600 So I find this-- if I were you, this is highly abstract. 540 00:37:37,600 --> 00:37:40,180 I think we need-- let's do an example of this 541 00:37:40,180 --> 00:37:43,610 and see how it works out. 542 00:37:43,610 --> 00:37:48,680 And since we were talking about that problem, 543 00:37:48,680 --> 00:37:51,840 I'll do this cart by this method. 544 00:39:13,610 --> 00:39:16,990 So I need a position vector to the point 545 00:39:16,990 --> 00:39:22,190 of application of this external non-conservative force. 546 00:39:22,190 --> 00:39:24,540 Because I'm calling this force one, 547 00:39:24,540 --> 00:39:26,880 I'll call that position vector R1. 548 00:39:26,880 --> 00:39:31,040 And it goes from here to there. 549 00:39:31,040 --> 00:39:34,530 But we know that the total motion of this point 550 00:39:34,530 --> 00:39:36,970 is made up of the motion of the main cart 551 00:39:36,970 --> 00:39:41,240 plus the motion of the wheel relative to the main cart. 552 00:39:41,240 --> 00:39:45,100 And so we fall back on our notation. 553 00:39:45,100 --> 00:39:48,080 So I'll say this is R1 and zero. 554 00:39:48,080 --> 00:39:51,970 Here's my point A. It's R is the vector. 555 00:39:51,970 --> 00:39:57,940 This is R of A with respect to o plus-- 556 00:39:57,940 --> 00:39:59,860 and we'll better give this a name. 557 00:39:59,860 --> 00:40:10,410 What have I-- so this is my point one here. 558 00:40:10,410 --> 00:40:15,410 So this is plus R1 with respect to A. 559 00:40:15,410 --> 00:40:17,730 So we've done that lots of times, this term. 560 00:40:17,730 --> 00:40:20,970 That's just how-- that point is the sum of this vector 561 00:40:20,970 --> 00:40:22,735 plus this vector. 562 00:40:22,735 --> 00:40:29,230 So you have an R-- this is R1 with respect to A from here 563 00:40:29,230 --> 00:40:30,250 to here. 564 00:40:30,250 --> 00:40:32,790 And the sum of those two is this one. 565 00:40:32,790 --> 00:40:36,170 So this is R1 here. 566 00:40:36,170 --> 00:40:41,070 OK, so let's see if we can come up with an expression for that. 567 00:40:41,070 --> 00:40:53,800 Well, this point A is just x in the I plus some Y in the J. 568 00:40:53,800 --> 00:40:56,500 That's this term. 569 00:40:56,500 --> 00:40:59,700 Then I need this one. 570 00:41:04,960 --> 00:41:07,500 So I want-- because I'm going to take some derivatives 571 00:41:07,500 --> 00:41:10,360 and things, I want to get everything in terms 572 00:41:10,360 --> 00:41:13,380 of unit vectors and one system. 573 00:41:13,380 --> 00:41:17,010 So I know this one, this is my x1, 574 00:41:17,010 --> 00:41:21,210 and it will have a unit vector i, lowercase i, 575 00:41:21,210 --> 00:41:23,390 in this direction. 576 00:41:23,390 --> 00:41:31,810 So the unit vector i here has components in the capital IJ 577 00:41:31,810 --> 00:41:33,590 system. 578 00:41:33,590 --> 00:41:44,790 And this is theta, and that's theta. 579 00:41:44,790 --> 00:41:51,830 So this has-- i has a component here, 580 00:41:51,830 --> 00:42:00,360 which is cosine theta cap I, and then this piece is 581 00:42:00,360 --> 00:42:20,430 minus sine theta J. So this should be x1, 582 00:42:20,430 --> 00:42:26,990 the distance this thing moves, broken into two pieces, 583 00:42:26,990 --> 00:42:35,330 cos theta I minus sine theta J. So that's now-- 584 00:42:35,330 --> 00:42:43,950 the position of this thing is the position of the cart at A 585 00:42:43,950 --> 00:42:49,940 plus the vector that goes from A to point one, 586 00:42:49,940 --> 00:42:53,330 which is the distance x1 to here, 587 00:42:53,330 --> 00:42:55,510 broken into two pieces, an I piece and a J piece. 588 00:43:01,820 --> 00:43:03,170 So now we're almost done. 589 00:43:03,170 --> 00:43:06,400 So I would like to find Qx. 590 00:43:10,900 --> 00:43:18,690 And Qx then should be the summation-- well, let's see. 591 00:43:37,900 --> 00:43:41,950 So I'm only going to compute the part of the generalized force 592 00:43:41,950 --> 00:43:46,030 in the Qx direction due to just this one force. 593 00:43:46,030 --> 00:43:48,260 Now, remember we have other forces, 594 00:43:48,260 --> 00:43:50,060 non-conservative forces, acting on this. 595 00:43:50,060 --> 00:43:52,050 We've got a bx dot too. 596 00:43:52,050 --> 00:43:55,650 But I'm just going to do the contribution to Qx 597 00:43:55,650 --> 00:43:58,360 that comes from this force F1. 598 00:44:01,040 --> 00:44:11,440 And so Qx delta x is the virtual work done, 599 00:44:11,440 --> 00:44:21,180 is F1I dot partial of R1 with respect to x delta x. 600 00:44:26,740 --> 00:44:35,760 But the derivative of R1 with respect to capital 601 00:44:35,760 --> 00:44:39,820 X, there's no capital X's over here. 602 00:44:39,820 --> 00:44:42,080 So nothing comes from that. 603 00:44:42,080 --> 00:44:43,690 There's one right here. 604 00:44:43,690 --> 00:44:46,530 So the derivative of x with respect to x gives me 1. 605 00:44:46,530 --> 00:44:49,610 I just get 1 times I back here. 606 00:44:55,520 --> 00:45:02,280 F1 I hat dot I hat delta x. 607 00:45:02,280 --> 00:45:09,800 So it's just F1 delta x, which we knew intuitively 608 00:45:09,800 --> 00:45:11,599 when we worked this problem earlier, when 609 00:45:11,599 --> 00:45:12,640 we were talking about it. 610 00:45:12,640 --> 00:45:16,410 The amount of virtual work that gets 611 00:45:16,410 --> 00:45:23,100 done by this particular force in a deflection, virtual 612 00:45:23,100 --> 00:45:26,870 displacement delta x, it's just F1 delta x. 613 00:45:26,870 --> 00:45:29,320 But we've proven it-- we've done it 614 00:45:29,320 --> 00:45:34,640 in a very precise, kinematic way where 615 00:45:34,640 --> 00:45:38,780 we found the position vector, worked the whole thing out. 616 00:45:38,780 --> 00:45:40,080 So that's the simple one. 617 00:45:40,080 --> 00:45:49,765 Let's now find the harder one, but not much now. 618 00:45:49,765 --> 00:46:09,600 So we'd like to find Qx1 due to just this F1 only delta x1. 619 00:46:09,600 --> 00:46:19,690 Well, that should be F1I dot partial of R1 620 00:46:19,690 --> 00:46:22,826 with respect to x1 delta x1. 621 00:46:27,030 --> 00:46:33,120 So the derivative of R1 with respect to x1-- this stuff 622 00:46:33,120 --> 00:46:35,670 has nothing to do with x1. 623 00:46:35,670 --> 00:46:37,520 The x1 only appears over here. 624 00:46:37,520 --> 00:46:40,830 And the derivative of this expression, just cosine I minus 625 00:46:40,830 --> 00:46:59,170 sine theta J. 626 00:46:59,170 --> 00:47:02,100 So I dot I is the only part you get back. 627 00:47:02,100 --> 00:47:06,700 This is-- and I need a delta x1. 628 00:47:06,700 --> 00:47:13,230 F1I dot I is cosine theta delta x1. 629 00:47:13,230 --> 00:47:30,170 So Qx1 and F1 only here equals F1 cosine theta. 630 00:47:30,170 --> 00:47:37,100 So this time the motion delta x1, only part of it 631 00:47:37,100 --> 00:47:40,530 is in the direction of F1. 632 00:47:40,530 --> 00:47:49,240 And that portion, by taking this derivative here, 633 00:47:49,240 --> 00:47:53,640 we get the contribution to this that comes from x1. 634 00:47:53,640 --> 00:47:57,300 And then dotted with the force, we only 635 00:47:57,300 --> 00:47:59,350 take that component of that motion 636 00:47:59,350 --> 00:48:01,650 in the direction of the force. 637 00:48:01,650 --> 00:48:03,550 And that gives us our total virtual work. 638 00:48:03,550 --> 00:48:06,650 So here is then the total virtual work 639 00:48:06,650 --> 00:48:11,860 done by F1 due to the little motion delta x1. 640 00:48:11,860 --> 00:48:14,120 So now we've got the contribution here 641 00:48:14,120 --> 00:48:23,300 to the generalized force that is associated with deflections 642 00:48:23,300 --> 00:48:24,405 of coordinate x1. 643 00:48:28,180 --> 00:48:31,980 Is that the total generalized force 644 00:48:31,980 --> 00:48:37,990 associated with generalized coordinate x1 in this problem? 645 00:48:37,990 --> 00:48:42,140 Are there Any other non-conservative forces 646 00:48:42,140 --> 00:48:46,516 in the problem that move when delta x1 is moved? 647 00:48:46,516 --> 00:48:48,310 AUDIENCE: What about friction? 648 00:48:48,310 --> 00:48:50,136 PROFESSOR: Well, let's see. 649 00:48:50,136 --> 00:48:50,635 Friction. 650 00:48:53,890 --> 00:48:56,090 Friction, you're presuming, on the wheel? 651 00:48:56,090 --> 00:48:56,850 OK. 652 00:48:56,850 --> 00:48:58,740 So he asked about friction on the wheel. 653 00:48:58,740 --> 00:49:05,340 Well, let's say that there's no slip in this case. 654 00:49:05,340 --> 00:49:07,960 Then does the friction at the point of contact 655 00:49:07,960 --> 00:49:10,021 with the wheel do any work? 656 00:49:10,021 --> 00:49:10,520 No. 657 00:49:10,520 --> 00:49:14,230 So do we have to account for it as a generalized force? 658 00:49:14,230 --> 00:49:15,930 No. 659 00:49:15,930 --> 00:49:18,115 So how about the dashpot? 660 00:49:20,740 --> 00:49:23,860 So a little virtual deflection, delta x1, 661 00:49:23,860 --> 00:49:27,090 does it make the big cart move? 662 00:49:27,090 --> 00:49:27,590 No. 663 00:49:27,590 --> 00:49:31,580 So are there any other forces in the problem 664 00:49:31,580 --> 00:49:37,293 that move when you cause a small movement in x1? 665 00:49:37,293 --> 00:49:38,160 No. 666 00:49:38,160 --> 00:49:45,040 So in this case, this is the total Qx1. 667 00:49:45,040 --> 00:49:48,580 Up here, we found Qx, the generalized force 668 00:49:48,580 --> 00:49:53,820 due to the motion of a cart, the contribution by F1. 669 00:49:53,820 --> 00:49:55,580 But is there another contribution? 670 00:49:55,580 --> 00:49:56,376 AUDIENCE: Yes. 671 00:49:56,376 --> 00:49:57,250 PROFESSOR: And it is? 672 00:49:57,250 --> 00:49:58,320 AUDIENCE: The dashpot. 673 00:49:58,320 --> 00:49:59,278 PROFESSOR: The dashpot. 674 00:49:59,278 --> 00:50:02,440 So we get additionally the total Qx 675 00:50:02,440 --> 00:50:07,650 here total would be the summation 676 00:50:07,650 --> 00:50:12,510 of two pieces, an F1 and an F2, which I'd call minus bx. 677 00:50:12,510 --> 00:50:15,000 It's in the same direction as delta x. 678 00:50:15,000 --> 00:50:20,700 So you're going to get a minus bx dot plus F1 would 679 00:50:20,700 --> 00:50:28,100 be the total generalized force in the capital X direction, 680 00:50:28,100 --> 00:50:29,630 the movement of the main cart. 681 00:50:37,010 --> 00:50:43,830 So any time you can actually specify a position vector 682 00:50:43,830 --> 00:50:47,380 to the point of application of an external non-conservative 683 00:50:47,380 --> 00:50:55,910 force, then you can just plug it into this. 684 00:50:55,910 --> 00:50:59,280 You do it at each force that's applied. 685 00:50:59,280 --> 00:51:00,780 You take the derivative with respect 686 00:51:00,780 --> 00:51:04,810 to that to coordinate qj, delta qj. 687 00:51:04,810 --> 00:51:09,570 That is the virtual work done by each of these forces. 688 00:51:09,570 --> 00:51:13,970 And you add up to get the total virtual work done 689 00:51:13,970 --> 00:51:19,120 due to a deflection at that particular coordinate, j. 690 00:51:19,120 --> 00:51:24,530 So in the case of capital of Qx, we had two contributions 691 00:51:24,530 --> 00:51:33,250 because we had two forces on the main cart, F1 and minus bx dot. 692 00:51:33,250 --> 00:51:36,120 And so the summation in that problem, 693 00:51:36,120 --> 00:51:40,020 when this is capital X, delta X, you 694 00:51:40,020 --> 00:51:41,770 have two contributions, F1 and F2. 695 00:51:45,100 --> 00:51:53,950 Now, what else can I do in the length of time? 696 00:51:53,950 --> 00:51:57,910 Actually, let me stop for a moment there, think about this. 697 00:51:57,910 --> 00:52:01,160 Would you have any questions about this? 698 00:52:01,160 --> 00:52:04,479 So we've described two ways of getting generalized forces. 699 00:52:04,479 --> 00:52:06,270 One's kind of the intuitive one, figure out 700 00:52:06,270 --> 00:52:07,686 how much it moves in the direction 701 00:52:07,686 --> 00:52:09,270 and do the dot product. 702 00:52:09,270 --> 00:52:11,990 The other one is straight mathematical way. 703 00:52:11,990 --> 00:52:12,620 Kinematics. 704 00:52:12,620 --> 00:52:17,520 Plug it in, take the derivative, same thing will come out. 705 00:52:17,520 --> 00:52:19,340 So while you're thinking about a question, 706 00:52:19,340 --> 00:52:21,560 I'll look and think what I was going to do next. 707 00:52:27,930 --> 00:52:32,160 I know what I'll do next, but do you have any questions on this? 708 00:52:32,160 --> 00:52:34,428 I'm going to do another example of this. 709 00:52:34,428 --> 00:52:36,404 AUDIENCE: I have a question. 710 00:52:36,404 --> 00:52:37,890 AUDIENCE: I have a question. 711 00:52:37,890 --> 00:52:39,236 PROFESSOR: Ah, Christina, yeah. 712 00:52:39,236 --> 00:52:42,722 AUDIENCE: So I still don't understand how 713 00:52:42,722 --> 00:52:46,706 if you're going to grab the wheel and move it, 714 00:52:46,706 --> 00:52:48,698 how that doesn't move the disk. 715 00:52:48,698 --> 00:52:52,190 Because they're attached, so I don't get it. 716 00:52:52,190 --> 00:52:57,200 PROFESSOR: Is the wheel-- it's all in how you specify 717 00:52:57,200 --> 00:52:59,830 your generalized coordinates. 718 00:52:59,830 --> 00:53:03,900 So in this problem, the two generalized coordinates 719 00:53:03,900 --> 00:53:07,990 are this capital X in the inertial system which 720 00:53:07,990 --> 00:53:11,630 describes the motion of the cart, 721 00:53:11,630 --> 00:53:15,470 and little x1 describes the motion 722 00:53:15,470 --> 00:53:18,990 of the wheel relative to the cart. 723 00:53:18,990 --> 00:53:22,960 And actually that allows you to write this statement. 724 00:53:22,960 --> 00:53:27,760 This is only relative to the cart. 725 00:53:27,760 --> 00:53:31,320 So the motion of the cart plus the motion of the point 726 00:53:31,320 --> 00:53:34,510 relative to the cart gives you the total motion. 727 00:53:34,510 --> 00:53:37,020 And you've picked two coordinates 728 00:53:37,020 --> 00:53:40,020 that allow you to describe those two things. 729 00:53:40,020 --> 00:53:45,070 So if you can-- in this problem, if this is the cart, 730 00:53:45,070 --> 00:53:51,050 this is the wheel, I even have a spring hooked to it here, 731 00:53:51,050 --> 00:53:56,130 if I move this a little bit, the cart doesn't have to move. 732 00:53:56,130 --> 00:53:59,870 This going back and forth accounts for x1 relative 733 00:53:59,870 --> 00:54:00,780 to this table. 734 00:54:00,780 --> 00:54:02,279 And the table's the cart. 735 00:54:02,279 --> 00:54:03,195 AUDIENCE: [INAUDIBLE]. 736 00:54:09,500 --> 00:54:12,140 PROFESSOR: You mean dynamically because you're putting forces 737 00:54:12,140 --> 00:54:12,990 on it? 738 00:54:12,990 --> 00:54:13,900 Yeah, well it might. 739 00:54:13,900 --> 00:54:20,370 But that's not-- in a way, you're asking the question, 740 00:54:20,370 --> 00:54:24,470 is the cart capable of moving because you 741 00:54:24,470 --> 00:54:25,990 put a force in the wheel? 742 00:54:25,990 --> 00:54:29,250 You move the wheel, which puts more spring force, which maybe 743 00:54:29,250 --> 00:54:30,835 that causes the cart to move. 744 00:54:30,835 --> 00:54:32,320 Yeah, that could happen. 745 00:54:32,320 --> 00:54:35,580 But that's not the problem you're 746 00:54:35,580 --> 00:54:39,100 solving when you're trying to find the generalized forces. 747 00:54:39,100 --> 00:54:44,830 You're, in fact, allowing a single motion at a time. 748 00:54:44,830 --> 00:54:47,760 So if you're talking about this motion, 749 00:54:47,760 --> 00:54:52,080 you have frozen the motion of the main cart. 750 00:54:52,080 --> 00:54:54,530 And you figure out what the consequence of that is. 751 00:54:54,530 --> 00:54:56,960 It does a little virtual work because there's a force. 752 00:54:56,960 --> 00:55:00,600 And you get one of the generalized forces. 753 00:55:00,600 --> 00:55:03,760 Then if you move the cart, you fix the wheel, 754 00:55:03,760 --> 00:55:05,036 and the whole cart moves. 755 00:55:08,180 --> 00:55:11,150 But the amount that this wheel moves 756 00:55:11,150 --> 00:55:14,140 is exactly equal to the amount that the cart 757 00:55:14,140 --> 00:55:19,070 moves because you've now fixed this relative position. 758 00:55:19,070 --> 00:55:22,040 And that's where you get the first-- that's where 759 00:55:22,040 --> 00:55:25,350 you get the capital Qx term. 760 00:55:25,350 --> 00:55:30,850 Even know this force F1 is applied here on the wheel, 761 00:55:30,850 --> 00:55:34,620 this wheel moves when the table moves. 762 00:55:34,620 --> 00:55:39,460 But the table doesn't move when this relative coordinate 763 00:55:39,460 --> 00:55:41,650 between the table and here changes. 764 00:55:41,650 --> 00:55:42,690 It doesn't have to. 765 00:55:42,690 --> 00:55:45,370 This is free to move when the table is frozen. 766 00:55:45,370 --> 00:55:48,370 Remember we talked about complete and independent 767 00:55:48,370 --> 00:55:50,500 coordinates? 768 00:55:50,500 --> 00:55:57,180 x1 is independent of capital X. If I freeze x1, 769 00:55:57,180 --> 00:55:59,360 and I make capital X change, just the whole thing 770 00:55:59,360 --> 00:56:00,980 moves like that. 771 00:56:00,980 --> 00:56:04,320 If I freeze capital X, x1 can still move. 772 00:56:08,220 --> 00:56:12,520 So you have to pick independent coordinates. 773 00:56:12,520 --> 00:56:13,337 Yeah. 774 00:56:13,337 --> 00:56:15,348 AUDIENCE: [INAUDIBLE] mass of the whole thing 775 00:56:15,348 --> 00:56:17,097 is much larger than the mass of the wheel? 776 00:56:17,097 --> 00:56:18,098 PROFESSOR: Not at all. 777 00:56:18,098 --> 00:56:21,270 AUDIENCE: Because [INAUDIBLE] if you have two massed connected 778 00:56:21,270 --> 00:56:22,978 by a spring, you pull the first one, 779 00:56:22,978 --> 00:56:24,442 they kind of pull each other along. 780 00:56:24,442 --> 00:56:27,074 So why don't you get the pull-along effect over here? 781 00:56:27,074 --> 00:56:28,490 PROFESSOR: That's a good question. 782 00:56:28,490 --> 00:56:30,890 It's similar-- it's essentially the same question 783 00:56:30,890 --> 00:56:32,710 that Christina asked. 784 00:56:32,710 --> 00:56:35,310 He asked basically-- let's think about it. 785 00:56:35,310 --> 00:56:37,030 Let's do a problem like that. 786 00:56:37,030 --> 00:56:41,440 Let's have two carts and a spring in between them, 787 00:56:41,440 --> 00:56:42,565 and they're both on wheels. 788 00:57:01,090 --> 00:57:03,230 This is a planar motion problem. 789 00:57:03,230 --> 00:57:05,780 Each of these bodies is capable in planar motion can 790 00:57:05,780 --> 00:57:08,460 have x and y and a rotation. 791 00:57:08,460 --> 00:57:11,990 But because of constraints, how many degrees of freedom 792 00:57:11,990 --> 00:57:13,030 does body one have? 793 00:57:17,140 --> 00:57:18,880 I hear one, right? 794 00:57:18,880 --> 00:57:20,020 I don't allow it to rotate. 795 00:57:20,020 --> 00:57:21,680 It's got two wheels. 796 00:57:21,680 --> 00:57:24,705 I don't allow it to go up because it's on the ground. 797 00:57:24,705 --> 00:57:26,919 I only allow it to move in this direction. 798 00:57:26,919 --> 00:57:28,210 Same thing to be said for this. 799 00:57:28,210 --> 00:57:29,310 Only one there. 800 00:57:29,310 --> 00:57:31,560 How many degrees of freedom do I have in this problem? 801 00:57:34,660 --> 00:57:37,555 One for each mass, right? 802 00:57:37,555 --> 00:57:38,930 So I have two degrees of freedom. 803 00:57:38,930 --> 00:57:42,530 How many generalized coordinates do I need? 804 00:57:42,530 --> 00:57:47,990 So my generalized coordinate for this one will be x1, 805 00:57:47,990 --> 00:57:51,595 and for this one will be some x2. 806 00:57:55,900 --> 00:58:14,130 If you're going to do this problem by Lagrange, and let's 807 00:58:14,130 --> 00:58:14,740 see. 808 00:58:14,740 --> 00:58:18,940 Let's put a force now to get back to your question. 809 00:58:18,940 --> 00:58:22,220 Let's put a force here on one. 810 00:58:22,220 --> 00:58:23,120 And we'll call it F1. 811 00:58:27,490 --> 00:58:45,375 And the potential energy-- actually, no, I 812 00:58:45,375 --> 00:58:46,558 don't want to do that. 813 00:59:04,860 --> 00:59:09,630 So the potential energy for this is some 1/2k times 814 00:59:09,630 --> 00:59:12,060 the amount that you stretch the springs, right? 815 00:59:12,060 --> 00:59:18,030 So you're going to-- the difference between x1 and x2 816 00:59:18,030 --> 00:59:20,210 minus the unstretched length. 817 00:59:20,210 --> 00:59:26,300 So x1 minus x2 minus the unstretched length. 818 00:59:26,300 --> 00:59:28,810 We'll call it l0. 819 00:59:28,810 --> 00:59:33,450 So this would be the stretch of the springs. 820 00:59:33,450 --> 00:59:37,120 If the spring had no length, then it 821 00:59:37,120 --> 00:59:41,640 would just be the difference in these two positions squared. 822 00:59:41,640 --> 00:59:44,781 So my potential energy looks something like that. 823 00:59:44,781 --> 00:59:46,280 Now we want to compute the general-- 824 00:59:46,280 --> 00:59:48,124 and you could use Lagrange, and you 825 00:59:48,124 --> 00:59:49,790 could figure out two equations of motion 826 00:59:49,790 --> 00:59:52,626 for this taking your derivatives. 827 00:59:52,626 --> 00:59:54,000 But the point of the question was 828 00:59:54,000 --> 00:59:57,840 about getting to the generalized forces, right? 829 00:59:57,840 --> 01:00:08,100 So now the generalized force Qx1 delta x1 830 01:00:08,100 --> 01:00:18,610 is F1 times the derivative of R1 with respect to x1 delta x1. 831 01:00:18,610 --> 01:00:21,480 So how much does the position vector 832 01:00:21,480 --> 01:00:27,120 from marking the position of this cart, which would be R1-- 833 01:00:27,120 --> 01:00:31,400 so R1 is in effect x1, right? 834 01:00:31,400 --> 01:00:32,930 It's a pretty trivial problem. 835 01:00:32,930 --> 01:00:36,300 So the derivative of R1 with respect to x1 is just 1, 836 01:00:36,300 --> 01:00:39,750 and the amount that it then moves is delta x1. 837 01:00:39,750 --> 01:00:44,910 So the virtual work done by this force on that first cart 838 01:00:44,910 --> 01:00:52,070 is just Qx1 equals F1. 839 01:00:52,070 --> 01:00:55,190 That's the generalized force caused 840 01:00:55,190 --> 01:00:57,480 by this first mass on the cart. 841 01:00:57,480 --> 01:00:59,885 What's the generalize force Qx2? 842 01:01:04,310 --> 01:01:09,830 Well, it's some F1. 843 01:01:09,830 --> 01:01:13,040 Actually, there's a dot here. 844 01:01:13,040 --> 01:01:26,870 This would be some x2 with respect to x1 delta x1. 845 01:01:26,870 --> 01:01:31,220 But how much does x2 move when you 846 01:01:31,220 --> 01:01:34,510 cause a little virtual deflection 847 01:01:34,510 --> 01:01:47,460 of-- the virtual work done if I cause 848 01:01:47,460 --> 01:01:50,310 a little deflection of this one is 849 01:01:50,310 --> 01:01:53,800 equal to the summation of the forces that 850 01:01:53,800 --> 01:01:56,040 act through delta x2. 851 01:01:56,040 --> 01:02:02,820 Now, if you move this a little delta x1 here, 852 01:02:02,820 --> 01:02:08,120 we figured out that that Qx1, the generalized force 853 01:02:08,120 --> 01:02:10,910 due to that, is indeed this. 854 01:02:10,910 --> 01:02:14,380 But what's the generalized force associated 855 01:02:14,380 --> 01:02:16,520 with motion delta x2? 856 01:02:16,520 --> 01:02:20,060 Let's move this one now a little bit. 857 01:02:20,060 --> 01:02:23,588 When you move that a little bit, how much work does F1 do? 858 01:02:29,940 --> 01:02:32,430 So we need to get two equations of motion, right? 859 01:02:32,430 --> 01:02:36,070 And you're going to get 1 by taking derivatives 860 01:02:36,070 --> 01:02:40,530 with respect to coordinate x1. 861 01:02:40,530 --> 01:02:45,630 You're going to get an equation of motion, which-- 862 01:02:45,630 --> 01:02:49,920 so EOM x1 associated with x1 double dot 863 01:02:49,920 --> 01:02:56,440 here is going to have on the right hand side some Qx1. 864 01:02:56,440 --> 01:02:57,990 And we figured out what that is. 865 01:02:57,990 --> 01:02:59,830 It's just F1. 866 01:02:59,830 --> 01:03:02,800 And we're going to get a second equation of motion associated 867 01:03:02,800 --> 01:03:05,630 with x2 double dot, the mass acceleration 868 01:03:05,630 --> 01:03:07,410 of the mass of the second one. 869 01:03:07,410 --> 01:03:11,585 And it's going to be equal to some external forces. 870 01:03:11,585 --> 01:03:13,210 And there's other terms in here, right? 871 01:03:13,210 --> 01:03:15,450 We have kx. 872 01:03:15,450 --> 01:03:19,191 You have your k terms and so forth in here. 873 01:03:19,191 --> 01:03:21,690 But on the right hand side are the external non-conservative 874 01:03:21,690 --> 01:03:22,190 forces. 875 01:03:22,190 --> 01:03:25,110 So are there any non-conservative forces 876 01:03:25,110 --> 01:03:26,760 on the second mass? 877 01:03:26,760 --> 01:03:27,260 None. 878 01:03:27,260 --> 01:03:32,670 So what do you expect Qx2 to be? 879 01:03:32,670 --> 01:03:43,610 So to get back to your point, when you're 880 01:03:43,610 --> 01:03:47,320 computing the generalized forces, 881 01:03:47,320 --> 01:03:50,650 you freeze all of the movements except one 882 01:03:50,650 --> 01:03:52,380 and figure out the work done. 883 01:03:52,380 --> 01:03:56,370 Even though in the real system, force 884 01:03:56,370 --> 01:03:59,782 will result in this whole system-- 885 01:03:59,782 --> 01:04:01,490 that whole system will move to the right. 886 01:04:01,490 --> 01:04:04,580 If I put a steady force F1 on there, 887 01:04:04,580 --> 01:04:06,810 the whole system will go off to the right hand side. 888 01:04:06,810 --> 01:04:10,500 That would be the solution to the equations of motion 889 01:04:10,500 --> 01:04:12,000 that you end up with. 890 01:04:12,000 --> 01:04:16,790 But for the purpose of computing the generalized force 891 01:04:16,790 --> 01:04:22,470 on each mass, you only fix where the masses 892 01:04:22,470 --> 01:04:24,810 are at some instant in time. 893 01:04:24,810 --> 01:04:27,380 And then one coordinate at a time 894 01:04:27,380 --> 01:04:28,930 cause a little virtual deflection 895 01:04:28,930 --> 01:04:31,650 and figure out how much work gets done. 896 01:04:31,650 --> 01:04:35,480 So see the distinction between the solution 897 01:04:35,480 --> 01:04:37,060 to the full equation to motion? 898 01:04:37,060 --> 01:04:37,600 Yes, indeed. 899 01:04:37,600 --> 01:04:39,940 Everything's going to move because of that force. 900 01:04:39,940 --> 01:04:41,550 And a little bit of work that gets 901 01:04:41,550 --> 01:04:45,690 done due to the motion of just one coordinate and then 902 01:04:45,690 --> 01:04:52,370 the other coordinate through all of the non-conservative forces 903 01:04:52,370 --> 01:04:55,150 that are applied. 904 01:04:55,150 --> 01:04:57,110 Did that get to your question? 905 01:04:57,110 --> 01:04:59,398 All right. 906 01:04:59,398 --> 01:05:07,410 Now, I'll set up-- I don't think I'll have time to finish this. 907 01:05:17,460 --> 01:05:21,410 So last time we had this problem, this is this rod. 908 01:05:21,410 --> 01:05:24,410 It's got a sleeve, and it's got a spring. 909 01:05:24,410 --> 01:05:27,760 And we figured out the potential and kinetic energy equation 910 01:05:27,760 --> 01:05:28,720 to motion. 911 01:05:28,720 --> 01:05:30,191 This is a planar motion problem. 912 01:05:30,191 --> 01:05:30,690 It pivots. 913 01:05:33,580 --> 01:05:35,220 Thing can slide up and down. 914 01:05:35,220 --> 01:05:39,420 Requires an angle theta and a deflection 915 01:05:39,420 --> 01:05:43,360 x with respect to the rod. 916 01:05:43,360 --> 01:05:46,610 And we figured out t and v, and we found 917 01:05:46,610 --> 01:05:49,810 the equations of motion for it. 918 01:05:49,810 --> 01:05:53,230 So here's my system. 919 01:05:53,230 --> 01:06:15,130 Point A here, spring, sleeve, theta, x1, y1. 920 01:06:22,630 --> 01:06:27,450 And the distance x1 was measured from here to G, 921 01:06:27,450 --> 01:06:28,510 to the center of mass. 922 01:06:31,820 --> 01:06:32,425 That's x1. 923 01:06:35,730 --> 01:06:38,030 This is in the direction here. 924 01:06:38,030 --> 01:06:42,730 This is the i1 unit vector direction. 925 01:06:42,730 --> 01:06:45,940 And the coordinate system is my x1, y1. 926 01:06:45,940 --> 01:06:50,180 So x1, it's down the axis like that. 927 01:06:50,180 --> 01:06:56,270 And this problem is a planar motion problem. 928 01:06:56,270 --> 01:07:00,180 There's two rigid bodies, the rod and the sleeve. 929 01:07:00,180 --> 01:07:02,850 And we can completely describe the motion of the system 930 01:07:02,850 --> 01:07:05,890 with it has two degrees of freedom. 931 01:07:05,890 --> 01:07:08,480 Theta defines the position of the rod, 932 01:07:08,480 --> 01:07:11,170 and x1 defines the position of the sleeve on the rod. 933 01:07:11,170 --> 01:07:13,100 So we have two degrees of freedom. 934 01:07:13,100 --> 01:07:16,610 And when we work this out, we end up 935 01:07:16,610 --> 01:07:20,050 with two equations of motion. 936 01:07:20,050 --> 01:07:24,690 And this was called M2. 937 01:07:24,690 --> 01:07:25,380 This was M1. 938 01:07:30,390 --> 01:07:34,140 So one equation of motion was M2 x1 double 939 01:07:34,140 --> 01:08:15,780 dot That was one equation of motion. 940 01:08:15,780 --> 01:08:19,300 And I'm just leaving the generalized force 941 01:08:19,300 --> 01:08:21,529 out of this for a minute. 942 01:08:21,529 --> 01:08:22,384 We'll figure it out. 943 01:08:22,384 --> 01:08:23,800 And the second equation of motion. 944 01:08:34,258 --> 01:08:35,752 This is the rod. 945 01:09:16,670 --> 01:09:19,410 So you get two equations of motion. 946 01:09:19,410 --> 01:09:21,630 We worked it out in the last lecture. 947 01:09:21,630 --> 01:09:27,010 And we can see this accounts for the mass moment 948 01:09:27,010 --> 01:09:29,250 inertia of the rod, mass moment of inertia 949 01:09:29,250 --> 01:09:33,910 of the sleeve with respect to G. So parallel axis theorem 950 01:09:33,910 --> 01:09:36,479 adds another piece to it. 951 01:09:36,479 --> 01:09:38,680 The whole thing, theta double dot. 952 01:09:38,680 --> 01:09:41,090 And then this is just due to gravity, 953 01:09:41,090 --> 01:09:44,560 gravity acting on the rod, gravity acting on the sleeve. 954 01:09:44,560 --> 01:09:46,109 And on the right hand side, you need 955 01:09:46,109 --> 01:09:50,439 to get these Qx, your generalized forces 956 01:09:50,439 --> 01:09:54,130 for generalized coordinate x1 and generalized 957 01:09:54,130 --> 01:09:55,850 coordinate theta. 958 01:09:55,850 --> 01:10:00,540 So we did it, in fact worked it out last time 959 01:10:00,540 --> 01:10:02,970 doing the intuitive approach. 960 01:10:02,970 --> 01:10:08,670 And what if we were to try to do this then 961 01:10:08,670 --> 01:10:11,890 by the kinematic approach that I described here? 962 01:10:21,880 --> 01:10:25,330 The force was applied here. 963 01:10:25,330 --> 01:10:26,450 It was horizontal. 964 01:10:26,450 --> 01:10:29,626 We called it F2 because it was applied to mass two. 965 01:10:37,930 --> 01:10:40,810 And in order to use this technique, 966 01:10:40,810 --> 01:10:42,940 we want now to compute Qx1. 967 01:10:50,880 --> 01:11:02,950 We have a force F2, and it is in the-- which 968 01:11:02,950 --> 01:11:05,890 way do I want to do it? 969 01:11:05,890 --> 01:11:07,203 We'll have an inertial system. 970 01:11:17,840 --> 01:11:28,170 So we need to describe a vector in-- I better not 971 01:11:28,170 --> 01:11:30,650 draw it out here. 972 01:11:30,650 --> 01:11:32,660 So I'll just set this problem up, 973 01:11:32,660 --> 01:11:34,670 and we'll finish it next time. 974 01:11:34,670 --> 01:11:41,210 I have an inertial system y and x. 975 01:11:44,480 --> 01:11:46,520 So that in this system, this force 976 01:11:46,520 --> 01:11:52,550 would be F2 capital J dotted with the derivative 977 01:11:52,550 --> 01:11:58,970 of some vector that runs here to this point. 978 01:11:58,970 --> 01:12:04,470 And I'll call that R2 in o. 979 01:12:04,470 --> 01:12:13,350 So the derivative of R2 with respect to x1 delta x1. 980 01:12:16,731 --> 01:12:21,840 And if you can work this out, then you're done. 981 01:12:21,840 --> 01:12:25,830 The key to this is figuring out what is R2, 982 01:12:25,830 --> 01:12:28,990 and doing it in unit vectors such 983 01:12:28,990 --> 01:12:30,930 that you can complete this dot product. 984 01:12:33,950 --> 01:12:42,330 So how would you describe what is this position to here, 985 01:12:42,330 --> 01:12:46,770 and what are its unit vectors that break it down? 986 01:12:46,770 --> 01:12:50,560 Take this R2, and you express it in terms 987 01:12:50,560 --> 01:12:57,300 of unit vectors in the inertial capital I capital J system. 988 01:12:57,300 --> 01:13:01,590 And once you've done that, you can take the derivative, 989 01:13:01,590 --> 01:13:05,600 dot it with that, and you're done. 990 01:13:05,600 --> 01:13:07,019 So we'll finish that next time. 991 01:13:07,019 --> 01:13:09,060 You might go off and think about it in your notes 992 01:13:09,060 --> 01:13:09,640 from last time. 993 01:13:09,640 --> 01:13:10,806 We already figured this out. 994 01:13:10,806 --> 01:13:12,910 We just did it the intuitive way. 995 01:13:12,910 --> 01:13:15,420 Drew the F and figured out which part's in the direction 996 01:13:15,420 --> 01:13:18,980 of x1, which part's in the direction of delta theta. 997 01:13:18,980 --> 01:13:19,900 And we figured it out. 998 01:13:19,900 --> 01:13:21,770 So you actually already have the answer. 999 01:13:21,770 --> 01:13:24,390 So go see if you can do that.