WEBVTT 00:00:00.070 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. 00:00:03.800 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.305 at ocw.mit.edu. 00:00:21.810 --> 00:00:23.960 PROFESSOR: All right, let's get started. 00:00:23.960 --> 00:00:28.420 Today is all about Lagrange method. 00:00:28.420 --> 00:00:31.200 We will talk a lot about what we really 00:00:31.200 --> 00:00:36.100 mean by generalized coordinates and generalized forces 00:00:36.100 --> 00:00:41.220 and then do a number of application examples. 00:00:41.220 --> 00:00:46.215 There's a set of notes on Stellar on the Lagrange method. 00:00:46.215 --> 00:00:54.340 It's about 10 pages long and I highly recommend you read them. 00:00:54.340 --> 00:00:56.590 They're not somehow up with the notes 00:00:56.590 --> 00:00:59.240 associated with lecture notes or [INAUDIBLE] 00:00:59.240 --> 00:01:00.210 way down at the bottom. 00:01:00.210 --> 00:01:02.670 So you have to scroll all the way down in the Stellar website 00:01:02.670 --> 00:01:03.211 to find them. 00:01:06.300 --> 00:01:09.260 Our second quiz is November 8. 00:01:09.260 --> 00:01:13.467 That's a week from next Tuesday. 00:01:13.467 --> 00:01:14.750 OK. 00:01:14.750 --> 00:01:18.960 Pretty much same format as the first one. 00:01:18.960 --> 00:01:20.570 OK. 00:01:20.570 --> 00:01:23.985 So let's talk about how to use Lagrange equations. 00:01:32.600 --> 00:01:37.200 So I defined what's called the Lagrangian last time. 00:01:37.200 --> 00:01:41.180 t minus v. The kinetic energy minus the potential energy 00:01:41.180 --> 00:01:43.870 of the entire system. 00:01:43.870 --> 00:01:48.940 Total kinetic and total potential energy expressions. 00:01:48.940 --> 00:01:51.770 Then we have some quantities. 00:01:51.770 --> 00:01:53.540 qj's. 00:01:53.540 --> 00:01:59.050 These are defined as the generalized forces. 00:01:59.050 --> 00:02:01.920 Generalized coordinates, I should say. 00:02:07.260 --> 00:02:12.150 And the capital Q sub j's are the generalized forces. 00:02:19.620 --> 00:02:24.550 And the Lagrange equation says that d 00:02:24.550 --> 00:02:32.130 by dt the time derivative of the partial of l with respect 00:02:32.130 --> 00:02:45.990 to the qj dots, the velocities, minus the partial derivative 00:02:45.990 --> 00:02:52.910 of l with respect to the generalized displacements 00:02:52.910 --> 00:02:56.670 equals the generalized forces. 00:02:56.670 --> 00:03:00.820 And for a typical system, you'll have a number of degrees 00:03:00.820 --> 00:03:02.812 of freedom, like say three. 00:03:02.812 --> 00:03:04.520 And if you have three degrees of freedom, 00:03:04.520 --> 00:03:06.700 you need three equations of motion. 00:03:06.700 --> 00:03:10.520 And so the j's will go from one to three in that case. 00:03:10.520 --> 00:03:13.470 So the j's here refer to an [? index ?] 00:03:13.470 --> 00:03:16.470 that gives you the number of equations that you need. 00:03:16.470 --> 00:03:19.840 So you do this calculation for coordinate one, 00:03:19.840 --> 00:03:22.610 again for coordinate two, again for coordinate three, 00:03:22.610 --> 00:03:26.060 and you get then three equations of motion. 00:03:26.060 --> 00:03:27.050 OK 00:03:27.050 --> 00:03:34.260 So this is a little obscure. 00:03:34.260 --> 00:03:35.520 Let's just plug in. 00:03:41.610 --> 00:03:45.680 For l equals t minus v. And just put it in here 00:03:45.680 --> 00:03:46.630 and see what happens. 00:03:51.600 --> 00:04:02.290 You get d by dt of the partial of t with respect 00:04:02.290 --> 00:04:11.600 to qj dot minus d by dt of the partial of v 00:04:11.600 --> 00:04:27.975 with respect to qj dot plus-- I'll organize it this way. 00:04:30.590 --> 00:04:33.770 Minus the partial of t with respect 00:04:33.770 --> 00:04:43.320 to qj plus the partial of v with respect to qj 00:04:43.320 --> 00:04:46.400 equals capital Qj. 00:04:46.400 --> 00:04:52.200 Now when we first talked about potential energy 00:04:52.200 --> 00:04:54.840 a few days ago, we said that for mechanical systems, 00:04:54.840 --> 00:05:01.070 the potential energy is not a function of time or-- anybody 00:05:01.070 --> 00:05:01.730 remember? 00:05:01.730 --> 00:05:04.200 Velocity. 00:05:04.200 --> 00:05:05.660 So if the potential energy is not 00:05:05.660 --> 00:05:10.420 a function of time nor velocity, what will happen to this term? 00:05:10.420 --> 00:05:11.830 This goes away. 00:05:11.830 --> 00:05:25.370 So this is 0 for mechanical systems. 00:05:25.370 --> 00:05:28.190 If you start getting into electrons moving 00:05:28.190 --> 00:05:31.550 and magnetic fields, then you start have a potential energies 00:05:31.550 --> 00:05:34.290 involving velocities. 00:05:34.290 --> 00:05:37.190 But for mechanical systems, this term's 0. 00:05:37.190 --> 00:05:40.120 And I think the bookkeeping. 00:05:40.120 --> 00:05:41.915 So this is the form of a Lagrange equations 00:05:41.915 --> 00:05:44.120 that I write down when I'm doing problems. 00:05:44.120 --> 00:05:44.960 I don't write this. 00:05:44.960 --> 00:05:49.530 Mathematicians like elegance. 00:05:49.530 --> 00:05:52.470 And this comes down to this is beautifully elegant simple 00:05:52.470 --> 00:05:53.730 looking formula. 00:05:53.730 --> 00:05:56.670 But I'm an engineer and I like it to be 00:05:56.670 --> 00:05:59.710 efficient and practical useful. 00:05:59.710 --> 00:06:03.360 This is the practical useful form of Lagrange equations. 00:06:03.360 --> 00:06:06.310 So you just use what you need. 00:06:06.310 --> 00:06:08.592 Kinetic energy here, kinetic energy there, 00:06:08.592 --> 00:06:09.550 potential energy there. 00:06:09.550 --> 00:06:11.360 And I number these. 00:06:11.360 --> 00:06:13.870 There's a lot of bookkeeping in Lagrange. 00:06:13.870 --> 00:06:19.045 So I call it term one, term two, term three, and term four. 00:06:19.045 --> 00:06:21.420 Because you have to grind through this quite a few times. 00:06:21.420 --> 00:06:23.480 And so when you do, basically you 00:06:23.480 --> 00:06:29.820 take one of the results of 1 plus 2 plus 3 equals 4. 00:06:29.820 --> 00:06:38.968 And you do that j times to get the equations you're after. 00:06:38.968 --> 00:06:39.468 OK. 00:06:46.230 --> 00:06:48.450 So now we need to talk a little bit 00:06:48.450 --> 00:06:55.391 about what we mean by generalized coordinates. 00:06:55.391 --> 00:06:55.890 qj. 00:07:11.630 --> 00:07:13.270 What's this word generalized mean? 00:07:13.270 --> 00:07:15.865 Generalized just means it doesn't have to be Cartesian. 00:07:23.810 --> 00:07:28.280 Not necessarily Cartesian as in xyz. 00:07:28.280 --> 00:07:30.910 You got a lot of liberty and how you choose coordinates. 00:07:30.910 --> 00:07:33.660 Not necessarily Cartesian. 00:07:33.660 --> 00:07:43.050 Not even inertial. 00:07:51.360 --> 00:07:55.110 They do have to satisfy certain requirements. 00:07:55.110 --> 00:08:02.106 The coordinates, they must be what we call independent. 00:08:09.830 --> 00:08:11.803 They must be complete. 00:08:19.280 --> 00:08:21.800 So it must be independent and complete 00:08:21.800 --> 00:08:31.820 and the system must be holonomic. 00:08:36.460 --> 00:08:38.600 I'll get to that in a minute. 00:08:38.600 --> 00:08:41.330 So you need to understand what it means to be independent, 00:08:41.330 --> 00:08:42.350 complete, and holonomic. 00:09:02.410 --> 00:09:04.740 So what do we mean by independent? 00:09:04.740 --> 00:09:09.150 So if you have a multiple degree of freedom system 00:09:09.150 --> 00:09:13.370 and you fix all but one of the coordinates, 00:09:13.370 --> 00:09:18.810 say the system can't move in all but one of its coordinates. 00:09:18.810 --> 00:09:20.980 That last degree of freedom still 00:09:20.980 --> 00:09:23.760 has to have a complete range of motion. 00:09:23.760 --> 00:09:27.310 So if you have a double pendulum and you grab the first mask, 00:09:27.310 --> 00:09:29.050 the second mask can still move. 00:09:29.050 --> 00:09:33.705 It takes two angles to define your double pendulum. 00:09:33.705 --> 00:09:34.330 So independent. 00:09:44.000 --> 00:09:58.600 When you fix all but one coordinate, 00:09:58.600 --> 00:10:17.470 still have a continuous range of movement 00:10:17.470 --> 00:10:20.890 essentially in the free coordinate. 00:10:26.720 --> 00:10:29.190 And that's independent. 00:10:29.190 --> 00:10:35.040 And we'll do this by example mostly. 00:10:35.040 --> 00:10:37.240 And complete. 00:10:37.240 --> 00:10:41.250 The complete really means it's capable of locating all parts 00:10:41.250 --> 00:10:44.385 of the system at all times. 00:11:12.020 --> 00:11:14.875 So let's look at a system here. 00:11:14.875 --> 00:11:15.840 It's a double pendulum. 00:11:28.535 --> 00:11:35.240 It's a simple one just made out of two particles and strings. 00:11:35.240 --> 00:11:38.340 I didn't bring one today. 00:11:38.340 --> 00:11:43.700 And I need to pick some coordinates to describe this. 00:11:43.700 --> 00:11:49.040 And we'll use some Cartesian coordinates. 00:11:49.040 --> 00:11:52.950 Here's an x and a y. 00:11:52.950 --> 00:11:54.030 And here's particle one. 00:11:56.790 --> 00:12:01.640 And I could choose to describe this system xy coordinates. 00:12:01.640 --> 00:12:06.900 And I'll specify the location in the system with coordinates x1 00:12:06.900 --> 00:12:08.050 and y1. 00:12:08.050 --> 00:12:11.290 Two values to specify the location of that. 00:12:11.290 --> 00:12:16.350 And down here I'm going to pick two more values, x2 and y2, 00:12:16.350 --> 00:12:22.170 just to describe the-- so x1 is a different coordinate from x2. 00:12:22.170 --> 00:12:25.380 x1 is the exposition of particle one. 00:12:25.380 --> 00:12:31.150 x2 is the x position of particle two. y1 and y2. 00:12:31.150 --> 00:12:34.670 So how many coordinates do I have to describe the system? 00:12:38.240 --> 00:12:40.330 How many have I used? 00:12:40.330 --> 00:12:41.530 Four, right? 00:12:41.530 --> 00:12:45.680 How many degrees of freedom do you think this problem has? 00:12:45.680 --> 00:12:46.297 Two. 00:12:46.297 --> 00:12:48.630 So there's something already a little out of whack here. 00:12:51.710 --> 00:12:55.930 But the point is these aren't independent, you'll find. 00:12:55.930 --> 00:12:56.961 You just do a test. 00:12:56.961 --> 00:12:58.710 You'll find that these aren't independent. 00:12:58.710 --> 00:13:03.330 If I fix x1 and x2, systems doesn't move. 00:13:07.066 --> 00:13:12.110 If I say this is going to be one and this has got to be three, 00:13:12.110 --> 00:13:15.330 this system is now frozen. 00:13:15.330 --> 00:13:18.530 So this system of core coordinates is not independent. 00:13:21.054 --> 00:13:21.720 What did we say? 00:13:21.720 --> 00:13:22.420 Independent. 00:13:22.420 --> 00:13:25.240 When you fix all but one coordinate, 00:13:25.240 --> 00:13:29.280 you still have continuous range of movement of the final one. 00:13:29.280 --> 00:13:30.960 I could fix only just two of these 00:13:30.960 --> 00:13:32.610 and I've frozen the system. 00:13:32.610 --> 00:13:36.390 I don't even have to go to the extent of fixing three. 00:13:36.390 --> 00:13:38.760 I'm assuming the strings are of fixed length. 00:13:38.760 --> 00:13:41.540 You can't change the string length. 00:13:41.540 --> 00:13:43.610 So this is not a very good choice of coordinates. 00:13:43.610 --> 00:13:45.510 And we had a hint that it might not be, 00:13:45.510 --> 00:13:47.590 because it's more than we ought to use. 00:13:47.590 --> 00:13:50.090 We only really need two. 00:13:50.090 --> 00:13:58.240 So and then if we choose these angles, v1 and v2, 00:13:58.240 --> 00:13:59.720 let's do the test with that. 00:13:59.720 --> 00:14:01.929 Are those independent? 00:14:01.929 --> 00:14:03.720 So those are the coordinates of the system. 00:14:03.720 --> 00:14:07.880 If you fix v1, is there still free and continuous movement 00:14:07.880 --> 00:14:09.880 of v2 of the system? 00:14:09.880 --> 00:14:10.540 Sure. 00:14:10.540 --> 00:14:14.680 And if you fixed v2, it means you can require this angle stay 00:14:14.680 --> 00:14:17.600 rigid like that, and move v2, well, the whole system 00:14:17.600 --> 00:14:18.820 will still move. 00:14:18.820 --> 00:14:24.160 So v1 and v2 are a system which satisfies the independence 00:14:24.160 --> 00:14:25.280 requirement. 00:14:25.280 --> 00:14:26.270 Complete. 00:14:26.270 --> 00:14:28.410 They're both systems that are complete. 00:14:28.410 --> 00:14:31.110 They're both capable locating all points at all times. 00:14:33.690 --> 00:14:39.400 But only the pair v1 and v2 in this example 00:14:39.400 --> 00:14:43.880 are both independent and complete. 00:14:43.880 --> 00:14:52.060 Now, the third requirement is a thing called holonomicity. 00:14:52.060 --> 00:14:56.360 And what it means to be holonomic 00:14:56.360 --> 00:15:01.860 is that the system, the number of degrees of freedom required 00:15:01.860 --> 00:15:07.170 is equal to the number of coordinates 00:15:07.170 --> 00:15:10.560 required to completely describe the motion. 00:15:10.560 --> 00:15:15.320 Now, every example we've ever done so far in this class 00:15:15.320 --> 00:15:16.570 satisfies that. 00:15:16.570 --> 00:15:21.470 We picked v1 and v2 and that's all the coordinates 00:15:21.470 --> 00:15:24.400 that we need to completely to describe the motion. 00:15:24.400 --> 00:15:28.495 Let me see if I can figure out a counter example. 00:15:34.690 --> 00:15:36.279 I didn't write down this definition. 00:15:36.279 --> 00:15:36.820 So holonomic. 00:16:14.280 --> 00:16:17.480 And if the answer to this question is no, 00:16:17.480 --> 00:16:21.330 you cannot use Lagrange equations. 00:16:21.330 --> 00:16:28.410 So let's see if I can show you an example of a system in which 00:16:28.410 --> 00:16:33.070 you need more coordinates than you have degrees of freedom. 00:16:33.070 --> 00:16:35.960 I've got a ball. 00:16:35.960 --> 00:16:38.740 This is an xy plane. 00:16:38.740 --> 00:16:44.744 And I'm not going to allow it to translate in z. 00:16:44.744 --> 00:16:49.910 And I'm not going to allow it to rotate about the z-axis. 00:16:49.910 --> 00:16:53.140 So those are two constraints. 00:16:53.140 --> 00:16:55.430 So this is one rigid body. 00:16:55.430 --> 00:16:58.720 In general how many degrees of freedom does it have? 00:16:58.720 --> 00:17:00.360 Six. 00:17:00.360 --> 00:17:03.040 I'm going to constrain it so no z motion. 00:17:03.040 --> 00:17:04.250 Five. 00:17:04.250 --> 00:17:06.020 No z rotation. 00:17:06.020 --> 00:17:08.710 Four. 00:17:08.710 --> 00:17:11.800 It's not going to allow it to slip. 00:17:11.800 --> 00:17:13.609 This is x and that's y. 00:17:13.609 --> 00:17:16.910 I'm not going to allow it to slip in the x. 00:17:16.910 --> 00:17:19.140 So now I've got another. 00:17:19.140 --> 00:17:21.520 Now I'm down to three. 00:17:21.520 --> 00:17:25.500 And I'm not going to allow it to slip in the y. 00:17:25.500 --> 00:17:26.520 Two. 00:17:26.520 --> 00:17:30.060 So by our calculus of how many degrees of freedom you need, 00:17:30.060 --> 00:17:31.600 we're down to two. 00:17:31.600 --> 00:17:33.720 We should be able to completely describe 00:17:33.720 --> 00:17:36.870 the motion of this system with two coordinates. 00:17:36.870 --> 00:17:37.650 OK. 00:17:37.650 --> 00:17:41.230 So I've put this piece of tape on the top. 00:17:41.230 --> 00:17:45.060 And it's pointing diagonally. 00:17:45.060 --> 00:17:46.490 That way. 00:17:46.490 --> 00:17:51.900 And I'm going to roll this ball like this 00:17:51.900 --> 00:17:53.104 until it shows up again. 00:17:53.104 --> 00:17:55.020 So it's right on top, just the way it started. 00:17:57.980 --> 00:18:01.540 Now start off same way again. 00:18:01.540 --> 00:18:03.580 I'm going to roll first this way. 00:18:06.260 --> 00:18:09.900 And then I'm going to roll this way to the same place. 00:18:12.470 --> 00:18:14.815 Where's the stripe? 00:18:14.815 --> 00:18:15.800 It's in the back. 00:18:18.450 --> 00:18:21.760 So I've gone to the same position 00:18:21.760 --> 00:18:25.550 but I've ended up with the ball not in the same orientation 00:18:25.550 --> 00:18:26.380 as it was. 00:18:26.380 --> 00:18:27.770 I went by two different paths. 00:18:35.680 --> 00:18:39.400 And the ball comes up over here rather than up there 00:18:39.400 --> 00:18:40.351 where it started. 00:18:40.351 --> 00:18:40.850 OK 00:18:40.850 --> 00:18:46.050 So to actually describe where the ball is at any place 00:18:46.050 --> 00:18:49.890 out here, having gotten there by rolling around, 00:18:49.890 --> 00:18:56.095 without slipping and without z rotation, how many coordinates 00:18:56.095 --> 00:18:59.290 do you think it'll take to actually specify where 00:18:59.290 --> 00:19:03.366 that stripe is at any arbitrary place 00:19:03.366 --> 00:19:04.740 that it's gotten to on the plane? 00:19:09.510 --> 00:19:10.558 Name them. 00:19:10.558 --> 00:19:11.474 AUDIENCE: [INAUDIBLE]. 00:19:16.460 --> 00:19:18.220 PROFESSOR: So. 00:19:18.220 --> 00:19:19.980 In order to actually fully describe it, 00:19:19.980 --> 00:19:21.670 you've got to say where it is x and y 00:19:21.670 --> 00:19:24.570 and you actually have to say some kind of theta and phi 00:19:24.570 --> 00:19:28.140 rotations that it's gone through so that you know where this is. 00:19:28.140 --> 00:19:32.750 So this system is not holonomic. 00:19:32.750 --> 00:19:38.167 And it has to be holonomic in order 00:19:38.167 --> 00:19:39.250 to use Lagrange equations. 00:19:43.310 --> 00:19:45.170 So when you go to do Lagrange problems, 00:19:45.170 --> 00:19:48.080 you need to test for your coordinates. 00:19:48.080 --> 00:19:51.100 Complete, independent, and holonomic. 00:19:51.100 --> 00:19:52.350 And you get pretty good at it. 00:20:12.330 --> 00:20:14.850 So here's my Lagrange equations. 00:20:14.850 --> 00:20:20.150 And I have itemized these four calculations you have to do. 00:20:20.150 --> 00:20:23.620 Call them one, two, three, and four. 00:20:23.620 --> 00:20:25.930 And what I'm going to write out is just 00:20:25.930 --> 00:20:30.540 to get you to adopt a systematic approach to doing Lagrange. 00:20:43.380 --> 00:20:44.670 Left hand side. 00:20:44.670 --> 00:20:47.780 To the left hand side of your equations of motion 00:20:47.780 --> 00:20:50.010 is everything with t and v. The right hand side 00:20:50.010 --> 00:20:52.480 has these generalized forces that you have to deal with. 00:20:52.480 --> 00:20:55.740 And generalized forces are the non conservative forces 00:20:55.740 --> 00:20:57.194 in the system. 00:20:57.194 --> 00:20:59.110 So this is going to get a little bit cookbook, 00:20:59.110 --> 00:21:02.900 but it's, I think, appropriate for the moment here. 00:21:02.900 --> 00:21:03.810 So step one. 00:21:10.550 --> 00:21:14.940 Determine the number of degrees of freedom that you need. 00:21:19.200 --> 00:21:26.960 And choose your delta j's. 00:21:26.960 --> 00:21:28.620 Not deltas, excuse me. 00:21:28.620 --> 00:21:30.640 qj's. 00:21:30.640 --> 00:21:31.880 Choose your coordinates. 00:21:31.880 --> 00:21:34.210 You find the number of degrees of freedom 00:21:34.210 --> 00:21:38.050 and choose the coordinates you're going to use, basically. 00:21:43.980 --> 00:21:57.010 Verify complete, independent, holonomic. 00:22:03.811 --> 00:22:04.310 Three. 00:22:08.270 --> 00:22:14.090 Compute t and v for every rigid body in the system. 00:22:14.090 --> 00:22:15.965 Compute your kinetic and potential energies. 00:22:24.680 --> 00:22:40.616 One, two, three for each qj. 00:22:40.616 --> 00:22:41.990 So for every coordinate you have, 00:22:41.990 --> 00:22:43.550 you have to go through these computations. 00:22:43.550 --> 00:22:45.383 One, two, three, four, for every coordinate. 00:22:48.080 --> 00:22:49.500 And this is your left hand side. 00:22:52.511 --> 00:22:54.760 And if you don't have any external forces and your non 00:22:54.760 --> 00:22:59.400 conservative external forces, then 1 plus 2 plus 3 equals 0. 00:22:59.400 --> 00:23:01.930 But if you have non conservative forces, 00:23:01.930 --> 00:23:04.330 then you have to compute the right hand side. 00:23:04.330 --> 00:23:05.480 So the right hand side. 00:23:12.470 --> 00:23:21.560 So for each qj, each generalized coordinate, 00:23:21.560 --> 00:23:31.920 you need to find the generalized force that 00:23:31.920 --> 00:23:33.120 potentially goes with it. 00:23:41.580 --> 00:24:00.450 And you do this by computing the virtual work delta w. 00:24:00.450 --> 00:24:02.360 I'll put the little nc up here to remind you 00:24:02.360 --> 00:24:04.430 these are for the non conservative forces. 00:24:04.430 --> 00:24:30.770 The delta w associated with the virtual displacement delta qj. 00:24:30.770 --> 00:24:33.710 So for every generalized coordinate you have, 00:24:33.710 --> 00:24:39.320 you're going to try out this little delta of motion 00:24:39.320 --> 00:24:47.040 in that coordinate and figure out how much virtual 00:24:47.040 --> 00:24:49.320 work you've done. 00:24:49.320 --> 00:24:56.950 So delta wj is going to be qj delta k. 00:24:56.950 --> 00:25:01.390 So this is the thing you're looking for. 00:25:01.390 --> 00:25:05.820 And it's going to be a function of all those external non 00:25:05.820 --> 00:25:09.940 conservative forces acting through a little virtual 00:25:09.940 --> 00:25:12.310 displacement, a little bit of work will be done. 00:25:12.310 --> 00:25:15.140 Mostly I'm going to teach you how to do this by example. 00:25:36.350 --> 00:25:43.120 So let's quickly do a really simple trivial system. 00:25:43.120 --> 00:25:49.880 Our mass spring dashpot system, single agree freedom mkb. 00:25:53.380 --> 00:25:55.520 It's going to take one coordinate 00:25:55.520 --> 00:25:57.840 to describe the motion. 00:25:57.840 --> 00:26:01.590 X happens to be Cartesian. 00:26:01.590 --> 00:26:04.580 There'll be one generalized coordinate. 00:26:04.580 --> 00:26:13.470 So qj equals q1 equals qx in this case. 00:26:13.470 --> 00:26:15.205 It's our x-coordinate. 00:26:15.205 --> 00:26:18.030 Actually I should just call it x. 00:26:18.030 --> 00:26:21.060 That's our generalized coordinates for this problem. 00:26:21.060 --> 00:26:22.190 Is it complete? 00:26:22.190 --> 00:26:23.100 Yeah. 00:26:23.100 --> 00:26:24.291 Is it independent? 00:26:24.291 --> 00:26:24.790 Yes. 00:26:24.790 --> 00:26:25.670 Is it holonomic? 00:26:25.670 --> 00:26:26.420 No problem. 00:26:29.360 --> 00:26:36.010 We need t 1/2 mx dot squared. 00:26:36.010 --> 00:26:42.920 We need v. And we have 1/2 kx squared 00:26:42.920 --> 00:26:53.490 for the spring minus mgx for the gravitational potential energy. 00:26:53.490 --> 00:26:55.880 And now we can start. 00:26:55.880 --> 00:26:59.587 And we have some external non conservative forces. 00:26:59.587 --> 00:27:00.170 What are they? 00:27:02.790 --> 00:27:04.060 fi non conservative. 00:27:08.110 --> 00:27:10.370 And I think I'm going to put an excitation up 00:27:10.370 --> 00:27:14.180 here too, some f of t. 00:27:14.180 --> 00:27:18.340 So what are the non conservative forces? 00:27:18.340 --> 00:27:19.380 Pardon? 00:27:19.380 --> 00:27:21.630 AUDIENCE: k x dot. 00:27:21.630 --> 00:27:22.970 PROFESSOR: It's not k. 00:27:27.090 --> 00:27:28.410 My mistake. 00:27:28.410 --> 00:27:31.190 You're correct. 00:27:31.190 --> 00:27:33.870 My brain is getting ahead of my writing here. 00:27:33.870 --> 00:27:35.663 That's normally b and this would be k. 00:27:35.663 --> 00:27:37.530 I'm not trying to really mess you up there. 00:27:37.530 --> 00:27:41.010 So would be bx dot, right. 00:27:41.010 --> 00:27:43.670 And is there anything else? 00:27:43.670 --> 00:27:45.900 Are there any other non conservative forces, 00:27:45.900 --> 00:27:48.762 things that could put energy into or out of the system? 00:27:48.762 --> 00:27:49.678 AUDIENCE: [INAUDIBLE]. 00:27:55.140 --> 00:27:58.015 PROFESSOR: So the damper can certainly extract energy. 00:27:58.015 --> 00:27:58.952 AUDIENCE: [INAUDIBLE]. 00:27:58.952 --> 00:28:00.160 PROFESSOR: Yeah, the force f. 00:28:00.160 --> 00:28:02.420 That external, it might be something that's 00:28:02.420 --> 00:28:03.670 making it vibrate or whatever. 00:28:03.670 --> 00:28:06.520 But it's an external force, and it could do work on the system. 00:28:06.520 --> 00:28:10.750 And it's not a potential. 00:28:10.750 --> 00:28:15.115 It's not a spring and it's not gravity. 00:28:15.115 --> 00:28:16.910 It's coming up and somebody's shaking 00:28:16.910 --> 00:28:18.330 it or something like that. 00:28:18.330 --> 00:28:19.960 So f is also non conservative. 00:28:19.960 --> 00:28:23.290 So the non conservative forces in this thing 00:28:23.290 --> 00:28:31.580 are f in the i direction and minus vx dot 00:28:31.580 --> 00:28:32.790 in the i direction. 00:28:32.790 --> 00:28:35.350 And we could, in our normal approach using Newton, 00:28:35.350 --> 00:28:39.080 we draw a free body diagram and we identify 00:28:39.080 --> 00:28:43.260 a bx dot on it and an f on it. 00:28:43.260 --> 00:28:47.710 But we'd also have our kx on it. 00:28:47.710 --> 00:28:50.480 That would be what our free body diagram would look like. 00:28:50.480 --> 00:28:51.830 That's a conservative force. 00:28:51.830 --> 00:28:55.580 Oops, and we need an mg. 00:28:55.580 --> 00:28:59.350 So we have two conservative forces, kx and mg, 00:28:59.350 --> 00:29:03.760 and we have two non conservative forces, bx dot and f. 00:29:03.760 --> 00:29:08.160 So in this case, some of the non conservative forces 00:29:08.160 --> 00:29:11.710 is that f in the i direction minus bx in the x 00:29:11.710 --> 00:29:14.990 dot in the i direction. 00:29:14.990 --> 00:29:19.940 So let's do our calculus here. 00:29:19.940 --> 00:29:25.520 So 1 d by dt of the partial of the t, which 00:29:25.520 --> 00:29:33.520 is 1/2 mx dot squared with respect to x dot. 00:29:33.520 --> 00:29:36.970 So that gives me the derivative of x dot squared with respect 00:29:36.970 --> 00:29:40.320 to x dot gives me 2 mx dot. 00:29:40.320 --> 00:29:45.690 So this is d by dt of mx dot. 00:29:45.690 --> 00:29:48.250 But that's mx double dot. 00:29:48.250 --> 00:29:49.942 And as you might expect when you're 00:29:49.942 --> 00:29:51.400 trying to drive equation of motion, 00:29:51.400 --> 00:29:53.608 you're probably going to end up with an mx double dot 00:29:53.608 --> 00:29:55.480 in the result. And it always comes out 00:29:55.480 --> 00:29:58.830 of these d by dt expressions. 00:29:58.830 --> 00:30:01.180 OK, so that's term one. 00:30:01.180 --> 00:30:04.190 Term two in this problem. 00:30:04.190 --> 00:30:09.690 Minus t with respect to x in this case. 00:30:09.690 --> 00:30:11.210 Is t a function of x? 00:30:18.560 --> 00:30:21.040 It's 1/2 mx dot squared. 00:30:21.040 --> 00:30:24.160 So is t a function of displacement x? 00:30:24.160 --> 00:30:26.170 It's a function of velocity in the x direction, 00:30:26.170 --> 00:30:28.940 but is it a function of displacement? 00:30:28.940 --> 00:30:29.490 No. 00:30:29.490 --> 00:30:34.850 So this term is 0. 00:30:34.850 --> 00:30:35.610 Three. 00:30:35.610 --> 00:30:38.090 Our third term. 00:30:38.090 --> 00:30:45.200 Partial of v with respect to x. 00:30:45.200 --> 00:30:47.640 Well, where's v 1/2 kx squared. 00:30:47.640 --> 00:30:51.075 The derivative of this is kx minus mg. 00:30:57.640 --> 00:30:58.595 And we sum those. 00:31:05.170 --> 00:31:17.620 So we get mx double dot plus kx minus mg equals. 00:31:17.620 --> 00:31:21.490 And on the right hand side this is 4. 00:31:21.490 --> 00:31:25.600 Now we need to do four for the right hand side. 00:31:25.600 --> 00:31:36.110 And four is really the summation of the fi's, 00:31:36.110 --> 00:31:43.880 the individual forces, dotted with dr. 00:31:43.880 --> 00:31:44.856 These are both vectors. 00:31:47.970 --> 00:31:50.226 dr is the movement. 00:31:50.226 --> 00:31:52.600 Little bit of work and it's going to be a delta quantity, 00:31:52.600 --> 00:31:54.390 like delta x. 00:31:54.390 --> 00:31:56.800 And f are the applied forces. 00:31:56.800 --> 00:31:59.010 And you need to sum these up. 00:31:59.010 --> 00:32:04.540 So this dr in general is going to be a function of the delta 00:32:04.540 --> 00:32:05.260 j's. 00:32:05.260 --> 00:32:08.780 The virtual displacements in all the possible degrees of freedom 00:32:08.780 --> 00:32:10.000 of the system. 00:32:10.000 --> 00:32:11.302 We do them one at a time. 00:32:11.302 --> 00:32:13.260 This case we'll only have one, so it's trivial. 00:32:13.260 --> 00:32:16.720 But this could be delta one, two, three, four. 00:32:16.720 --> 00:32:20.710 And each one of them might do some work 00:32:20.710 --> 00:32:24.470 when f moves through it. 00:32:24.470 --> 00:32:28.200 But work is f dot the displacement. 00:32:28.200 --> 00:32:30.770 So it's the component of the force 00:32:30.770 --> 00:32:34.360 in the direction of the movement, the dot product, that 00:32:34.360 --> 00:32:36.870 gives you this little bit of virtual work. 00:32:36.870 --> 00:32:43.740 OK, so in this problem, this is going to be equal to-- we 00:32:43.740 --> 00:32:47.570 actually have an f of t of some function of time in the i 00:32:47.570 --> 00:32:58.930 direction minus bx dot in the i direction dotted width delta 00:32:58.930 --> 00:33:01.070 x, which is our virtual displacement 00:33:01.070 --> 00:33:03.420 in our single generalized coordinate. 00:33:03.420 --> 00:33:10.130 And this whole thing is going to be equal to Qx delta x. 00:33:15.350 --> 00:33:18.160 So you figure out the virtual work that's done. 00:33:21.930 --> 00:33:23.390 So if you do this dot product, this 00:33:23.390 --> 00:33:25.020 is also in the i hat direction. 00:33:25.020 --> 00:33:27.090 So i dot i, i dot i. 00:33:27.090 --> 00:33:28.480 You just get ones. 00:33:28.480 --> 00:33:30.890 Because the forces are in the same direction 00:33:30.890 --> 00:33:34.340 as the displacement. 00:33:34.340 --> 00:33:36.500 You're going to get an ft. 00:33:36.500 --> 00:33:40.320 f of t delta x is one of the little bits of virtual work. 00:33:40.320 --> 00:33:44.870 And you'll get a minus bx dot delta x. 00:33:44.870 --> 00:33:48.800 And that together, those two pieces added together, 00:33:48.800 --> 00:33:53.800 are the generalized force times delta x. 00:33:53.800 --> 00:33:58.950 This total here gives you delta w non 00:33:58.950 --> 00:34:05.460 conservative for in this case coordinate x. 00:34:05.460 --> 00:34:10.909 So we're trying to solve for what 00:34:10.909 --> 00:34:12.219 goes on the right hand side. 00:34:12.219 --> 00:34:13.160 We need the qx. 00:34:17.400 --> 00:34:20.469 You notice what'll happen, it'll cancel out 00:34:20.469 --> 00:34:24.460 the delta x is the result. And in this case, what 00:34:24.460 --> 00:34:40.430 you're left with is Qx equals f of t minus bx dot. 00:34:40.430 --> 00:34:41.745 So this is number four. 00:34:45.489 --> 00:34:46.429 So Qx. 00:34:46.429 --> 00:34:50.120 Delta x is the bit of virtual work that's done. 00:34:50.120 --> 00:34:55.040 What goes into our equation of motion is the Qx part. 00:34:55.040 --> 00:34:58.330 And we got it by computing the virtual work done 00:34:58.330 --> 00:35:01.230 by the applied external non conservative forces 00:35:01.230 --> 00:35:06.440 as we imagine them going through delta x. 00:35:06.440 --> 00:35:07.570 And we're done. 00:35:07.570 --> 00:35:11.005 You have the complete equation of motion for a single degree 00:35:11.005 --> 00:35:11.630 freedom system. 00:35:11.630 --> 00:35:13.300 You could rearrange it a little bit. 00:35:13.300 --> 00:35:18.830 mx double dot plus bx dot plus kx equals mg plus f of t 00:35:18.830 --> 00:35:20.260 if you well. 00:35:20.260 --> 00:35:24.230 So it's the same thing you would have gotten from using Newton. 00:35:24.230 --> 00:35:27.950 In a trivial kind of example, but it 00:35:27.950 --> 00:35:31.767 helps to find each of the steps, things that we said 00:35:31.767 --> 00:35:32.350 were required. 00:35:36.470 --> 00:35:43.740 OK, so we're going to go from there to a much harder problem. 00:35:43.740 --> 00:35:51.900 So any issues or questions about definitions, procedure? 00:35:51.900 --> 00:35:54.520 So we start getting into multiple degrees of freedom. 00:35:54.520 --> 00:35:58.570 You need is set up a careful bookkeeping. 00:35:58.570 --> 00:36:01.550 So I just do this myself. 00:36:01.550 --> 00:36:07.250 The top of the page, I identify my coordinates, write down t, 00:36:07.250 --> 00:36:11.840 write down v. Then I say, OK, coordinate one. 00:36:11.840 --> 00:36:14.250 One, two, three, four. 00:36:14.250 --> 00:36:15.020 Equation. 00:36:15.020 --> 00:36:17.020 Then I started the coordinate two. 00:36:17.020 --> 00:36:20.840 Calculus for one, two, three, and four and so forth until you 00:36:20.840 --> 00:36:22.431 get to the end. 00:36:22.431 --> 00:36:23.290 OK, questions? 00:36:23.290 --> 00:36:24.962 Yeah. 00:36:24.962 --> 00:36:28.834 AUDIENCE: On the [INAUDIBLE] what's 00:36:28.834 --> 00:36:30.931 that thing after the [INAUDIBLE] It's 00:36:30.931 --> 00:36:33.690 like an open parentheses [INAUDIBLE]. 00:36:33.690 --> 00:36:39.130 PROFESSOR: Oh, these are functions of the delta j's. 00:36:39.130 --> 00:36:42.790 This dr, where it comes from, the work 00:36:42.790 --> 00:36:45.480 that's being done in a virtual displacement 00:36:45.480 --> 00:36:50.810 around a dynamic equilibrium position for the system 00:36:50.810 --> 00:36:53.000 is a little movement of the system. 00:36:53.000 --> 00:36:56.460 dr. And we express it. 00:36:56.460 --> 00:37:02.160 It's expressed in terms of a virtual displacement 00:37:02.160 --> 00:37:05.770 of the generalized coordinates of the system. 00:37:05.770 --> 00:37:08.540 So where the dr comes from is going to be delta. 00:37:08.540 --> 00:37:10.160 In this case, it's only delta x. 00:37:12.740 --> 00:37:15.020 And in the next problem, we're going 00:37:15.020 --> 00:37:17.370 to do the force in the problem is 00:37:17.370 --> 00:37:22.970 not in exactly the same direction as the delta x's 00:37:22.970 --> 00:37:24.730 and delta theta's and so forth. 00:37:24.730 --> 00:37:28.070 So when you do the dot product, only that complement 00:37:28.070 --> 00:37:29.930 of the force that's in the direction 00:37:29.930 --> 00:37:33.090 of the virtual displacement does work. 00:37:33.090 --> 00:37:34.960 And you account for that. 00:37:34.960 --> 00:37:41.330 So let's look into a more difficult problem. 00:37:41.330 --> 00:37:42.830 So the problem is this. 00:37:42.830 --> 00:37:45.460 I tried to fix assessed before I came to class. 00:37:45.460 --> 00:37:51.070 I didn't really quite have the parts and pieces I needed. 00:37:51.070 --> 00:37:55.720 But this a piece of steel pipe here. 00:37:55.720 --> 00:37:59.630 It's a sleeve on the outside of this rod. 00:37:59.630 --> 00:38:06.340 And I've got a spring that's on the outside connected 00:38:06.340 --> 00:38:07.390 to this piece. 00:38:07.390 --> 00:38:08.397 And so it can do this. 00:38:12.640 --> 00:38:16.200 And it's also, though, a pendulum. 00:38:16.200 --> 00:38:19.430 So the system I really want to look at is this system. 00:38:24.330 --> 00:38:27.960 So this swings back and forth, the thing slides up and down. 00:38:27.960 --> 00:38:32.960 So this has multiple sources of kinetic energy, multiple forms 00:38:32.960 --> 00:38:36.520 of potential energy. 00:38:36.520 --> 00:38:39.730 And for the purpose of the problem, 00:38:39.730 --> 00:38:44.500 I'm going to say that there's a force that's always horizontal 00:38:44.500 --> 00:38:49.640 acting on this mass pushing this system back and forth. 00:38:49.640 --> 00:38:53.050 Some f cosine omega t, always horizontal. 00:38:53.050 --> 00:38:54.560 And I want to drive the equations 00:38:54.560 --> 00:38:55.600 of motion of the system. 00:38:58.390 --> 00:39:02.170 So is it a planar motion problem? 00:39:06.060 --> 00:39:08.020 How many rigid bodies involved? 00:39:13.260 --> 00:39:15.250 There's two rigid bodies. 00:39:15.250 --> 00:39:18.440 Each could have possibly six degrees of freedom. 00:39:18.440 --> 00:39:21.740 But when you say it's a planar motion, 00:39:21.740 --> 00:39:24.460 you're actually immediately confining each rigid body 00:39:24.460 --> 00:39:25.830 to three. 00:39:25.830 --> 00:39:29.780 Each rigid body can move x and y and rotate in z. 00:39:29.780 --> 00:39:33.220 So when you [? spread ?] out and say this is planar motion, 00:39:33.220 --> 00:39:35.940 you've just said each rigid body has max three. 00:39:35.940 --> 00:39:39.705 So this is a maximum of six possible. 00:39:39.705 --> 00:39:41.080 Where the other three disappeared 00:39:41.080 --> 00:39:46.330 to is no z deflection and no rotation in the x or y. 00:39:46.330 --> 00:39:53.360 OK, so we have a possible maximum six. 00:39:53.360 --> 00:39:56.640 How many degrees of freedom does this problem have? 00:39:56.640 --> 00:40:01.609 How many coordinates will we need to completely describe 00:40:01.609 --> 00:40:02.650 the motion of the system? 00:40:02.650 --> 00:40:04.160 So think about that. 00:40:04.160 --> 00:40:05.170 Talk to a neighbor. 00:40:05.170 --> 00:40:08.730 Decide on the coordinates that we need to use for this system 00:40:08.730 --> 00:40:09.900 while I'm drawing it. 00:41:27.180 --> 00:41:28.547 OK. 00:41:28.547 --> 00:41:29.380 What did you decide? 00:41:29.380 --> 00:41:31.090 How many? 00:41:31.090 --> 00:41:31.850 Two. 00:41:31.850 --> 00:41:33.350 All right, what would you recommend? 00:41:37.994 --> 00:41:38.910 What would you choose? 00:41:41.742 --> 00:41:42.242 Pardon? 00:41:42.242 --> 00:41:44.630 AUDIENCE: The angle and how far down it is. 00:41:44.630 --> 00:41:46.480 PROFESSOR: An angle and a deflection 00:41:46.480 --> 00:41:49.860 of what I'm calling m2 here. 00:41:49.860 --> 00:41:51.740 So this is m2. 00:41:51.740 --> 00:41:52.738 The rod is m1. 00:41:55.366 --> 00:42:06.950 And he's suggesting an angle theta and a deflection 00:42:06.950 --> 00:42:09.280 which I'll call x1. 00:42:09.280 --> 00:42:14.450 And I've attached to this bar, the rod I'm calling it, 00:42:14.450 --> 00:42:20.660 a rotating coordinate system x1 y1. 00:42:20.660 --> 00:42:25.000 About point A. So A x1 y1's my rotating coordinate system 00:42:25.000 --> 00:42:28.490 attached to this rod. 00:42:28.490 --> 00:42:29.240 OK. 00:42:29.240 --> 00:42:31.520 So I'm going to locate the position of this 00:42:31.520 --> 00:42:35.510 by some value x1 measured from point A. 00:42:35.510 --> 00:42:40.910 And locate the position of the rod itself by an angle theta. 00:42:40.910 --> 00:42:42.700 Good. 00:42:42.700 --> 00:42:44.330 Is it complete? 00:42:44.330 --> 00:42:48.380 So if you freeze one, do you still have-- the complete. 00:42:48.380 --> 00:42:52.522 [INAUDIBLE] describe the motion at any possible position. 00:42:52.522 --> 00:42:53.230 Those two things. 00:42:53.230 --> 00:42:53.729 Yes. 00:42:53.729 --> 00:42:55.750 Is it independent? 00:42:55.750 --> 00:42:59.510 If you freeze x, can theta still move? 00:42:59.510 --> 00:43:02.310 If you freeze theta, can the x still move? 00:43:02.310 --> 00:43:05.360 OK, is it holonomic? 00:43:05.360 --> 00:43:06.550 Right. 00:43:06.550 --> 00:43:08.240 You need two, we got two and they're 00:43:08.240 --> 00:43:09.620 independent and complete. 00:43:09.620 --> 00:43:10.680 Good. 00:43:10.680 --> 00:43:13.990 Now the harder work starts. 00:43:13.990 --> 00:43:17.790 So I'm going to give us the mass of the rod, the mass moment 00:43:17.790 --> 00:43:22.920 of inertia the rod about the z-axis but with respect to A. 00:43:22.920 --> 00:43:25.440 The length of the rod is l1. 00:43:25.440 --> 00:43:31.840 The sleeve mass m2 izz with respect to its g. 00:43:31.840 --> 00:43:32.860 So it has a g. 00:43:32.860 --> 00:43:36.370 There's also and I'd better call it g2. 00:43:36.370 --> 00:43:38.180 That's the g of the sleeve. 00:43:38.180 --> 00:43:41.810 There's also a g1. 00:43:41.810 --> 00:43:45.000 A center of mass for the rod and a center 00:43:45.000 --> 00:43:47.260 of mass for the sleeve. 00:43:47.260 --> 00:43:48.910 Those are properties we'll need to know 00:43:48.910 --> 00:43:52.000 and I'll give them to you. 00:43:52.000 --> 00:43:53.480 OK. 00:43:53.480 --> 00:43:57.910 So we need to come up with expressions 00:43:57.910 --> 00:44:03.000 for potential energy and kinetic energy. 00:44:03.000 --> 00:44:06.920 So this problem, the potential energy's a little messy. 00:44:06.920 --> 00:44:09.610 Because you have to pick references. 00:44:09.610 --> 00:44:12.100 You have to account for the unstretched length 00:44:12.100 --> 00:44:12.695 of the spring. 00:44:15.930 --> 00:44:31.360 So call l 0 is the unstretched spring length. 00:44:35.240 --> 00:44:37.490 We know that also. 00:44:37.490 --> 00:44:42.880 So I propose that the potential energy look like 1/2, 00:44:42.880 --> 00:44:45.940 for the spring, anyway, 1/2 the amount that it 00:44:45.940 --> 00:44:50.200 stretches in a movement x1. 00:44:50.200 --> 00:44:51.610 The amount that it stretches then 00:44:51.610 --> 00:44:55.260 should be whatever that x1 position is. 00:44:55.260 --> 00:45:03.880 And that x1 position, and I drew it slightly incorrectly. 00:45:03.880 --> 00:45:07.390 I'm going to use x1 to locate the center of mass, which 00:45:07.390 --> 00:45:08.570 is always a good practice. 00:45:08.570 --> 00:45:10.000 So here's the center of mass. 00:45:10.000 --> 00:45:15.190 So my x1 goes to the center of mass. 00:45:15.190 --> 00:45:16.960 That's x1. 00:45:16.960 --> 00:45:20.050 So that's the total distance. 00:45:20.050 --> 00:45:23.000 And from that, we need to subtract 00:45:23.000 --> 00:45:25.940 l0, the unstretched length of the spring. 00:45:25.940 --> 00:45:31.240 And we need to subtract 1/2 the length of the body 00:45:31.240 --> 00:45:35.820 because that's that extra bit here. 00:45:35.820 --> 00:45:42.990 So this is the amount that the string is actually stretched 00:45:42.990 --> 00:45:44.759 when the coordinate is x1. 00:45:44.759 --> 00:45:45.800 And you've got to square. 00:45:45.800 --> 00:45:51.320 And that'll be the potential energy stored in the spring. 00:45:51.320 --> 00:45:55.980 Then we got to do the same thing for the potential energy. 00:45:55.980 --> 00:45:58.755 We have two sources of potential energy due to gravity. 00:45:58.755 --> 00:45:59.516 And they are? 00:46:03.410 --> 00:46:05.350 Two objects, right? 00:46:05.350 --> 00:46:06.430 Two potential energy. 00:46:06.430 --> 00:46:08.230 So why don't you take a minute and tell me 00:46:08.230 --> 00:46:10.210 the potential energy associated with the rod. 00:46:13.800 --> 00:46:16.180 So the rod has a center of mass. 00:46:16.180 --> 00:46:17.490 It's a pendulum basically. 00:46:17.490 --> 00:46:19.320 So it's the same as all the pendulum 00:46:19.320 --> 00:46:20.403 problems you've ever seen. 00:46:22.620 --> 00:46:28.230 And I would recommend that we use as our reference position 00:46:28.230 --> 00:46:32.290 its equilibrium position hanging straight down. 00:46:32.290 --> 00:46:34.260 And I'll tell you in advance, I'm 00:46:34.260 --> 00:46:37.170 going to use the unstretched spring position this time. 00:46:37.170 --> 00:46:38.110 Just stay with that. 00:46:38.110 --> 00:46:39.860 That's where it's going to start from. 00:46:39.860 --> 00:46:41.870 That's my reference for potential energy. 00:46:41.870 --> 00:46:46.470 But does the unstretched spring position 00:46:46.470 --> 00:46:50.130 have anything to do with the potential energy of the rod? 00:46:50.130 --> 00:46:50.650 No. 00:46:50.650 --> 00:46:51.150 OK. 00:46:51.150 --> 00:46:55.540 So its reference position is just hanging straight down. 00:46:55.540 --> 00:46:56.300 So figure it out. 00:46:56.300 --> 00:46:58.280 What's the potential energy expression 00:46:58.280 --> 00:47:01.572 for just the rod part? 00:47:01.572 --> 00:47:02.280 Think about that. 00:47:11.070 --> 00:47:12.920 So I'm going to remind you about something 00:47:12.920 --> 00:47:14.510 about potential energy. 00:47:14.510 --> 00:47:16.720 Potential energy, one of the requirements about it 00:47:16.720 --> 00:47:19.820 is the change in potential energy from one position 00:47:19.820 --> 00:47:23.940 to another is path independent. 00:47:23.940 --> 00:47:28.950 So you don't actually ever have to do the integral of minus mg 00:47:28.950 --> 00:47:30.140 dot dr. 00:47:30.140 --> 00:47:31.670 You don't have to do the integral. 00:47:31.670 --> 00:47:33.294 You just have to account for the change 00:47:33.294 --> 00:47:35.590 in height between its starting position 00:47:35.590 --> 00:47:38.220 and its some other position. 00:47:42.250 --> 00:47:43.940 Spend a minute or two, think about that. 00:47:43.940 --> 00:47:44.440 Work it out. 00:47:44.440 --> 00:47:45.935 You got a question? 00:47:45.935 --> 00:47:46.435 OK. 00:48:14.210 --> 00:48:14.850 Can you talk? 00:48:14.850 --> 00:48:17.105 Talk to a neighbor, check your ideas. 00:48:54.650 --> 00:48:56.950 So you have a suggestion for me? 00:48:56.950 --> 00:48:58.484 Ladies? 00:48:58.484 --> 00:48:59.400 AUDIENCE: [INAUDIBLE]. 00:49:06.825 --> 00:49:09.300 l1/2 [INAUDIBLE]. 00:49:16.127 --> 00:49:16.710 PROFESSOR: OK. 00:49:19.440 --> 00:49:21.600 Anybody want to make an improvement on that? 00:49:21.600 --> 00:49:23.010 Or they like it? 00:49:23.010 --> 00:49:24.212 Improvement? 00:49:24.212 --> 00:49:26.810 AUDIENCE: [INAUDIBLE]. 00:49:26.810 --> 00:49:29.280 PROFESSOR: 1 minus cosine theta. 00:49:29.280 --> 00:49:35.550 So let's put that up and let's figure out if we need that. 00:49:35.550 --> 00:49:43.140 We have a bid for cosine theta and 1 minus cosine theta. 00:49:43.140 --> 00:49:46.870 So you need to have a potential energy at the reference 00:49:46.870 --> 00:49:49.680 and you need to have a potential energy at the final point. 00:49:49.680 --> 00:49:51.330 And the difference between the two 00:49:51.330 --> 00:49:54.430 is a change in potential energy here. 00:49:54.430 --> 00:49:57.450 So what's the reference potential energy 00:49:57.450 --> 00:50:01.850 is mg l1 over 2 when it hangs straight down. 00:50:01.850 --> 00:50:04.610 And then when it moves up to this other position, 00:50:04.610 --> 00:50:10.000 this is the l1 over 2 times this is a delta h. 00:50:10.000 --> 00:50:13.920 This is the change in height that it goes through. 00:50:13.920 --> 00:50:14.935 So you need the 1 minus. 00:50:22.030 --> 00:50:24.585 Do we have the sines right? 00:50:24.585 --> 00:50:25.085 Yeah. 00:50:30.030 --> 00:50:30.530 OK. 00:50:37.970 --> 00:50:39.940 So now we need another term. 00:50:39.940 --> 00:50:42.240 And I'll write this one down. 00:50:42.240 --> 00:50:43.970 This one's a little messier. 00:50:43.970 --> 00:50:49.360 We need a potential energy term due to gravity for the sleeve. 00:50:49.360 --> 00:50:51.590 And that's going to mimic this. 00:50:51.590 --> 00:50:54.495 You're going to have a term here plus m2g. 00:50:58.190 --> 00:50:59.720 And it's reference, I'm just going 00:50:59.720 --> 00:51:02.950 to do it as a reference amount minus the final amount. 00:51:02.950 --> 00:51:07.610 The reference will be at the initial location 00:51:07.610 --> 00:51:20.740 of its center of mass, which is l0 plus l2 over 2 minus m2 g 00:51:20.740 --> 00:51:24.520 x1 cosine theta. 00:51:27.970 --> 00:51:29.420 Because this one is a little messy 00:51:29.420 --> 00:51:30.670 because you've got this thing. 00:51:30.670 --> 00:51:34.200 It can move up and down the sleeve. 00:51:34.200 --> 00:51:37.010 And if that moves, you've lost your reference. 00:51:37.010 --> 00:51:40.070 So you can't do this as a concise little term like this. 00:51:40.070 --> 00:51:43.220 You have to separate out the reference 00:51:43.220 --> 00:51:46.220 and then this is the final. 00:51:46.220 --> 00:51:49.790 And the l0 plus l2, this quantity 00:51:49.790 --> 00:51:52.170 here is the starting height. 00:51:52.170 --> 00:51:56.074 This x1 cosine theta is the finishing height. 00:51:56.074 --> 00:51:57.490 And the difference between the two 00:51:57.490 --> 00:51:59.760 gives you the change in the potential energy. 00:51:59.760 --> 00:52:02.170 So this is your potential energy expression. 00:52:02.170 --> 00:52:04.965 This plus this plus these. 00:52:04.965 --> 00:52:05.465 All right. 00:52:11.910 --> 00:52:13.180 So what about t? 00:52:13.180 --> 00:52:14.880 We got to be able write it. 00:52:14.880 --> 00:52:16.877 Kinetic energy is generally easier. 00:52:16.877 --> 00:52:18.710 Got to account for all the parts and pieces. 00:52:18.710 --> 00:52:20.190 So we have to chunks. 00:52:20.190 --> 00:52:25.090 And we're going to have rotational kinetic energy 00:52:25.090 --> 00:52:28.750 associated with the rod, rotational kinetic energy 00:52:28.750 --> 00:52:33.190 associated with the sleeve. 00:52:33.190 --> 00:52:35.450 But also some translational kinetic energy 00:52:35.450 --> 00:52:37.292 associated with the sleeve. 00:52:37.292 --> 00:52:38.625 And I'll write these terms down. 00:52:41.860 --> 00:52:43.845 Make the problem go a little faster here. 00:52:48.390 --> 00:52:54.655 1/2 izz about A. That's the rod. 00:52:57.930 --> 00:53:08.055 Plus 1/2 izz for the sleeve about g. 00:53:10.620 --> 00:53:13.920 We'll discuss why the difference here. 00:53:13.920 --> 00:53:18.190 And that's theta dot squared. 00:53:20.960 --> 00:53:30.870 Now for the kinetic energy that comes from translation 00:53:30.870 --> 00:53:33.580 of the center of mass. 00:53:33.580 --> 00:53:34.990 Because I'm broken up. 00:53:38.930 --> 00:53:40.690 Let me start over. 00:53:40.690 --> 00:53:47.680 This system is pinned about A. And the rod is just 00:53:47.680 --> 00:53:52.370 simply pinned at A. And the last lecture 00:53:52.370 --> 00:53:55.450 I put up these different conditions and simplifications. 00:53:55.450 --> 00:54:00.470 You can account for a something about a fixed pin 00:54:00.470 --> 00:54:04.405 by computing maximum inertia about A. 00:54:04.405 --> 00:54:06.950 It's basically a parallel axis theorem argument. 00:54:06.950 --> 00:54:09.700 Times 1/2 times that times theta dot squared. 00:54:09.700 --> 00:54:12.140 So this gives you all the kinetic energy in one 00:54:12.140 --> 00:54:14.460 go with the rod. 00:54:14.460 --> 00:54:19.000 But for the sliding mass, because its position 00:54:19.000 --> 00:54:21.580 is changing, you can't do that. 00:54:21.580 --> 00:54:26.080 You have to account for the two components of kinetic energy 00:54:26.080 --> 00:54:27.060 separately. 00:54:27.060 --> 00:54:29.990 This accounts for rotation about g. 00:54:29.990 --> 00:54:32.380 Even though g is moving. 00:54:32.380 --> 00:54:33.715 That accounts for that energy. 00:54:36.140 --> 00:54:37.890 Because it's only a function of theta dot. 00:54:37.890 --> 00:54:40.860 It's not a function of that position x. 00:54:40.860 --> 00:54:44.060 This term is going to account for the kinetic energy 00:54:44.060 --> 00:54:46.890 associated with the movement of the center of mass. 00:54:46.890 --> 00:54:53.405 So we need a vg2 in the inertial frame dot vg2. 00:54:58.300 --> 00:54:59.620 These be in vectors. 00:54:59.620 --> 00:55:02.580 And does that get everything? 00:55:02.580 --> 00:55:04.270 I think that does. 00:55:04.270 --> 00:55:16.350 So vgo is it certainly has a component that 00:55:16.350 --> 00:55:18.800 is its speed sliding up and down the rod, right? 00:55:18.800 --> 00:55:20.950 And that's in the i hat direction. 00:55:23.760 --> 00:55:28.080 But it has another component due to what? 00:55:28.080 --> 00:55:30.370 Can you tell me what it is? 00:55:30.370 --> 00:55:34.435 Its contribution to its speed due to its rotation. 00:55:44.113 --> 00:55:46.617 AUDIENCE: [INAUDIBLE]. 00:55:46.617 --> 00:55:47.950 PROFESSOR: It's got a theta dot. 00:55:47.950 --> 00:55:49.500 Yep. 00:55:49.500 --> 00:55:51.485 It needs an r, right? 00:55:51.485 --> 00:55:52.401 AUDIENCE: [INAUDIBLE]. 00:55:58.714 --> 00:55:59.380 PROFESSOR: Yeah. 00:55:59.380 --> 00:56:01.246 So this would be an x1 plus. 00:56:04.710 --> 00:56:12.210 No actually, I made x1 go right to the-- so just x1 theta 00:56:12.210 --> 00:56:15.308 dot in one direction. 00:56:15.308 --> 00:56:18.400 Yeah, so j hat here. 00:56:18.400 --> 00:56:20.980 Actually that's the moving coordinate system unit 00:56:20.980 --> 00:56:22.690 vector in the y direction. 00:56:22.690 --> 00:56:26.310 And so we do the dot product. 00:56:26.310 --> 00:56:32.810 You get this times itself. i dot i and j dot j. 00:56:32.810 --> 00:56:42.190 This quantity here is 1/2 m2 x dot squared 00:56:42.190 --> 00:56:46.820 plus x, this next one I guess. 00:56:46.820 --> 00:56:51.880 x1 squared of theta dot squared. 00:56:51.880 --> 00:56:55.180 That's the kinetic energy of accounting for the velocity 00:56:55.180 --> 00:56:56.204 of the center of mass. 00:56:56.204 --> 00:56:58.370 So now we have our entire kinetic energy expression. 00:57:08.230 --> 00:57:12.330 So now we have how many coordinates? 00:57:12.330 --> 00:57:12.949 Two, right? 00:57:12.949 --> 00:57:14.740 How many times do we have to turn the crank 00:57:14.740 --> 00:57:17.390 and go through the Lagrangian? 00:57:17.390 --> 00:57:19.160 Got to go through it twice. 00:57:19.160 --> 00:57:25.640 So let's apply Lagrange here. 00:57:28.720 --> 00:57:30.980 And we'll just do number one first. 00:57:30.980 --> 00:57:32.720 So and let's see. 00:57:32.720 --> 00:57:34.860 Which one do I have on my paper first? 00:57:34.860 --> 00:57:38.120 I guess we'll do the x1 equation. 00:57:38.120 --> 00:57:40.260 This is delta x1. 00:57:40.260 --> 00:57:42.920 So this generalized coordinate x1. 00:57:42.920 --> 00:57:45.520 And we need to do term one, which 00:57:45.520 --> 00:58:00.210 is in d by dt of partial of t with respect to x1 dot. 00:58:00.210 --> 00:58:01.830 OK. 00:58:01.830 --> 00:58:06.500 So we look at this and say, well, is this a function? 00:58:06.500 --> 00:58:09.415 Is this term a function of x? 00:58:09.415 --> 00:58:10.010 Nothing. 00:58:10.010 --> 00:58:12.180 You get nothing from there. 00:58:12.180 --> 00:58:14.370 Is this term a function of x? 00:58:14.370 --> 00:58:16.840 Yeah, it's down here. 00:58:16.840 --> 00:58:19.300 We only have to take the derivative of this. 00:58:19.300 --> 00:58:22.100 We have to do that job. 00:58:22.100 --> 00:58:25.200 So the derivative of this with respect to x 00:58:25.200 --> 00:58:32.152 dot, you get a 2x dot here. 00:58:32.152 --> 00:58:34.610 Do you get anything from here when you do this with respect 00:58:34.610 --> 00:58:35.197 to x dot? 00:58:35.197 --> 00:58:36.780 You only get a contribution from here. 00:58:36.780 --> 00:58:38.580 The two cancels that. 00:58:38.580 --> 00:58:48.930 And so this should look like m2 x1 dot but d by dt. 00:58:48.930 --> 00:58:53.400 Do this once in two steps here so you see what happens. 00:58:53.400 --> 00:58:57.610 You get an m2 x1 double dot out of that. 00:59:08.200 --> 00:59:10.300 So we've gotten the first piece of this. 00:59:10.300 --> 00:59:11.690 We've got a couple to go. 00:59:11.690 --> 00:59:15.184 But you know a lot about Newton's laws 00:59:15.184 --> 00:59:17.100 and you know a lot about calculating equations 00:59:17.100 --> 00:59:21.250 of motion now using sum of torques and all that stuff, 00:59:21.250 --> 00:59:22.070 right? 00:59:22.070 --> 00:59:27.500 So this is just something moving, has a circular motion, 00:59:27.500 --> 00:59:29.090 has translational motion. 00:59:29.090 --> 00:59:32.720 What other accelerations had better 00:59:32.720 --> 00:59:35.860 appear in this equation of motion? 00:59:35.860 --> 00:59:39.320 And which equation are we getting? 00:59:39.320 --> 00:59:40.695 There's going to be two equations 00:59:40.695 --> 00:59:42.790 and it has physical significance to it. 00:59:42.790 --> 00:59:46.830 What equation does this begin to look like? 00:59:46.830 --> 00:59:50.800 Just physically, what movement is being accounted for here? 00:59:50.800 --> 00:59:52.970 Looks like translation in the x direction. 00:59:52.970 --> 00:59:54.800 It's this thing sliding. 00:59:54.800 --> 00:59:57.200 It's this part of the motion sliding up and down. 00:59:57.200 --> 00:59:59.100 You're writing an equation of motion and mx 00:59:59.100 --> 01:00:00.405 double dot has units of what? 01:00:04.930 --> 01:00:06.540 Torque? 01:00:06.540 --> 01:00:07.400 Force. 01:00:07.400 --> 01:00:08.490 So it's a force equation. 01:00:08.490 --> 01:00:11.869 This is just f equals ma is what this is going to show us. 01:00:11.869 --> 01:00:13.660 Remember, the direct method has to give you 01:00:13.660 --> 01:00:16.920 the same answer as Lagrange. 01:00:16.920 --> 01:00:20.810 So we're getting a force equation. 01:00:20.810 --> 01:00:23.460 It's describing mx double dot. 01:00:23.460 --> 01:00:26.040 What other acceleration terms do you 01:00:26.040 --> 01:00:27.900 expect to appear in this from what you know? 01:00:27.900 --> 01:00:28.400 Yeah. 01:00:28.400 --> 01:00:28.895 AUDIENCE: [INAUDIBLE]. 01:00:28.895 --> 01:00:30.340 PROFESSOR: A centripetal term. 01:00:30.340 --> 01:00:32.464 Do you believe there ought to be a centripetal term 01:00:32.464 --> 01:00:34.150 in this answer? 01:00:34.150 --> 01:00:37.030 Why? 01:00:37.030 --> 01:00:39.550 Because it's got circular motion involved. 01:00:39.550 --> 01:00:41.380 For sure. 01:00:41.380 --> 01:00:42.040 Any others? 01:00:42.040 --> 01:00:44.780 Is there any Coriolis in this? 01:00:44.780 --> 01:00:45.890 In this direction. 01:00:45.890 --> 01:00:49.690 Which direction are we working in? 01:00:49.690 --> 01:00:53.680 Is there Coriolis acceleration in the x direction? 01:00:53.680 --> 01:00:58.530 By the way, these equations, do we have any ijk's in here? 01:00:58.530 --> 01:01:00.440 These are pure scalar equations. 01:01:03.090 --> 01:01:05.170 No unit vectors involved. 01:01:05.170 --> 01:01:07.810 This equation only described motion in the x. 01:01:11.730 --> 01:01:15.270 So will there be a Coriolis force in this acceleration 01:01:15.270 --> 01:01:16.220 in this problem? 01:01:16.220 --> 01:01:18.050 Will there be an Eulerian acceleration 01:01:18.050 --> 01:01:19.360 in this equation of motion? 01:01:19.360 --> 01:01:21.151 The reason I'm going through this with you, 01:01:21.151 --> 01:01:23.590 I want you to start developing your own intuition about 01:01:23.590 --> 01:01:25.950 whether or not when you get it at the end 01:01:25.950 --> 01:01:27.450 it's got everything it ought to have 01:01:27.450 --> 01:01:30.590 and doesn't have stuff it shouldn't have. 01:01:30.590 --> 01:01:31.150 OK. 01:01:31.150 --> 01:01:35.190 So your forecasting, then we better get a centripetal term. 01:01:35.190 --> 01:01:36.550 Well, let's see what happens. 01:01:36.550 --> 01:01:39.030 So that was number one. 01:01:39.030 --> 01:01:56.010 Number two here is our dt by minus the derivative 01:01:56.010 --> 01:01:58.970 with respect to x, in this case. 01:01:58.970 --> 01:02:00.460 So we go here. 01:02:00.460 --> 01:02:02.770 x1 we've been calling it. 01:02:02.770 --> 01:02:04.500 Is this a function of x? 01:02:04.500 --> 01:02:05.000 This piece? 01:02:05.000 --> 01:02:06.290 Nope, it's x dot. 01:02:06.290 --> 01:02:07.391 How about this one? 01:02:07.391 --> 01:02:07.890 Right. 01:02:07.890 --> 01:02:10.550 Take this derivative, you get 2x. 01:02:10.550 --> 01:02:12.580 So this fellow is going to give us 01:02:12.580 --> 01:02:22.535 minus m2 x1 theta dot squared. 01:02:25.090 --> 01:02:27.620 What's that look like? 01:02:27.620 --> 01:02:28.590 There it is. 01:02:28.590 --> 01:02:30.980 There's your centripetal term you're expecting to get. 01:02:30.980 --> 01:02:31.790 OK. 01:02:31.790 --> 01:02:39.900 And step three is plus partial of v with respect to x. 01:02:42.450 --> 01:02:44.280 In this case with respect to x. 01:02:44.280 --> 01:02:46.900 And where's our potential energy expression? 01:02:46.900 --> 01:02:48.530 Well, it's up here. 01:02:48.530 --> 01:02:55.910 And where the x dependency is in it. 01:02:55.910 --> 01:03:00.480 There is no x in that term and no x in that term. 01:03:00.480 --> 01:03:04.220 But we have x's in both of these other terms. 01:03:04.220 --> 01:03:07.840 So when we run through this, I'll 01:03:07.840 --> 01:03:11.190 write down what we come up with. 01:03:11.190 --> 01:03:13.670 We get certainly a spring term. 01:03:13.670 --> 01:03:22.950 k x1 minus l0 minus l2 over 2. 01:03:22.950 --> 01:03:29.420 So that's the spring piece when you take the derivative. 01:03:29.420 --> 01:03:32.010 The two cancels the 1/2 and the derivative of parts 01:03:32.010 --> 01:03:33.670 inside just gives you 1. 01:03:33.670 --> 01:03:36.790 So that's the first piece of the potential energy expression. 01:03:36.790 --> 01:03:41.940 And the second piece is only going to come from here. 01:03:41.940 --> 01:03:43.780 The derivative of this with respect to x1 01:03:43.780 --> 01:03:47.915 is just m2g cosine theta minus. 01:03:56.280 --> 01:03:58.920 And you add those bits together, you 01:03:58.920 --> 01:04:11.230 end up with m2 x1 double dot minus m2 x1 theta 01:04:11.230 --> 01:04:21.200 dot squared plus k x1 minus l0 minus l2 01:04:21.200 --> 01:04:29.440 over 2 minus m2g cosine theta. 01:04:29.440 --> 01:04:31.800 So those are the three terms, 1 plus 2 01:04:31.800 --> 01:04:34.160 plus 3, that go on the left hand side. 01:04:34.160 --> 01:04:39.940 And they're going to equal my qx that I find. 01:04:39.940 --> 01:04:42.570 I still have to find what the generalized force is 01:04:42.570 --> 01:04:43.386 in the x direction. 01:04:55.070 --> 01:04:57.190 So all that's left to do for this problem 01:04:57.190 --> 01:05:01.440 is to find q sub x, the generalized force that 01:05:01.440 --> 01:05:03.540 goes on the right hand side. 01:05:03.540 --> 01:05:10.520 So now let's draw a little diagram here of my system. 01:05:10.520 --> 01:05:13.330 And at the end of the sleeve. 01:05:13.330 --> 01:05:17.490 So here's my sleeve. 01:05:17.490 --> 01:05:18.960 I've applied this force. 01:05:21.710 --> 01:05:25.110 This is f of t. 01:05:25.110 --> 01:05:30.030 And maybe it's some f not cosine omega t. 01:05:30.030 --> 01:05:33.290 It's an oscillatory force, external force. 01:05:33.290 --> 01:05:34.190 Make it vibrate. 01:05:37.730 --> 01:05:42.550 And I need to know the virtual work done 01:05:42.550 --> 01:05:47.405 making that force go through a displacement in what direction? 01:05:54.200 --> 01:05:57.340 So this equation is the x1 equation, right? 01:05:57.340 --> 01:06:02.430 And so the virtual displacement I'm talking about is delta x1. 01:06:02.430 --> 01:06:04.660 And the amount of work that it does 01:06:04.660 --> 01:06:09.340 is delta x1 times the component of this force that's 01:06:09.340 --> 01:06:11.420 in its direction. 01:06:11.420 --> 01:06:19.380 So I'm going to take this force and break it up 01:06:19.380 --> 01:06:22.690 into two components. 01:06:22.690 --> 01:06:29.228 And if this is my theta, this is also theta. 01:06:37.700 --> 01:06:41.505 So this will be f0 product. 01:06:41.505 --> 01:06:46.255 And I'll leave out the cosine omega t here. 01:06:46.255 --> 01:06:47.690 It's a function of time. 01:06:47.690 --> 01:06:51.770 But this side then is cosine theta i. 01:06:54.500 --> 01:06:56.460 No, hey, I got this wrong. 01:06:56.460 --> 01:06:57.700 I drew this wrong, I'm sorry. 01:06:57.700 --> 01:06:59.120 This is theta. 01:06:59.120 --> 01:07:00.165 This is going to be sine. 01:07:02.700 --> 01:07:08.500 This side is sine theta in the i direction. 01:07:08.500 --> 01:07:18.025 And this piece is f0 of t cosine theta in the j. 01:07:18.025 --> 01:07:21.260 So I break it up in two parts. 01:07:21.260 --> 01:07:26.440 And the virtual work associated with x1 01:07:26.440 --> 01:07:33.810 is the thing I'm looking for, qx, dotted with delta x1. 01:07:33.810 --> 01:07:40.910 And that is f of t here, the vector 01:07:40.910 --> 01:07:45.990 dotted with dr, my little displacement. 01:07:45.990 --> 01:07:48.540 But in this case, this then all works out 01:07:48.540 --> 01:08:00.080 to be f0 cosine omega t. 01:08:00.080 --> 01:08:12.750 And it has sine theta i plus cos theta j components 01:08:12.750 --> 01:08:17.890 dotted width delta x in the i. 01:08:17.890 --> 01:08:21.620 So you're only going to get i dot j gives you 0, i dot i 01:08:21.620 --> 01:08:22.920 gets you 1. 01:08:22.920 --> 01:08:26.149 So you're going to get one piece out of this. 01:08:26.149 --> 01:08:41.190 This says in the qx equals f0 cosine omega t sine theta. 01:08:45.109 --> 01:08:50.250 And start with you have a delta x here and a delta x here 01:08:50.250 --> 01:08:52.560 and that gives you the delta virtual work. 01:08:56.581 --> 01:08:58.080 Personally when I do these problems, 01:08:58.080 --> 01:09:01.390 I have to think in terms of that little virtual deflection. 01:09:01.390 --> 01:09:04.670 I've actually figure out what's the virtual work done. 01:09:04.670 --> 01:09:07.770 And then at the end I take this out 01:09:07.770 --> 01:09:11.660 and this is the qx that I'm looking for. 01:09:11.660 --> 01:09:15.100 So my final equation of motion says, 01:09:15.100 --> 01:09:26.520 this equals f0 cosine omega t sine theta. 01:09:26.520 --> 01:09:30.054 And that's your equation of motion in the x1 direction. 01:09:35.680 --> 01:09:38.490 So when you finish one of these, you need to ask yourself, 01:09:38.490 --> 01:09:39.479 does this make sense? 01:09:39.479 --> 01:09:41.859 Does this jive with my understanding 01:09:41.859 --> 01:09:43.115 of Newtonian physics? 01:09:46.540 --> 01:09:48.189 Better have a linear acceleration term, 01:09:48.189 --> 01:09:50.180 because that's what it's doing. 01:09:50.180 --> 01:09:52.960 You have another acceleration term in the same direction 01:09:52.960 --> 01:09:54.810 due to centripetal. 01:09:54.810 --> 01:09:57.060 A spring force for sure. 01:09:57.060 --> 01:10:01.470 And a component of gravity in the direction of motion, 01:10:01.470 --> 01:10:04.700 up and down the slide, equal to any external forces 01:10:04.700 --> 01:10:05.840 in that direction. 01:10:05.840 --> 01:10:08.090 So it makes pretty good sense. 01:10:08.090 --> 01:10:09.220 OK. 01:10:09.220 --> 01:10:12.850 Now, also another test you can do 01:10:12.850 --> 01:10:17.690 is does it satisfy the laws of statics? 01:10:17.690 --> 01:10:19.350 That's another check you could perform. 01:10:19.350 --> 01:10:22.840 Does this thing at static equilibrium tell the truth? 01:10:22.840 --> 01:10:25.230 A static equilibrium all time derivative is 0. 01:10:25.230 --> 01:10:27.430 So this would be 0, this would be 0. 01:10:27.430 --> 01:10:29.620 You know its static equilibrium hangs down, 01:10:29.620 --> 01:10:31.800 so cosine theta is 1. 01:10:31.800 --> 01:10:34.320 Static you don't have any time dependent forces. 01:10:34.320 --> 01:10:35.610 That's 0. 01:10:35.610 --> 01:10:50.630 So the static part of this says that k x1 minus l0 minus l2 01:10:50.630 --> 01:10:57.690 over 2 equals m2g cosine. 01:10:57.690 --> 01:11:02.000 And that's cosine theta is 1, so it's m2g. 01:11:02.000 --> 01:11:06.680 And you could figure out then this must be k times something. 01:11:06.680 --> 01:11:08.480 This is the x. 01:11:08.480 --> 01:11:12.180 This is the amount of the spring stretches, the static stretch 01:11:12.180 --> 01:11:16.730 of the spring, so the spring has an equal and opposite force 01:11:16.730 --> 01:11:18.690 to the weight of the thing m2g. 01:11:18.690 --> 01:11:23.220 So that's another check you can do when doing the problems. 01:11:23.220 --> 01:11:31.070 OK, I'll write up the final one. 01:11:31.070 --> 01:11:32.530 We have one more question to go. 01:11:32.530 --> 01:11:36.090 Got to do all the derivatives with respect to theta. 01:11:36.090 --> 01:11:43.390 So you take a minute to decide how many acceleration terms 01:11:43.390 --> 01:11:45.040 and what acceleration terms do you 01:11:45.040 --> 01:11:48.670 expect to see come out of this second equation of motion. 01:11:48.670 --> 01:11:51.870 Because now we're talking about which motion? 01:11:51.870 --> 01:11:53.870 Swinging motion. 01:11:53.870 --> 01:11:55.550 And what's its direction? 01:11:55.550 --> 01:11:58.250 In Newtonian sense, it would have a vector direction. 01:11:58.250 --> 01:12:01.530 It's in what we call j here. 01:12:01.530 --> 01:12:05.680 OK, so you're about to get the j equation. 01:12:05.680 --> 01:12:08.042 What terms do you expect to find in it? 01:12:11.020 --> 01:12:12.730 Talk to your neighbors and sort this out. 01:12:12.730 --> 01:12:15.550 And basically tell me what the answer's going to be. 01:13:02.765 --> 01:13:03.690 What do you think? 01:13:09.660 --> 01:13:12.836 What are we going to get? 01:13:12.836 --> 01:13:15.306 AUDIENCE: We were debating about whether or not 01:13:15.306 --> 01:13:22.240 it was going to be like speeding up in the theta [INAUDIBLE]. 01:13:22.240 --> 01:13:25.330 PROFESSOR: So it is a pendulum, just a weird pendulum. 01:13:25.330 --> 01:13:33.940 So does the theta change speed? 01:13:33.940 --> 01:13:34.460 Sure. 01:13:34.460 --> 01:13:36.680 When he gets up the top of the swing at 0. 01:13:36.680 --> 01:13:38.620 All the way down, it's maximum speed. 01:13:38.620 --> 01:13:41.900 So what term does that imply that you're going to get? 01:13:41.900 --> 01:13:43.960 AUDIENCE: [INAUDIBLE]. 01:13:43.960 --> 01:13:46.110 PROFESSOR: Well, maybe, maybe not. 01:13:46.110 --> 01:13:46.610 Yeah? 01:13:46.610 --> 01:13:47.527 AUDIENCE: [INAUDIBLE]. 01:13:47.527 --> 01:13:49.818 PROFESSOR: Going to get an Eulerian, which means you've 01:13:49.818 --> 01:13:51.140 got a theta double dot term. 01:13:51.140 --> 01:13:53.640 You're expecting a theta double dot term to show up. 01:13:53.640 --> 01:13:54.140 OK. 01:13:54.140 --> 01:13:55.120 What else? 01:13:59.040 --> 01:14:02.860 Will you get a Coriolis term? 01:14:02.860 --> 01:14:04.580 Do you expect a Coriolis term? 01:14:04.580 --> 01:14:07.556 Something that looks like x dot theta dot. 01:14:07.556 --> 01:14:10.074 AUDIENCE: [INAUDIBLE]. 01:14:10.074 --> 01:14:10.740 PROFESSOR: Yeah. 01:14:10.740 --> 01:14:12.573 The thing is sliding up and down the sleeve. 01:14:12.573 --> 01:14:15.640 It has a non 0 value of x dot. 01:14:15.640 --> 01:14:19.650 Any time you got things moving radially 01:14:19.650 --> 01:14:24.020 while something is swinging in a circle, 01:14:24.020 --> 01:14:25.605 you will get Coriolis forces. 01:14:25.605 --> 01:14:28.220 It means the angular momentum of that thing is changing 01:14:28.220 --> 01:14:30.280 and it takes forces to make that happen. 01:14:30.280 --> 01:14:34.515 So here's what this answer looks like. 01:14:52.340 --> 01:14:55.850 That's the one term. 01:14:55.850 --> 01:15:00.530 The two piece gives you 0. 01:15:03.435 --> 01:15:09.210 It's not a function of x in the three piece. 01:15:09.210 --> 01:15:19.290 The potential energy pace gives you m2g x1 sine theta 01:15:19.290 --> 01:15:27.980 plus m1 g l1 over 2 sine theta. 01:15:27.980 --> 01:15:32.930 And the fourth piece, the q theta, 01:15:32.930 --> 01:15:35.536 well, that's just going to be the virtual work done. 01:15:38.660 --> 01:15:40.260 There's a tricky bit to this one. 01:15:40.260 --> 01:15:43.240 Now there's virtual work, but which direction? 01:15:43.240 --> 01:15:49.800 So we have an f dot dr. The only f we have is this. 01:15:49.800 --> 01:15:51.519 What's the dr? 01:15:51.519 --> 01:15:52.517 What direction is it? 01:15:58.010 --> 01:15:59.310 This is the theta coordinate. 01:15:59.310 --> 01:16:01.810 What direction does that give you displacements? 01:16:01.810 --> 01:16:05.616 f dot dr's a displacement, not an angle. 01:16:05.616 --> 01:16:07.490 To get the work done, you got to move a force 01:16:07.490 --> 01:16:08.975 through a distance. 01:16:08.975 --> 01:16:11.100 So the distance, first of all, is in what direction 01:16:11.100 --> 01:16:14.256 when theta moves? 01:16:14.256 --> 01:16:14.756 j. 01:16:14.756 --> 01:16:16.910 Little j hat, right? 01:16:16.910 --> 01:16:21.050 And now if you get a virtual deflection 01:16:21.050 --> 01:16:25.670 of delta theta, what's the virtual displacement? 01:16:25.670 --> 01:16:28.310 You had a virtual change in [? angle ?] delta to put theta. 01:16:28.310 --> 01:16:29.880 But is that the virtual displacement? 01:16:32.970 --> 01:16:39.970 What's the displacement of this point here in the j direction, 01:16:39.970 --> 01:16:42.576 given a virtual displacement delta theta? 01:16:42.576 --> 01:16:43.568 Think that out. 01:16:53.984 --> 01:16:56.067 AUDIENCE: [INAUDIBLE]. 01:16:56.067 --> 01:16:57.400 PROFESSOR: Can't quite hear you. 01:16:57.400 --> 01:16:59.240 AUDIENCE: [INAUDIBLE]. 01:16:59.240 --> 01:17:01.340 PROFESSOR: x1 delta theta will give you 01:17:01.340 --> 01:17:04.710 the motion to displacement at the center of mass 01:17:04.710 --> 01:17:07.560 in that direction. 01:17:07.560 --> 01:17:09.700 x1 comes from here to here. 01:17:09.700 --> 01:17:12.220 So x1 delta theta will give you a little displacement 01:17:12.220 --> 01:17:13.340 in that direction. 01:17:13.340 --> 01:17:17.140 But is that the displacement we care about? 01:17:17.140 --> 01:17:18.600 We need the displacement here. 01:17:18.600 --> 01:17:19.540 So you're close. 01:17:24.240 --> 01:17:32.670 So we're going to get some force dot a displacement dr. 01:17:32.670 --> 01:17:36.370 And that's going to be our force, this guy, 01:17:36.370 --> 01:17:38.530 with its i and j components. 01:17:38.530 --> 01:17:39.780 i and j terms. 01:17:39.780 --> 01:17:47.270 But this term out here is x1 plus l2 over 2 01:17:47.270 --> 01:17:48.950 to get to the end. 01:17:48.950 --> 01:17:52.100 And it's in the j direction. 01:17:52.100 --> 01:17:55.700 So it's a length times A. And you need the delta. 01:17:59.710 --> 01:18:01.640 This quantity. 01:18:01.640 --> 01:18:05.230 And you need a delta theta. 01:18:05.230 --> 01:18:07.490 Delta theta. 01:18:07.490 --> 01:18:10.150 This is the term. 01:18:10.150 --> 01:18:14.590 This is the dr for the system. 01:18:14.590 --> 01:18:18.020 An angle, a virtual deflection in angle times the moment arm 01:18:18.020 --> 01:18:19.650 gives you a distance. 01:18:19.650 --> 01:18:27.350 It's in the j hat direction dotted with the same force 01:18:27.350 --> 01:18:30.950 breaking the force up into its i and j components. 01:18:30.950 --> 01:18:33.680 It had a sine theta i cos theta j. 01:18:33.680 --> 01:18:36.840 So this is going to give me a f cosine omega t 01:18:36.840 --> 01:18:38.960 cos theta j dot j. 01:18:45.912 --> 01:19:02.230 f0 cosine omega t cos theta x1 plus l2 over 2 delta theta 01:19:02.230 --> 01:19:03.930 is the delta w. 01:19:03.930 --> 01:19:07.160 That's the work and the virtual. 01:19:10.030 --> 01:19:15.820 The generalized force q theta is this part of it. 01:19:15.820 --> 01:19:22.580 So this plus this plus this equals 01:19:22.580 --> 01:19:24.800 that on the right hand side. 01:19:24.800 --> 01:19:27.940 So this is part four. 01:19:27.940 --> 01:19:30.750 And look at it. 01:19:30.750 --> 01:19:31.620 Yeah? 01:19:31.620 --> 01:19:32.536 AUDIENCE: [INAUDIBLE]. 01:19:37.690 --> 01:19:41.230 PROFESSOR: So f of t, I didn't want to write it all out. 01:19:41.230 --> 01:19:43.760 This thing breaks into an i and a j piece, 01:19:43.760 --> 01:19:45.310 which is written over there. 01:19:45.310 --> 01:19:48.230 This is the sine theta i cos theta j term. 01:19:51.140 --> 01:19:54.040 Which I brought back from over there. 01:19:54.040 --> 01:19:57.190 And we dot it with the dr that we care about, 01:19:57.190 --> 01:20:02.620 which is this length times that angle in the j direction. 01:20:02.620 --> 01:20:03.650 So j dot. 01:20:03.650 --> 01:20:06.620 We only pick up the j piece of this. 01:20:06.620 --> 01:20:10.740 And that gives us this cosine theta term. 01:20:10.740 --> 01:20:13.070 OK. 01:20:13.070 --> 01:20:14.640 Let's look quickly. 01:20:14.640 --> 01:20:17.730 This is a rotational thing. 01:20:17.730 --> 01:20:19.515 It has units of is it force? 01:20:19.515 --> 01:20:21.125 Is this a force equation? 01:20:24.050 --> 01:20:26.780 i theta double dot has units of what? 01:20:26.780 --> 01:20:27.280 Torque. 01:20:27.280 --> 01:20:28.900 This is a torque equation. 01:20:28.900 --> 01:20:36.080 This is the total mass moment of inertia izz 01:20:36.080 --> 01:20:40.190 with respect to A for this system such that the Eulerian 01:20:40.190 --> 01:20:41.030 acceleration. 01:20:41.030 --> 01:20:44.030 The torque it takes to make that happen 01:20:44.030 --> 01:20:48.280 is the sum of the mass moment of inertia of the rod 01:20:48.280 --> 01:20:51.910 plus the mass moment of inertia of g plus m2 x1 01:20:51.910 --> 01:20:55.340 squared, which looks a lot like the parallel axis theorem. 01:20:55.340 --> 01:20:59.550 This is izz A for the moving mass. 01:20:59.550 --> 01:21:00.710 There's your Coriolis term. 01:21:03.730 --> 01:21:06.120 And here's your potential terms and there's 01:21:06.120 --> 01:21:08.371 your external force. 01:21:08.371 --> 01:21:08.870 OK. 01:21:12.570 --> 01:21:15.560 Talk more about these things in recitation.