WEBVTT 00:00:00.100 --> 00:00:02.450 The following content is provided under a Creative 00:00:02.450 --> 00:00:03.830 Commons license. 00:00:03.830 --> 00:00:06.080 Your support will help MIT OpenCourseWare 00:00:06.080 --> 00:00:10.170 continue to offer high quality educational resources for free. 00:00:10.170 --> 00:00:12.710 To make a donation or to view additional materials 00:00:12.710 --> 00:00:16.620 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.620 --> 00:00:17.325 at ocw.mit.edu. 00:00:22.084 --> 00:00:25.800 PROFESSOR: And I have put on the second board 00:00:25.800 --> 00:00:31.260 just a quick review of the key equations that we 00:00:31.260 --> 00:00:34.690 use to do the direct method. 00:00:34.690 --> 00:00:40.140 So straight from Newton's second law, the first one, 00:00:40.140 --> 00:00:45.850 second one-- the summarization of torques about some point a. 00:00:45.850 --> 00:00:51.370 a may be g, in which case things get much simplified, g 00:00:51.370 --> 00:00:53.270 being the center of mass. 00:00:53.270 --> 00:00:59.210 And if point a is moving, then you have this extra term. 00:00:59.210 --> 00:01:01.150 And this is a new equation over here, 00:01:01.150 --> 00:01:05.060 which I got frustrated with this equation a few days ago 00:01:05.060 --> 00:01:10.170 and derive something, which I think is easier to use. 00:01:10.170 --> 00:01:15.260 So I'll do a problem today using that expression. 00:01:15.260 --> 00:01:17.530 So I won't talk more about it now-- show it to you 00:01:17.530 --> 00:01:18.610 in a minute. 00:01:18.610 --> 00:01:22.450 And then finally, it's convenient at times 00:01:22.450 --> 00:01:24.800 to express angular momentum with respect 00:01:24.800 --> 00:01:28.400 to a point a as angular momentum about g 00:01:28.400 --> 00:01:36.580 plus the position vector from a to g times the linear momentum. 00:01:36.580 --> 00:01:39.340 So that's kind of the covered on the quiz. 00:01:39.340 --> 00:01:41.200 And we're going to use these equations. 00:01:41.200 --> 00:01:47.250 Today, what I'm going to do is look at a few problems 00:01:47.250 --> 00:01:50.730 from the point of view both equations of motion, 00:01:50.730 --> 00:01:52.800 by the direct method and the Lagrange method, 00:01:52.800 --> 00:01:59.150 and kind of talk about and which one's easier, when should 00:01:59.150 --> 00:02:01.580 you choose this one, when to choose that one. 00:02:01.580 --> 00:02:04.370 That's the nature of today's lecture to sort of get 00:02:04.370 --> 00:02:06.460 you prepped for the quiz. 00:02:09.400 --> 00:02:11.355 And a little bit more on generalized forces. 00:02:23.467 --> 00:02:25.800 You have questions about what's going to be on the quiz? 00:02:29.744 --> 00:02:31.979 AUDIENCE: What is that last word-- balancing of? 00:02:31.979 --> 00:02:33.270 PROFESSOR: Balancing of rotors. 00:02:37.070 --> 00:02:41.060 There's a rotor-- any thing that we spin about an axis. 00:02:41.060 --> 00:02:42.980 A rigid body spinning about an axis, 00:02:42.980 --> 00:02:45.040 you can call a rotor of some kind. 00:02:45.040 --> 00:02:50.050 And balancing is usually-- the axes are fixed. 00:02:50.050 --> 00:02:54.970 And so is this one-- this axis passes 00:02:54.970 --> 00:02:59.365 through the center of mass-- so is 00:02:59.365 --> 00:03:04.771 this rotor statically balanced? 00:03:04.771 --> 00:03:05.270 Right. 00:03:05.270 --> 00:03:07.790 Is this rotor dynamically balance? 00:03:07.790 --> 00:03:11.120 Would you expect to when this is spinning, 00:03:11.120 --> 00:03:15.335 it's direction of spin is vertical, right. 00:03:18.930 --> 00:03:22.820 Is the direction of the angular momentum of this 00:03:22.820 --> 00:03:28.000 object in the same direction as the angle of rotation? 00:03:28.000 --> 00:03:28.500 No. 00:03:28.500 --> 00:03:32.420 And so that's an indication of dynamic imbalance. 00:03:32.420 --> 00:03:34.370 Angular momentum is not in the same direction 00:03:34.370 --> 00:03:36.720 as angular rotation, right. 00:03:36.720 --> 00:03:38.090 All right. 00:03:38.090 --> 00:03:40.170 And, well, while we're on that subject, 00:03:40.170 --> 00:03:41.730 we'll just hit it briefly here. 00:03:44.740 --> 00:03:47.115 One other prop I need. 00:03:50.180 --> 00:03:54.960 So another rigid body, and I don't know 00:03:54.960 --> 00:03:56.755 where g is on this rigid body. 00:04:04.260 --> 00:04:07.760 So does-- this is rotating about an axis. 00:04:11.010 --> 00:04:14.050 Is this object statically balanced, 00:04:14.050 --> 00:04:15.246 just doing this experiment? 00:04:15.246 --> 00:04:16.329 We're doing an experiment. 00:04:16.329 --> 00:04:19.089 We're sticking this on an axis, and we're 00:04:19.089 --> 00:04:20.130 letting it do it's thing. 00:04:20.130 --> 00:04:22.540 And I'm asking you is it statically balance. 00:04:22.540 --> 00:04:23.290 How do you know? 00:04:23.290 --> 00:04:27.736 AUDIENCE: Because if you were to flip it upside down and do it, 00:04:27.736 --> 00:04:29.220 it would still come to a stop. 00:04:29.220 --> 00:04:31.240 PROFESSOR: So it goes to a low point, right. 00:04:31.240 --> 00:04:35.190 And that tells you what about the position of this axis? 00:04:35.190 --> 00:04:37.686 AUDIENCE: [INAUDIBLE] 00:04:37.686 --> 00:04:39.060 PROFESSOR: The center of mass has 00:04:39.060 --> 00:04:42.200 got gravity, puts a force at the center of mass, 00:04:42.200 --> 00:04:45.610 creates a torque which brings it to the low point. 00:04:45.610 --> 00:04:54.100 Does this object have any symmetries 00:04:54.100 --> 00:04:55.230 that you could point out? 00:04:55.230 --> 00:04:59.960 Or in other words, can you without doing any math tell me 00:04:59.960 --> 00:05:02.400 one principal axis of this object? 00:05:11.824 --> 00:05:15.098 AUDIENCE: Where the dotted line is, that's a plane [INAUDIBLE]. 00:05:15.098 --> 00:05:17.480 PROFESSOR: So you say I've got a dotted line around here 00:05:17.480 --> 00:05:18.920 and that's a plane of what? 00:05:18.920 --> 00:05:19.570 AUDIENCE: Plane of symmetry. 00:05:19.570 --> 00:05:21.180 PROFESSOR: So there's a plane of symmetry cutting 00:05:21.180 --> 00:05:22.310 this thing through here. 00:05:22.310 --> 00:05:25.840 So if you can identify one plane of symmetry, 00:05:25.840 --> 00:05:28.350 then where can you tell me that for sure you 00:05:28.350 --> 00:05:33.090 know that you can make a principal axis? 00:05:33.090 --> 00:05:36.550 He says perpendicular to that plane. 00:05:36.550 --> 00:05:41.390 Any place on that plane perpendicular to it? 00:05:41.390 --> 00:05:43.240 What do you think? 00:05:43.240 --> 00:05:46.570 For now, we've identified it has a plane of symmetry 00:05:46.570 --> 00:05:57.100 and that an axis perpendicular to it-- a principle axis should 00:05:57.100 --> 00:05:58.440 be perpendicular to that plane. 00:05:58.440 --> 00:06:00.430 But the question is where do I put it? 00:06:00.430 --> 00:06:02.750 Where, perpendicular to that plane 00:06:02.750 --> 00:06:05.385 can this axis be and be a principal axis? 00:06:13.320 --> 00:06:14.530 She says through g. 00:06:14.530 --> 00:06:17.160 Do you agree with that? 00:06:17.160 --> 00:06:18.510 So certainly through g. 00:06:18.510 --> 00:06:22.490 Let's say that that really is g and I put it through here. 00:06:22.490 --> 00:06:24.647 And if it was exactly through g, it 00:06:24.647 --> 00:06:25.980 wouldn't rotate to the low spot. 00:06:25.980 --> 00:06:26.900 This is pretty close. 00:06:26.900 --> 00:06:28.760 It struggles to get there. 00:06:28.760 --> 00:06:32.350 That's definitely a principal axis, but are there more? 00:06:32.350 --> 00:06:34.390 This is kind of the real point of the question. 00:06:34.390 --> 00:06:37.900 Are there other allowable principal axes 00:06:37.900 --> 00:06:38.675 in this direction? 00:06:41.770 --> 00:06:43.520 So we haven't talked about this very much. 00:06:43.520 --> 00:06:46.710 Any axis-- if this is a principal axis, 00:06:46.710 --> 00:06:51.200 any axis parallel to it is a principal axis. 00:06:51.200 --> 00:06:52.860 It's no longer through g. 00:06:52.860 --> 00:06:56.430 It's no longer statically balanced about that point. 00:06:56.430 --> 00:06:58.460 But it's still a principal axis. 00:06:58.460 --> 00:07:01.830 And one measure of whether or not something 00:07:01.830 --> 00:07:06.360 is a principal axis-- if you know the principal axes 00:07:06.360 --> 00:07:10.820 of an object and you write the mass moment of inertia 00:07:10.820 --> 00:07:14.410 matrix of it with respect to g, what 00:07:14.410 --> 00:07:18.150 does that matrix look like? 00:07:18.150 --> 00:07:21.190 Where does it have 0s and non-0s? 00:07:21.190 --> 00:07:23.870 It's a diagonal matrix. 00:07:23.870 --> 00:07:26.800 So if i about g is diagonal, that 00:07:26.800 --> 00:07:30.420 means you're computed the moment, mass moment 00:07:30.420 --> 00:07:32.980 properties with respect to three principal axes. 00:07:32.980 --> 00:07:36.800 If it's diagonal, and those axes all pass through g, 00:07:36.800 --> 00:07:39.950 if I now take one of those axes and say, 00:07:39.950 --> 00:07:42.560 I want to move it over here, and you've 00:07:42.560 --> 00:07:47.480 re-computed the inertia matrix, would it still be diagonal? 00:07:50.110 --> 00:07:55.580 So you've moved just one axis to another spot, parallel 00:07:55.580 --> 00:07:57.090 to its original one. 00:07:57.090 --> 00:07:59.040 You know this one is a principal axis. 00:07:59.040 --> 00:08:01.480 I'm going to move it just parallel. 00:08:01.480 --> 00:08:04.200 Only that one axis-- the other two are saying where they are. 00:08:04.200 --> 00:08:06.600 I've just moved it parallel. 00:08:06.600 --> 00:08:09.410 Is that also a principal axis, or another way of saying it, 00:08:09.410 --> 00:08:13.990 does that mass matrix, mass moment of inertia matrix stay 00:08:13.990 --> 00:08:16.130 diagonal? 00:08:16.130 --> 00:08:19.910 Do any of the terms in it change? 00:08:19.910 --> 00:08:23.500 Let's say this is the z principal axis and I move away. 00:08:23.500 --> 00:08:29.380 So now you have this new mass moment inertia matrix. 00:08:29.380 --> 00:08:31.680 Does the Izz term change? 00:08:36.580 --> 00:08:39.564 Yeah, by how much? 00:08:39.564 --> 00:08:42.059 AUDIENCE: [INAUDIBLE] 00:08:42.059 --> 00:08:44.340 PROFESSOR: He said ml squared, l being what? 00:08:44.340 --> 00:08:46.300 AUDIENCE: The length you moved. 00:08:46.300 --> 00:08:47.370 PROFESSOR: Here. 00:08:47.370 --> 00:08:51.690 And that's this thing though as parallel axis there, right? 00:08:51.690 --> 00:08:57.920 If you move one axis, any axis parallel to one of those axes 00:08:57.920 --> 00:08:59.940 is also a principal axis. 00:08:59.940 --> 00:09:01.360 So that's a subtle point. 00:09:01.360 --> 00:09:03.390 We don't often talking about it. 00:09:03.390 --> 00:09:06.730 Any axis parallel to a principal axis is also principal axis. 00:09:16.460 --> 00:09:18.230 Well, let's think about a problem. 00:09:18.230 --> 00:09:19.760 We've done this problem before. 00:09:19.760 --> 00:09:22.090 But now we have all the full tools. 00:09:22.090 --> 00:09:25.870 We have Lagrange, we have direct method. 00:09:25.870 --> 00:09:31.835 So that's this problem of the wheel on a horizontal plane. 00:09:37.690 --> 00:09:42.080 Kind of a complicated distribution of mass 00:09:42.080 --> 00:09:51.320 for which we know Izz about g is some mass radius and gyration 00:09:51.320 --> 00:09:51.820 squared. 00:09:54.590 --> 00:09:57.540 Maybe we measured it rather than computed it. 00:09:57.540 --> 00:10:03.630 So this thing is sliding on ice, and it has cord 00:10:03.630 --> 00:10:05.190 wrapped around it. 00:10:05.190 --> 00:10:08.360 I'm pulling with a force F, call it F1. 00:10:12.050 --> 00:10:15.815 I'll provide the inertial frame. 00:10:21.910 --> 00:10:29.600 And it's got some radius R. And we'll 00:10:29.600 --> 00:10:31.233 say the surface is frictionless. 00:10:44.530 --> 00:10:46.200 So I don't have a string wrapped around. 00:10:46.200 --> 00:10:47.730 But we're really talking about this problem. 00:10:47.730 --> 00:10:49.425 I'm pulling on the string and this thing 00:10:49.425 --> 00:10:51.070 can slide and rotate. 00:10:59.360 --> 00:11:00.270 Take a second. 00:11:00.270 --> 00:11:01.980 Think about how many different ways 00:11:01.980 --> 00:11:07.810 can you think of solving for the equations of motion of this? 00:11:07.810 --> 00:11:10.180 So if this on a quiz, I want you to find 00:11:10.180 --> 00:11:13.057 the equations of motion. 00:11:13.057 --> 00:11:14.140 What method would you use? 00:11:16.870 --> 00:11:17.640 And what's this? 00:11:17.640 --> 00:11:19.624 And then you're confronted with, well, 00:11:19.624 --> 00:11:20.790 I know more than one method. 00:11:20.790 --> 00:11:22.280 What's the simplest method? 00:11:26.483 --> 00:11:28.290 And before that, you need to decide 00:11:28.290 --> 00:11:30.399 how many coordinates you're going to need, 00:11:30.399 --> 00:11:32.690 how many equations you're going-- so let's start there. 00:11:32.690 --> 00:11:36.470 How many independent coordinates do you need to do this problem? 00:11:42.390 --> 00:11:44.220 We'll back up one more question, then. 00:11:44.220 --> 00:11:47.680 Is this a planar motion problem? 00:11:47.680 --> 00:11:49.430 So it's confined to one plane. 00:11:49.430 --> 00:11:50.860 I see a lot of heads nodding. 00:11:50.860 --> 00:11:55.250 And it rotates only with one degree of freedom of rotation, 00:11:55.250 --> 00:11:57.100 axis perpendicular to the plane. 00:11:57.100 --> 00:11:58.599 So this is a planar motion problem. 00:11:58.599 --> 00:12:00.140 So in general, planar motion problems 00:12:00.140 --> 00:12:04.420 have found many degrees of freedom per rigid body? 00:12:04.420 --> 00:12:06.580 So I see a bunch of 3's. 00:12:06.580 --> 00:12:09.259 Does this one have any constraints on those three 00:12:09.259 --> 00:12:10.050 degrees of freedom? 00:12:10.050 --> 00:12:11.860 So this has three degrees of freedom. 00:12:11.860 --> 00:12:15.680 You conceivably need three equations of motion. 00:12:19.060 --> 00:12:21.267 What are the choices for getting them? 00:12:21.267 --> 00:12:23.100 I want you to think about this for a minute. 00:12:23.100 --> 00:12:25.360 You decide what way you would do, and let's 00:12:25.360 --> 00:12:26.900 take a poll in a minute. 00:12:26.900 --> 00:12:28.650 You think about it, and then let's take 00:12:28.650 --> 00:12:30.865 a poll for how people would do this problem. 00:13:14.020 --> 00:13:17.390 So what's your favorite? 00:13:17.390 --> 00:13:19.720 How would you do it? 00:13:19.720 --> 00:13:22.440 He'd use the direct method. 00:13:22.440 --> 00:13:25.780 Do you need a torque equation? 00:13:25.780 --> 00:13:27.630 About what point? 00:13:27.630 --> 00:13:28.630 He'd use center of mass. 00:13:28.630 --> 00:13:30.979 So he'd say, direct method and for the torque equation, 00:13:30.979 --> 00:13:32.270 do it about the center of mass. 00:13:32.270 --> 00:13:36.340 Anybody have a different way they would do this problem? 00:13:36.340 --> 00:13:39.400 Is there another way to do it? 00:13:39.400 --> 00:13:41.610 He wants to do Lagrange. 00:13:41.610 --> 00:13:44.960 Everybody agree it could be done that way? 00:13:44.960 --> 00:13:50.620 So I agree, and let's quickly do this problem. 00:13:50.620 --> 00:14:00.100 And so direct method-- I'm going to make it easier on myself 00:14:00.100 --> 00:14:00.600 here. 00:14:09.610 --> 00:14:12.970 So the three coordinates, you're going to need an x, a y, 00:14:12.970 --> 00:14:16.070 and a theta is what I've chosen for the three coordinates 00:14:16.070 --> 00:14:18.000 to do this direct method. 00:14:18.000 --> 00:14:19.810 And to make it easy on myself, I'm 00:14:19.810 --> 00:14:24.120 going to align the force with my coordinates. 00:14:24.120 --> 00:14:26.085 So I'm going to essentially make theta 0. 00:14:39.593 --> 00:14:43.820 Essentially aligning F1, my force F1 with the x-axis. 00:14:43.820 --> 00:14:49.770 And if I do that, then I can say the sum of the forces in the y, 00:14:49.770 --> 00:14:54.440 the external forces in the y direction are nonexistent, 0. 00:14:54.440 --> 00:14:59.820 So that must be my double dot, so y double dot is 0. 00:14:59.820 --> 00:15:03.000 So that simplifies the problem. 00:15:03.000 --> 00:15:07.040 Summation of forces in the x direction, external forces 00:15:07.040 --> 00:15:07.660 are F1I. 00:15:10.930 --> 00:15:15.350 And they must equal to the mass times the acceleration 00:15:15.350 --> 00:15:16.280 in the x direction. 00:15:16.280 --> 00:15:19.080 And I'm basically done with that equation. 00:15:19.080 --> 00:15:20.876 AUDIENCE: Is that a rotating [INAUDIBLE] 00:15:26.288 --> 00:15:28.730 PROFESSOR: Well, I'm doing this. 00:15:28.730 --> 00:15:30.230 I'm saying this is my inertia frame. 00:15:38.690 --> 00:15:40.950 You add an inertial frame? 00:15:40.950 --> 00:15:43.000 Sure. 00:15:43.000 --> 00:15:46.670 Now I can-- I don't have to break this force 00:15:46.670 --> 00:15:52.255 into components, F1 cosine theta in the x direction 00:15:52.255 --> 00:15:55.350 and F1 sine theta in the y direction 00:15:55.350 --> 00:16:00.810 and get two-- get a finite amount, a nonzero amount 00:16:00.810 --> 00:16:03.710 for y double dot and non-0 amount for x double dot. 00:16:03.710 --> 00:16:05.860 I end up with three equations I have to work with. 00:16:05.860 --> 00:16:09.670 Just by realigning it, I can make this problem even simpler. 00:16:09.670 --> 00:16:11.660 So just to do it quickly, that's what I did. 00:16:14.650 --> 00:16:17.800 So I've got two of the three equations done. 00:16:17.800 --> 00:16:23.340 And finally, the sum of the torques about g. 00:16:32.750 --> 00:16:35.390 And we know the angular momentum of this 00:16:35.390 --> 00:16:41.120 would be some izz g times omega z 00:16:41.120 --> 00:16:43.190 would be the angular momentum. 00:16:43.190 --> 00:16:51.150 And it's d by dt of that izz about g, omega z dot, 00:16:51.150 --> 00:16:53.790 which is theta double dot, right? 00:16:53.790 --> 00:16:55.240 And what are the external torques? 00:17:02.852 --> 00:17:03.560 So figure it out. 00:17:03.560 --> 00:17:05.440 What's the external torque in this problem? 00:17:05.440 --> 00:17:07.190 What is its size and what's its direction? 00:17:28.410 --> 00:17:33.460 So, somebody give me-- what's the external torque 00:17:33.460 --> 00:17:36.200 with respect to where, to start with? 00:17:36.200 --> 00:17:40.870 So we're summing torques about g, the center of mass. 00:17:40.870 --> 00:17:44.355 R cross F, right, which comes out in what direction? 00:17:49.830 --> 00:17:51.170 k direction, right? 00:17:51.170 --> 00:17:54.150 And minus or plus? 00:17:54.150 --> 00:17:55.210 Do your right hand rule. 00:17:55.210 --> 00:17:58.040 And notice I've drawn x and y such 00:17:58.040 --> 00:18:00.250 that positive z is into the board. 00:18:03.460 --> 00:18:04.710 So this is positive. 00:18:04.710 --> 00:18:07.250 I put a positive z into the board on purpose 00:18:07.250 --> 00:18:13.660 so that this would just work out R F1 in the k direction. 00:18:13.660 --> 00:18:22.360 And that's equal to Izz about g theta double dot. 00:18:22.360 --> 00:18:24.380 And since it's all with respect to k, 00:18:24.380 --> 00:18:27.000 you can drop the unit vectors here. 00:18:27.000 --> 00:18:31.170 And so we have one equation. 00:18:31.170 --> 00:18:35.100 We have two equations. 00:18:35.100 --> 00:18:38.730 And we have three equations. 00:18:38.730 --> 00:18:40.461 And that's all we need to complete this, 00:18:40.461 --> 00:18:42.960 is three degrees of freedom to completely define the motion. 00:18:42.960 --> 00:18:43.459 Yes? 00:18:43.459 --> 00:18:46.965 AUDIENCE: Do you want us to have a trivial case? 00:18:46.965 --> 00:18:51.240 PROFESSOR: Do you mean the my double dot equals 0? 00:18:51.240 --> 00:18:57.600 Well, in general-- when I started this one, when 00:18:57.600 --> 00:19:01.952 I didn't align them, then you're forced to use the three, right? 00:19:01.952 --> 00:19:03.410 So the only difference between what 00:19:03.410 --> 00:19:09.390 I did and this probe making a problem slightly more messy, 00:19:09.390 --> 00:19:14.120 is if I had let theta, this angle, be other than 0, I 00:19:14.120 --> 00:19:16.940 would have had to have come up with three 00:19:16.940 --> 00:19:22.400 non-zero acceleration equations. 00:19:22.400 --> 00:19:28.110 So I'd say technically, or I'd just say I think it's prudent. 00:19:28.110 --> 00:19:32.280 If you're ever not sure, if you can 00:19:32.280 --> 00:19:36.630 look at it and by inspection write down, my double dot is 0, 00:19:36.630 --> 00:19:39.230 you've essentially just written an equation of motion. 00:19:39.230 --> 00:19:42.720 Rather than say let's just ignore it, 00:19:42.720 --> 00:19:45.630 I'd keep it in your-- I'd just put it on the paper, 00:19:45.630 --> 00:19:47.670 my double dot equals 0. 00:19:47.670 --> 00:19:50.100 And that's your first trivial equation of motion. 00:19:50.100 --> 00:19:52.880 But it is an equation of motion, because this body 00:19:52.880 --> 00:19:55.480 has a true degree of freedom in that direction. 00:19:57.911 --> 00:19:58.910 That's really the point. 00:19:58.910 --> 00:20:02.080 Don't confuse trivial results because there's 00:20:02.080 --> 00:20:06.860 no external forces with constraints. 00:20:06.860 --> 00:20:09.260 It's unconstrained in the x direction, 00:20:09.260 --> 00:20:12.190 or in the y direction the way I've set it up. 00:20:12.190 --> 00:20:17.640 So I think you're safer if you do just-- 00:20:17.640 --> 00:20:20.195 say the torques or the forces in that direction are 0 00:20:20.195 --> 00:20:22.640 and you write it down. 00:20:22.640 --> 00:20:25.390 Let's look at this problem from the point of view of Lagrange. 00:20:25.390 --> 00:20:28.930 It's also equally simple, the Lagrange, 00:20:28.930 --> 00:20:33.370 except for the generalized forces. 00:20:33.370 --> 00:20:36.102 Actually, Lagrange is harder because you 00:20:36.102 --> 00:20:38.310 have to think your way through the generalized forces 00:20:38.310 --> 00:20:39.420 in this problem. 00:20:39.420 --> 00:20:41.690 And I got myself in a pickle with this problem 00:20:41.690 --> 00:20:44.110 and I spent hours trying to figure out 00:20:44.110 --> 00:20:47.210 a good explanation of a conundrum I 00:20:47.210 --> 00:20:49.800 ran into with the generalized forces. 00:20:49.800 --> 00:20:50.644 So let's look. 00:20:50.644 --> 00:20:52.310 I'll tell you what that was in a second. 00:20:52.310 --> 00:20:53.268 Let's look at Lagrange. 00:21:00.480 --> 00:21:07.310 So you need T. And we're pretty good at this now, 00:21:07.310 --> 00:21:10.900 I with respect to g zz, theta dot squared. 00:21:10.900 --> 00:21:13.220 That's your rotational kinetic energy. 00:21:13.220 --> 00:21:19.430 1/2 mx dot squared plus y dot squared. 00:21:21.940 --> 00:21:23.950 Translational kinetic energy associated 00:21:23.950 --> 00:21:25.080 with the center of mass. 00:21:25.080 --> 00:21:27.360 Notice this is with respect to the center 00:21:27.360 --> 00:21:29.260 of mass. 00:21:29.260 --> 00:21:30.290 How about v? 00:21:35.096 --> 00:21:36.470 Potential energy in this problem? 00:21:39.520 --> 00:21:40.710 Any? 00:21:40.710 --> 00:21:41.810 It's all 0's, right? 00:21:41.810 --> 00:21:42.810 There's no springs. 00:21:42.810 --> 00:21:46.820 There's no gravity in this plane, operating in this plane. 00:21:46.820 --> 00:21:48.337 So this is 0. 00:21:48.337 --> 00:21:50.295 This problem's going to be particularly simple. 00:22:17.550 --> 00:22:23.340 So in the theta direction here, this derivative 00:22:23.340 --> 00:22:28.760 with respect to Q dot gives us Iz theta dot 00:22:28.760 --> 00:22:34.910 and the time derivative of that zz theta double dot. 00:22:34.910 --> 00:22:40.930 T with respect to theta-- nothing. 00:22:40.930 --> 00:22:42.850 V with respect to theta? 00:22:42.850 --> 00:22:43.675 Nothing. 00:22:43.675 --> 00:22:50.710 And on the right hand side, we need to get Q theta here. 00:22:50.710 --> 00:22:58.940 And the other equation, it's the x equation. 00:22:58.940 --> 00:23:00.930 Well, actually, we can have x and y. 00:23:00.930 --> 00:23:02.930 So what about the x equation? 00:23:02.930 --> 00:23:04.570 Derivative with respect to x dot? 00:23:04.570 --> 00:23:08.170 I get an mx dot, time derivative, mx double dot. 00:23:15.100 --> 00:23:16.420 And with respect to y. 00:23:16.420 --> 00:23:20.900 The thing about Lagrange, if you can do Lagrange, just write 00:23:20.900 --> 00:23:24.140 down the total expression and just crank it out, 00:23:24.140 --> 00:23:27.380 because if you say, oh, well, I know this is trivial, 00:23:27.380 --> 00:23:29.410 then you're actually employing some information 00:23:29.410 --> 00:23:30.554 from the direct method. 00:23:30.554 --> 00:23:33.220 If you're saying, I know the sum of the forces in this direction 00:23:33.220 --> 00:23:34.310 is 0. 00:23:34.310 --> 00:23:37.440 So when you straight out in applying Lagrange, 00:23:37.440 --> 00:23:41.340 you can do it without any reference at all to Newton. 00:23:41.340 --> 00:23:45.830 So in this case, we just blindly clunk along and we say, well, 00:23:45.830 --> 00:23:49.190 what's the derivative now with respect to y dot? 00:23:49.190 --> 00:23:55.810 Will I get an my dot time derivative y double dot. 00:23:55.810 --> 00:23:57.490 This derivative would be 0. 00:23:57.490 --> 00:23:58.910 This derivative is 0. 00:23:58.910 --> 00:24:02.760 And over here I get Q1. 00:24:02.760 --> 00:24:06.620 Now I'm left-- so the left hand side of Lagrange equations-- 00:24:06.620 --> 00:24:08.855 remember, you just add these terms together. 00:24:13.790 --> 00:24:16.225 And these are going to be our three equations of motion 00:24:16.225 --> 00:24:19.430 that result. And now the remaining work 00:24:19.430 --> 00:24:21.640 is to get these three generalized forces. 00:24:28.450 --> 00:24:30.980 Let's go back to our picture. 00:24:30.980 --> 00:24:35.480 If we had lined it up like this, and now we 00:24:35.480 --> 00:24:39.470 give it a general little virtual displacement delta 00:24:39.470 --> 00:24:42.940 y, how much work gets done? 00:24:42.940 --> 00:24:43.680 None. 00:24:43.680 --> 00:24:47.780 And so you suddenly realize, oh, I went to a lot of work 00:24:47.780 --> 00:24:48.680 for nothing here. 00:24:48.680 --> 00:24:50.900 This equation is trivial. 00:24:50.900 --> 00:24:51.980 Nothing happens there. 00:24:51.980 --> 00:24:56.370 And so now it's reduced to doing the Qx and Qy. 00:24:59.400 --> 00:25:03.700 So just to give us some practice, 00:25:03.700 --> 00:25:10.350 let's do Q theta and Qx the rigorous kinematic way. 00:25:37.050 --> 00:25:39.420 I'm going to draw the general picture here, 00:25:39.420 --> 00:25:42.225 because this is where I got myself into trouble. 00:25:42.225 --> 00:25:43.600 You're working along on a problem 00:25:43.600 --> 00:25:46.310 and then something doesn't quite work for you. 00:25:46.310 --> 00:25:49.647 I had set this problem to do generalized-- to compute 00:25:49.647 --> 00:25:50.605 the generalized forces. 00:25:50.605 --> 00:26:00.030 And I set it up looking like set it up like this so theta is 0. 00:26:00.030 --> 00:26:02.540 It's lined up like that. 00:26:02.540 --> 00:26:05.430 And I started thinking about, then generalized, 00:26:05.430 --> 00:26:08.470 my little virtual displacements, and I 00:26:08.470 --> 00:26:11.426 got into a conceptual problem that I couldn't sort out 00:26:11.426 --> 00:26:11.925 for a while. 00:26:11.925 --> 00:26:17.699 And I'll work it into the explanation here. 00:26:17.699 --> 00:26:19.490 So I'm going to set this up more generally. 00:26:19.490 --> 00:26:23.290 I have an angle theta here, and this is my F1. 00:26:23.290 --> 00:26:25.400 So to do this by this kinematic way, 00:26:25.400 --> 00:26:28.020 I need to find the position vector. 00:26:28.020 --> 00:26:30.720 Here's a point-- I'm going to call it D, 00:26:30.720 --> 00:26:35.800 and this is my point G, and here's O. 00:26:35.800 --> 00:26:39.580 So I have a position vector going to here-- 00:26:39.580 --> 00:26:44.490 it's RG; and a position vector going to here-- that's 00:26:44.490 --> 00:26:54.000 R of D with respect to G; and a position vector from here, 00:26:54.000 --> 00:27:07.000 and that's RD in O. And RD in O is RG in O plus RD with respect 00:27:07.000 --> 00:27:08.639 to G. I can write that. 00:27:08.639 --> 00:27:09.305 They're vectors. 00:27:12.820 --> 00:27:15.510 So I'm going to need those. 00:27:15.510 --> 00:27:17.830 So what is RG in O? 00:27:17.830 --> 00:27:19.900 Well, I have this coordinate system. 00:27:19.900 --> 00:27:27.560 It's going to be some XI plus YJ in the inertial frame. 00:27:27.560 --> 00:27:33.380 And then RD with respect to G-- I 00:27:33.380 --> 00:27:37.500 need to have some kind of a coordinate system that 00:27:37.500 --> 00:27:40.200 rotates with the body. 00:27:40.200 --> 00:27:47.240 So I'll call this y1, and this x1-- how 00:27:47.240 --> 00:27:49.380 it's rotating with the body. 00:27:49.380 --> 00:27:58.170 So now, this one down here is minus R in the j1 direction. 00:28:02.370 --> 00:28:07.400 So that's the position of this point D with respect to O, 00:28:07.400 --> 00:28:09.040 written as the sum of two vectors. 00:28:25.150 --> 00:28:27.690 And I'm eventually going to need to be able to express 00:28:27.690 --> 00:28:33.480 this j1 in an inertial frame. 00:28:33.480 --> 00:28:37.500 So here's just this little vector j1, 00:28:37.500 --> 00:28:46.810 and I need its components in the J and I directions. 00:28:46.810 --> 00:28:51.220 So it's made up of a piece like that, a piece like this, 00:28:51.220 --> 00:28:57.120 and this angle here is theta. 00:28:57.120 --> 00:29:13.365 So little j1-- so I can converge. 00:29:13.365 --> 00:29:17.790 I can move from this unit vector system to that one 00:29:17.790 --> 00:29:19.239 by this transformation. 00:29:19.239 --> 00:29:20.697 I'm going to need that in a minute. 00:29:26.898 --> 00:29:29.020 Actually, I'll invoke it right now. 00:29:29.020 --> 00:29:34.555 So my position vector here is-- the x component is x. 00:29:48.190 --> 00:29:49.270 Can't read my plus. 00:29:49.270 --> 00:29:52.620 I crossed out a plus and minus, so I got to sort this out 00:29:52.620 --> 00:29:53.575 carefully here. 00:30:04.890 --> 00:30:05.545 Is this right? 00:30:15.077 --> 00:30:17.701 What do you think? 00:30:17.701 --> 00:30:18.950 Laura, what have I done wrong? 00:30:22.950 --> 00:30:25.480 So this is the plus I direction. 00:30:25.480 --> 00:30:32.400 This piece here is-- so minus sine 00:30:32.400 --> 00:30:37.015 theta in the I plus cosine theta in the J are the components. 00:30:37.015 --> 00:30:39.441 So I'm missing a minus sign there. 00:30:39.441 --> 00:30:39.940 OK. 00:30:42.640 --> 00:30:45.070 So I want to collect the I pieces together. 00:30:45.070 --> 00:30:48.250 So I have an X in the I, and I have 00:30:48.250 --> 00:30:52.690 a minus R minus sine theta. 00:30:52.690 --> 00:31:00.660 So, plus R sine theta-- this is the I contribution. 00:31:00.660 --> 00:31:09.600 And then over here I have a Y and a minus R cosine 00:31:09.600 --> 00:31:15.720 theta in the J. Then I have this vector all worked out 00:31:15.720 --> 00:31:16.780 in my inertial frame. 00:31:20.410 --> 00:31:29.810 So now I can say that-- remember, Qj dot 00:31:29.810 --> 00:31:38.290 dR-- it's a sort of virtual deflection for the jth 00:31:38.290 --> 00:31:39.920 contribution. 00:31:39.920 --> 00:31:44.965 So in this case, I'm interested in, say, the 1 in the-- which 00:31:44.965 --> 00:31:48.195 one should we do first-- The x direction. 00:31:52.260 --> 00:32:02.530 So this is the F1 in the I dot derivative of RD with respect 00:32:02.530 --> 00:32:08.900 to x delta x. 00:32:08.900 --> 00:32:11.100 But now I have an-- I can work this out. 00:32:11.100 --> 00:32:13.150 I can figure this one out directly now, 00:32:13.150 --> 00:32:17.690 because I have an expression for R-- 00:32:17.690 --> 00:32:22.780 that vector-- in terms of one set of unit vectors, 00:32:22.780 --> 00:32:26.160 and I can take the derivative with respect to x. 00:32:26.160 --> 00:32:27.812 And the only x that shows up in here 00:32:27.812 --> 00:32:29.645 is this, so the derivative is pretty simple. 00:32:29.645 --> 00:32:31.490 It's just 1. 00:32:31.490 --> 00:32:38.710 So this is F1I dot I delta x. 00:32:38.710 --> 00:32:45.670 So this tells us that Qx equals F1. 00:32:45.670 --> 00:32:46.610 No great surprise. 00:32:46.610 --> 00:32:49.920 We knew the answer to that from before. 00:32:49.920 --> 00:32:57.760 And I-- oh, I made a mistake. 00:33:01.740 --> 00:33:04.620 What didn't I-- I was mixing the two. 00:33:04.620 --> 00:33:07.085 I did a simplification when I did the direct one. 00:33:07.085 --> 00:33:08.186 Is F1 in the I direction? 00:33:17.090 --> 00:33:20.390 So F1 has components. 00:33:20.390 --> 00:33:33.420 Vector F1 has a magnitude F1 times cosine theta I plus sine 00:33:33.420 --> 00:33:37.000 theta J. And so down here, I need 00:33:37.000 --> 00:33:40.020 to put those in and take the dot product. 00:33:40.020 --> 00:33:44.980 So the only dot product that will matter is the I component. 00:33:44.980 --> 00:33:50.372 This is going to be F1 cosine theta I 00:33:50.372 --> 00:33:54.460 dot this, which we know this gives us just and I. 00:33:54.460 --> 00:33:56.830 So I'm leaving out the J piece. 00:33:56.830 --> 00:34:02.010 So I end up here with F1 cosine theta. 00:34:02.010 --> 00:34:04.630 Does that look right? 00:34:04.630 --> 00:34:06.990 So here is your theta. 00:34:06.990 --> 00:34:10.680 Here is your-- this is F1 cosine theta. 00:34:13.360 --> 00:34:18.159 So for our coordinate system that's squared up like this, 00:34:18.159 --> 00:34:21.730 it's only that term that's in the F1 direction, 00:34:21.730 --> 00:34:24.610 and this would come out correctly. 00:34:24.610 --> 00:34:25.149 Now 00:34:25.149 --> 00:34:28.790 And if we let-- if we now reorient our coordinate system 00:34:28.790 --> 00:34:36.199 so that theta is 0, then this would just turn out to be F1. 00:34:36.199 --> 00:34:38.719 If you can see that this is F1 cosine theta without going 00:34:38.719 --> 00:34:42.489 through all the work, what's Qy for this system? 00:34:47.090 --> 00:34:47.900 Sure. 00:34:47.900 --> 00:34:50.359 F1 sine theta. 00:34:50.359 --> 00:34:51.734 We won't go through all the work, 00:34:51.734 --> 00:34:55.300 but you go through the same taking 00:34:55.300 --> 00:34:57.500 the derivative-- in this case, the derivative of R 00:34:57.500 --> 00:35:01.880 with respect to y, and then dot it with F, 00:35:01.880 --> 00:35:05.900 and you get F1 sine theta in the J direction. 00:35:05.900 --> 00:35:08.020 So there's Qy. 00:35:08.020 --> 00:35:22.540 Now, the problem I ran into, conceptually, 00:35:22.540 --> 00:35:25.570 is around this other piece. 00:35:25.570 --> 00:35:31.000 So Q theta, the virtual work done in the theta direction, 00:35:31.000 --> 00:35:34.360 we know has to do with rotation. 00:35:34.360 --> 00:35:41.460 And that is going to be sum F1 dot-- now 00:35:41.460 --> 00:35:46.421 this is the derivative of RD with respect to theta delta 00:35:46.421 --> 00:35:46.920 theta. 00:35:53.620 --> 00:36:02.632 So RD is here, and it's indeed a function of theta, right? 00:36:02.632 --> 00:36:04.840 And so I can take a derivative with respect to theta, 00:36:04.840 --> 00:36:08.010 and I'll get derivative of that R sine theta term, 00:36:08.010 --> 00:36:10.996 and I'll get a derivative of the R cosine theta term. 00:36:10.996 --> 00:36:12.620 But here's how I got myself in trouble. 00:36:12.620 --> 00:36:14.190 It took me-- and this is the sort of thing 00:36:14.190 --> 00:36:15.273 that happens to all of us. 00:36:15.273 --> 00:36:18.407 You're working a problem, throw in a simplification, 00:36:18.407 --> 00:36:19.740 and then something doesn't work. 00:36:19.740 --> 00:36:31.290 Well, what if you had oriented the axis, initially, 00:36:31.290 --> 00:36:38.720 so that the force is aligned with x and perpendicular to y? 00:36:38.720 --> 00:36:45.780 Then this theta angle is 0, right? 00:36:45.780 --> 00:36:48.670 And you know, when you set it up, 00:36:48.670 --> 00:36:53.980 you don't end up with-- the cosine of 0 is 1, and sine of 0 00:36:53.980 --> 00:36:55.672 is 0. 00:36:55.672 --> 00:36:58.260 You don't end up with the cosine and sines in there 00:36:58.260 --> 00:37:00.140 to take derivatives of. 00:37:00.140 --> 00:37:02.450 You just jump to the simplification, 00:37:02.450 --> 00:37:13.060 and you just find out that you just have R in the J. 00:37:13.060 --> 00:37:14.480 There's no R sine theta term. 00:37:14.480 --> 00:37:17.380 It's just vanished on you. 00:37:17.380 --> 00:37:21.400 And now you need to take that derivative to do this step. 00:37:26.300 --> 00:37:29.010 And I do count with it as huh? 00:37:29.010 --> 00:37:31.600 What have I done wrong? 00:37:31.600 --> 00:37:33.550 In order to do this method, you need 00:37:33.550 --> 00:37:37.040 to leave it in the sort of general formulation, 00:37:37.040 --> 00:37:40.010 and then let theta be 0 at the end, if you want to do that. 00:37:40.010 --> 00:37:45.340 So if I had then finished it, I can 00:37:45.340 --> 00:37:47.690 do this for this general problem, 00:37:47.690 --> 00:37:50.520 and then let theta go to 0, and the answer 00:37:50.520 --> 00:37:51.520 will come out all right. 00:37:51.520 --> 00:37:53.140 So should we do this? 00:37:53.140 --> 00:37:56.460 So this is F1 dot this. 00:37:56.460 --> 00:38:09.920 So, F1 dotted with the derivative of R 00:38:09.920 --> 00:38:12.010 with respect to theta. 00:38:12.010 --> 00:38:21.957 So I get an R cosine theta in the I, 00:38:21.957 --> 00:38:23.915 and the derivative of this term will be-- well, 00:38:23.915 --> 00:38:26.150 the cosine theta is minus sine theta 00:38:26.150 --> 00:38:36.550 times a minus is a plus R sine theta in the J delta theta. 00:38:36.550 --> 00:38:40.810 And I do this dot product, so I get 00:38:40.810 --> 00:38:46.715 the I times the I's, and the J terms times the J terms. 00:39:11.980 --> 00:39:13.510 And it all works out nicely. 00:39:13.510 --> 00:39:16.090 Sine squared plus cosine squared-- you get 1. 00:39:16.090 --> 00:39:19.390 And you find out that the torque doesn't care about the angle. 00:39:19.390 --> 00:39:20.839 And does that make sense? 00:39:20.839 --> 00:39:21.755 AUDIENCE: [INAUDIBLE]? 00:39:25.434 --> 00:39:26.350 PROFESSOR: Absolutely. 00:39:26.350 --> 00:39:30.160 I did this on purpose to give us a little more practice using 00:39:30.160 --> 00:39:33.330 this kinematic grind it out method. 00:39:33.330 --> 00:39:36.870 But, for sure, the intuitive method 00:39:36.870 --> 00:39:41.390 would have yielded result a lot faster here, right? 00:39:41.390 --> 00:39:45.500 We would've said, let's have a little deflection delta x. 00:39:45.500 --> 00:39:46.870 What's the virtual work, then? 00:39:46.870 --> 00:39:50.700 Well, would have been F1 cosine theta delta x. 00:39:50.700 --> 00:39:53.740 It would be Qx delta x. 00:39:53.740 --> 00:40:00.140 Qy delta y would be F1 delta y sine theta, and we're done. 00:40:00.140 --> 00:40:04.660 And the rotation one-- we would have looked at this 00:40:04.660 --> 00:40:08.130 and said, how much-- in a little motion delta theta-- how 00:40:08.130 --> 00:40:09.370 much does this move? 00:40:09.370 --> 00:40:15.565 It moves R delta theta crossed with the F 00:40:15.565 --> 00:40:19.440 that it moves through, or dot product with that distance, 00:40:19.440 --> 00:40:22.550 would have given you F1 R delta theta equals 00:40:22.550 --> 00:40:24.130 Q theta delta theta. 00:40:24.130 --> 00:40:26.215 So we could have done it in a minute. 00:40:28.850 --> 00:40:30.840 And sometimes, doing it the hard way 00:40:30.840 --> 00:40:32.360 is what will get you in trouble. 00:40:32.360 --> 00:40:35.869 I didn't have the cosine theta and sine thetas 00:40:35.869 --> 00:40:38.160 working out in this and I couldn't take the derivative. 00:40:38.160 --> 00:40:40.540 What did I do wrong? 00:40:40.540 --> 00:40:43.084 And I finally realized, I made the problem too simple, 00:40:43.084 --> 00:40:46.840 simplified it too soon. 00:40:46.840 --> 00:40:50.810 So I wanted to use this as a stepping stone 00:40:50.810 --> 00:40:55.147 to something I meant to cover perhaps weeks ago. 00:40:55.147 --> 00:40:56.605 And there's a general little lesson 00:40:56.605 --> 00:40:59.860 that we can learn from this, which you've probably 00:40:59.860 --> 00:41:02.990 been taught before, but I'm going to remind you. 00:41:02.990 --> 00:41:05.840 Here's a rigid body. 00:41:05.840 --> 00:41:08.440 Got some center of mass. 00:41:08.440 --> 00:41:13.840 And I've got a force acting on it here, 00:41:13.840 --> 00:41:15.880 just some arbitrary position and angle. 00:41:19.600 --> 00:41:22.410 So that force has a line of action, 00:41:22.410 --> 00:41:25.680 and perpendicular to that line of action 00:41:25.680 --> 00:41:28.825 is a distance, which I'll call d here. 00:41:31.450 --> 00:41:34.580 And we know that in R cross F, we 00:41:34.580 --> 00:41:44.290 can-- this force exerts a moment about G that would be R cross 00:41:44.290 --> 00:41:47.160 F, where R goes from here to here, 00:41:47.160 --> 00:41:50.830 but it's only the component perpendicular to it. 00:41:50.830 --> 00:41:54.060 So we know that this is going to create a moment. 00:41:54.060 --> 00:41:56.570 And I won't draw pictures yet here for a moment. 00:41:56.570 --> 00:42:00.555 This diagram is the same as-- well, 00:42:00.555 --> 00:42:03.510 I could draw the same peanut here. 00:42:03.510 --> 00:42:09.740 It's the same as the following-- a force here, that's F, 00:42:09.740 --> 00:42:12.540 and another force at G, that's F. Equal and opposite, 00:42:12.540 --> 00:42:15.340 so it's like adding 0 to the problem. 00:42:15.340 --> 00:42:19.310 So I can do that with impunity, and I still 00:42:19.310 --> 00:42:23.970 have my force F down here. 00:42:23.970 --> 00:42:25.882 So this problem's equal to that problem. 00:42:29.530 --> 00:42:36.260 But now, this force and this force, equal and opposite, 00:42:36.260 --> 00:42:37.860 cancel one another, but create what's 00:42:37.860 --> 00:42:42.440 called a couple-- a moment that we can compute about my point G 00:42:42.440 --> 00:42:43.340 here. 00:42:43.340 --> 00:42:48.110 So these two forces create just a pure torque, 00:42:48.110 --> 00:42:51.670 leaving this as a force. 00:42:51.670 --> 00:42:53.550 They're equal and opposite, so no net force. 00:42:53.550 --> 00:42:56.280 But they create a torque in this direction, 00:42:56.280 --> 00:42:59.910 and there's a remaining force on the center of mass. 00:42:59.910 --> 00:43:04.110 So this whole thing is also, then, 00:43:04.110 --> 00:43:12.220 equal to the equivalent of a torque 00:43:12.220 --> 00:43:16.400 around the center of mass and a force 00:43:16.400 --> 00:43:19.620 applied at the center of mass. 00:43:19.620 --> 00:43:23.520 And in this case, the torque would be dF. 00:43:28.310 --> 00:43:30.750 This problem is easier to do if you're 00:43:30.750 --> 00:43:34.200 trying to compute generalized forces than the one we started 00:43:34.200 --> 00:43:39.600 with, because now, the position vector that we need 00:43:39.600 --> 00:43:41.035 goes only to the center of mass. 00:43:45.630 --> 00:43:50.890 And the general rule for this-- you have a body, 00:43:50.890 --> 00:43:59.590 you have several forces-- F1, F2, F3. 00:44:03.410 --> 00:44:15.800 That's equal to the body with G, with a F total on it, 00:44:15.800 --> 00:44:20.700 and a moment-- a torque-- total. 00:44:20.700 --> 00:44:27.430 And F total-- it's a vector-- is a summation of the Fi's. 00:44:27.430 --> 00:44:32.890 And T, the torque total, is the summation 00:44:32.890 --> 00:44:34.630 of each of these guys. 00:44:34.630 --> 00:44:36.920 Here's G. This has an R1. 00:44:39.560 --> 00:44:40.490 Here is an R2. 00:44:43.800 --> 00:44:49.020 Here's an Ri to the ith force. 00:44:49.020 --> 00:44:55.380 So this is a summation of the Ri cross Fi's. 00:44:55.380 --> 00:44:58.240 So if you sum all the forces and compute the equivalent torque 00:44:58.240 --> 00:45:00.555 about the center of mass, sum all the forces 00:45:00.555 --> 00:45:03.030 and apply it at the center of mass, 00:45:03.030 --> 00:45:07.440 then this problem is equal to that problem. 00:45:07.440 --> 00:45:09.540 So this can come in handy when you're 00:45:09.540 --> 00:45:10.785 trying to simplify problems. 00:45:22.330 --> 00:45:38.800 And for this particular problem, let's draw our hockey puck. 00:45:46.745 --> 00:45:51.400 Going to put my force in the simplest direction here. 00:45:51.400 --> 00:45:53.260 But there's my force. 00:45:53.260 --> 00:46:00.520 This problem is equal, doing the same thing here. 00:46:00.520 --> 00:46:04.925 It's the same thing as if I had drew forces like this. 00:46:12.570 --> 00:46:15.060 And that gives me an equivalent problem, 00:46:15.060 --> 00:46:22.420 where I have-- this couple here produces a torque, 00:46:22.420 --> 00:46:28.200 and there is a net force acting on the center of mass. 00:46:28.200 --> 00:46:32.980 So this problem is equal to this problem now. 00:46:32.980 --> 00:46:36.410 A torque and a force both acting at the center of mass, 00:46:36.410 --> 00:46:43.250 and the torque is RF1 in the k. 00:46:43.250 --> 00:46:50.530 And the force is just F1 in the positive I direction. 00:46:53.620 --> 00:46:58.500 And R-- but now, what is the little R vector? 00:46:58.500 --> 00:47:02.650 If we're going to do this the kinematic way? 00:47:02.650 --> 00:47:08.210 From here to G, this is R1, I'll call it. 00:47:08.210 --> 00:47:09.470 And that's it. 00:47:09.470 --> 00:47:13.700 I only-- now I have everything acting here. 00:47:13.700 --> 00:47:26.740 And Qx delta x is F1 dot derivative of R 00:47:26.740 --> 00:47:30.610 with respect to x delta x. 00:47:30.610 --> 00:47:33.375 But what is R? 00:47:38.690 --> 00:47:41.610 It's XI plus YJ. 00:47:41.610 --> 00:47:49.310 The derivative of R1 with respect to x is just 1 00:47:49.310 --> 00:47:52.940 in the I, and the derivative of R with respect to Y 00:47:52.940 --> 00:47:58.460 is just J. So this becomes a trivial calculation. 00:47:58.460 --> 00:48:07.060 F1I dot I-- I get Qx equals F1 delta x. 00:48:07.060 --> 00:48:15.910 Then I get Qy is F1, which is in the I direction dot J is 0. 00:48:15.910 --> 00:48:19.600 And I get Q theta-- ah, Q theta. 00:48:19.600 --> 00:48:23.270 Now, do I use this expression? 00:48:23.270 --> 00:48:27.100 So, I'm not dealing with little changes in distance anymore. 00:48:27.100 --> 00:48:29.400 When you can put rotation-- when you put moments 00:48:29.400 --> 00:48:31.500 about the centers of mass, it's just easier. 00:48:31.500 --> 00:48:34.790 This one-- you call intuitive, but this is just 00:48:34.790 --> 00:48:38.960 going to be the sum delta theta is 00:48:38.960 --> 00:48:46.070 equal to the total torques acting 00:48:46.070 --> 00:48:53.070 at the center of mass dot delta theta. 00:48:53.070 --> 00:48:56.110 And in this case, the torque is in the k direction, 00:48:56.110 --> 00:48:58.210 delta theta's in the k direction, 00:48:58.210 --> 00:49:12.910 and this is RF1k dot k delta theta, which just gives us 00:49:12.910 --> 00:49:16.740 RF1 equals Q theta. 00:49:16.740 --> 00:49:20.252 So the torque equations you can derive straight out. 00:49:20.252 --> 00:49:21.710 You don't have to take derivatives. 00:49:21.710 --> 00:49:24.140 It doesn't have anything to do with change in position. 00:49:24.140 --> 00:49:26.554 It's just a little rotation, delta theta. 00:49:29.210 --> 00:49:31.920 So it simplifies this problem and, actually, totally avoids 00:49:31.920 --> 00:49:35.180 the conundrum I got into. 00:49:35.180 --> 00:49:37.060 If I work everything about the center of mass 00:49:37.060 --> 00:49:40.010 of that rigid body, the R gets easier, 00:49:40.010 --> 00:49:42.580 the derivatives get easier, and the torque 00:49:42.580 --> 00:49:45.050 becomes obvious how it applies. 00:50:16.950 --> 00:50:21.330 We'll talk about-- I have a pendulum in kind 00:50:21.330 --> 00:50:22.060 of a weird shape. 00:50:29.260 --> 00:50:34.400 But the pendulum has the following properties. 00:50:34.400 --> 00:50:38.760 So that weird shape-- that may be this. 00:50:38.760 --> 00:50:45.100 It's got one plane of symmetry like this does. 00:50:45.100 --> 00:50:49.650 And I'm going to put-- what did I do with my axle? 00:50:49.650 --> 00:50:53.780 I'm going to put the axis of rotation perpendicular 00:50:53.780 --> 00:50:57.510 to that plane of symmetry, somewhere not at G, 00:50:57.510 --> 00:51:01.280 not at the center of mass-- so, just what I've done here. 00:51:01.280 --> 00:51:07.790 I can make any object that has a plane of symmetry-- 00:51:07.790 --> 00:51:13.940 if I have it rotate about an axis that's 00:51:13.940 --> 00:51:16.030 perpendicular to that plane of symmetry, 00:51:16.030 --> 00:51:19.440 it becomes a simple pendulum. 00:51:19.440 --> 00:51:27.210 And you can write down with my inspection 00:51:27.210 --> 00:51:28.930 what the equation of motion is. 00:51:28.930 --> 00:51:33.720 So, how many degrees of freedom does this system have? 00:51:37.120 --> 00:51:40.870 So first of all, then, is it planar motion? 00:51:40.870 --> 00:51:45.760 So it has, at most, 3, and how many-- 00:51:45.760 --> 00:51:47.900 and it has a pin through it that doesn't allow it 00:51:47.900 --> 00:51:50.590 to move in x or y. 00:51:50.590 --> 00:51:51.945 How many constraints is that? 00:51:51.945 --> 00:51:52.599 AUDIENCE: 2. 00:51:52.599 --> 00:51:53.140 PROFESSOR: 2. 00:51:53.140 --> 00:51:55.240 So how many you got left? 00:51:55.240 --> 00:51:55.740 AUDIENCE: 1. 00:51:55.740 --> 00:51:56.281 PROFESSOR: 1. 00:51:56.281 --> 00:51:59.580 So this is a single degree of freedom problem, 00:51:59.580 --> 00:52:04.070 and I've chosen to have it move about an axis that 00:52:04.070 --> 00:52:08.450 is-- is this axis a principal axis of this body? 00:52:08.450 --> 00:52:10.390 Absolutely. 00:52:10.390 --> 00:52:13.330 And so, if I ask you, what's the mass moment of inertia of this? 00:52:13.330 --> 00:52:18.217 If I gave you Izz about G for this thing, and this distance, 00:52:18.217 --> 00:52:19.800 what would you tell me the mass moment 00:52:19.800 --> 00:52:21.380 of inertia about the pivot is? 00:52:26.330 --> 00:52:27.925 I'll give you this. 00:52:34.534 --> 00:52:35.450 What's the rest of it? 00:52:38.670 --> 00:52:41.155 Not a G. You've got some distance here, 00:52:41.155 --> 00:52:46.860 which you're given-- L, we'll call it. 00:52:46.860 --> 00:52:47.390 Correct. 00:52:47.390 --> 00:52:50.060 So, parallel axis-- remember, ML squared. 00:52:50.060 --> 00:52:55.350 So we know that that's true, and the G is over here someplace, 00:52:55.350 --> 00:53:00.120 and this is this distance L. And what's the equation of motion 00:53:00.120 --> 00:53:00.990 for this? 00:53:00.990 --> 00:53:03.860 And gravity acts. 00:53:03.860 --> 00:53:05.660 No damping for now. 00:53:05.660 --> 00:53:08.204 What's the equation of motion? 00:53:08.204 --> 00:53:09.370 Right off the top your head. 00:53:09.370 --> 00:53:10.602 You've done enough of these. 00:53:14.670 --> 00:53:16.960 I'll give you one minute think about it. 00:53:16.960 --> 00:53:19.690 We'll figure out the equation of motion for this problem. 00:53:30.050 --> 00:53:32.205 I highly recommend you don't do the Lagrange. 00:54:35.420 --> 00:54:37.170 Somebody give me their equation of motion. 00:54:54.090 --> 00:54:58.210 So, what method would you choose to use? 00:54:58.210 --> 00:55:01.440 Simplest one you could think of. 00:55:01.440 --> 00:55:02.130 Direct? 00:55:02.130 --> 00:55:03.576 About what point? 00:55:03.576 --> 00:55:04.390 AUDIENCE: Around A. 00:55:04.390 --> 00:55:05.848 PROFESSOR: About A. And what do you 00:55:05.848 --> 00:55:09.940 say is-- what equation do you apply? 00:55:09.940 --> 00:55:11.487 General equation. 00:55:11.487 --> 00:55:16.927 AUDIENCE: [INAUDIBLE] It's just AIzz [INAUDIBLE]. 00:55:16.927 --> 00:55:19.010 PROFESSOR: Yeah, that would be part of the answer. 00:55:19.010 --> 00:55:21.468 And the equation-- remember, I wrote a list of them-- 1, 2, 00:55:21.468 --> 00:55:22.770 3 over there. 00:55:22.770 --> 00:55:23.860 Which one do you use? 00:55:27.077 --> 00:55:28.660 AUDIENCE: You just use the second one. 00:55:28.660 --> 00:55:30.250 PROFESSOR: Just a second one, right. 00:55:30.250 --> 00:55:33.086 And do the nuisance terms go away? 00:55:33.086 --> 00:55:34.410 Point A's not moving. 00:55:34.410 --> 00:55:37.010 So some of the torque is equal to dHdt, 00:55:37.010 --> 00:55:38.890 and it's H with respect to A, right? 00:55:57.740 --> 00:55:59.670 So that's H, right? 00:56:03.170 --> 00:56:12.880 d by dt Izz with respect to A theta double dot k 00:56:12.880 --> 00:56:14.650 equals the sum of the external torques. 00:56:14.650 --> 00:56:18.320 And the rest of the problem is finding the external torques. 00:56:18.320 --> 00:56:20.130 What's the free body diagram look like? 00:56:24.730 --> 00:56:39.400 So, here's the object, possibly A and Rx, and an Ry, and an Mg. 00:56:39.400 --> 00:56:41.650 Those are your possible-- that will set your free body 00:56:41.650 --> 00:56:43.280 diagram, right? 00:56:43.280 --> 00:56:47.530 So here's the Mg down. 00:56:47.530 --> 00:56:51.800 What's the torque that gravitational force 00:56:51.800 --> 00:56:55.570 puts on this object with respect to point A? 00:56:58.100 --> 00:56:59.500 R cross F, right? 00:57:02.340 --> 00:57:05.582 R cross F into the board. 00:57:05.582 --> 00:57:06.790 Is that positive or negative? 00:57:10.090 --> 00:57:11.510 And what's the link to this R? 00:57:16.830 --> 00:57:19.520 Well, from here to here is, I guess-- oh, that 00:57:19.520 --> 00:57:23.395 is L. So the length of this side and that triangle? 00:57:26.350 --> 00:57:28.820 L sine theta? 00:57:28.820 --> 00:57:31.890 L sine theta cross Mg. 00:57:31.890 --> 00:57:36.260 And it is positive or negative-- what did we decide? 00:57:36.260 --> 00:57:42.790 Looks like it's-- this is positive theta xy system. 00:57:42.790 --> 00:57:51.180 It gets negative minus MgL sine theta, and we're done. 00:57:51.180 --> 00:57:59.780 So, generically-- and this is also in the k direction, 00:57:59.780 --> 00:58:01.200 so we can drop the k's now. 00:58:01.200 --> 00:58:08.680 We have our equation of motion Izz theta double dot plus MgL 00:58:08.680 --> 00:58:10.530 sine theta. 00:58:15.843 --> 00:58:24.520 Is there any then single degree of freedom pendulum made out 00:58:24.520 --> 00:58:30.320 of a rigid body, and rotating about an axis 00:58:30.320 --> 00:58:35.161 that is a principal axis that has that equation of motion? 00:58:40.240 --> 00:58:43.400 Any of them, no matter what the shape. 00:58:43.400 --> 00:58:46.090 If you can rotate it about a principal axis, 00:58:46.090 --> 00:58:48.790 and not through G, because if you put-- 00:58:48.790 --> 00:58:53.680 what's the natural frequency if you run the axis through G? 00:58:53.680 --> 00:58:55.516 What's the torque? 00:58:55.516 --> 00:58:56.015 0. 00:58:56.015 --> 00:58:58.999 Nothing happens, right? 00:58:58.999 --> 00:59:00.290 It doesn't want to do anything. 00:59:00.290 --> 00:59:03.980 But as soon as it's not through G, then the things oscillates. 00:59:03.980 --> 00:59:06.860 And that's because this is now a differential equation. 00:59:06.860 --> 00:59:11.320 And I need to say this equals 0 here. 00:59:11.320 --> 00:59:12.340 There's no other forces. 00:59:12.340 --> 00:59:14.548 If you've got a damping force, then it would show up. 00:59:14.548 --> 00:59:18.060 But this is a generic, undamped equation 00:59:18.060 --> 00:59:22.250 of motion for any single degree of freedom pendulum 00:59:22.250 --> 00:59:27.400 rotating about one of its principal axes. 00:59:27.400 --> 00:59:29.980 And if the body you're shown or given 00:59:29.980 --> 00:59:33.280 has a single plane of symmetry, then you 00:59:33.280 --> 00:59:35.010 can figure out one way, immediately, 00:59:35.010 --> 00:59:38.790 that you know this will-- that equation applies. 00:59:38.790 --> 00:59:42.540 An axis perpendicular to that plane of symmetry is a pendulum 00:59:42.540 --> 00:59:45.210 and it is that equation. 00:59:45.210 --> 00:59:50.970 And the i respect to A you can get from simple parallel axis, 00:59:50.970 --> 00:59:54.960 if you're told what i with respect to G is. 01:00:00.221 --> 01:00:00.720 Questions? 01:00:13.490 --> 01:00:16.670 I'm going to put up two additional problems. 01:00:16.670 --> 01:00:18.659 And I'm not going to solve them. 01:00:18.659 --> 01:00:19.950 We're going to talk about them. 01:00:19.950 --> 01:00:22.366 You're going to tell me how you would go about doing them. 01:00:22.366 --> 01:00:24.930 And they wouldn't be bad problems for you 01:00:24.930 --> 01:00:25.980 to practice on. 01:00:35.590 --> 01:00:40.200 And you've actually seen both of them before. 01:00:40.200 --> 01:00:51.935 So, the first one-- pulley rotating about that fixed axis. 01:00:54.800 --> 01:00:59.895 And this is that thing we call Atwood's Machine. 01:01:02.825 --> 01:01:04.075 How did I draw my coordinates? 01:01:10.240 --> 01:01:21.250 So I have my Izz about G. It's sum M3 kappa squared. 01:01:21.250 --> 01:01:23.790 You're given the radius of gyration of this pulley. 01:01:23.790 --> 01:01:27.090 So that's its mass moment of inertia about the center. 01:01:27.090 --> 01:01:34.755 And you know R. You could specify a theta. 01:01:38.780 --> 01:01:48.500 And we have a pair of masses-- start off like this, M1, M2. 01:01:48.500 --> 01:01:51.440 And this, now-- so I've going to say, no slip. 01:01:54.550 --> 01:01:59.390 And this rope goes over the pulley, no slip, initially 01:01:59.390 --> 01:02:00.425 stationary. 01:02:00.425 --> 01:02:02.580 I let go. 01:02:02.580 --> 01:02:05.310 I want an equation of motion for this system. 01:02:05.310 --> 01:02:07.490 So how many is it-- first of all, 01:02:07.490 --> 01:02:10.800 is it a planar motion problem? 01:02:10.800 --> 01:02:11.530 OK. 01:02:11.530 --> 01:02:12.500 How many rigid bodies? 01:02:15.280 --> 01:02:17.980 How many potential degrees of freedom? 01:02:17.980 --> 01:02:18.750 AUDIENCE: None. 01:02:18.750 --> 01:02:20.360 PROFESSOR: None. 01:02:20.360 --> 01:02:21.810 How many do you think they really 01:02:21.810 --> 01:02:23.270 are going to end up with here? 01:02:27.982 --> 01:02:29.590 How many do you expect to find? 01:02:29.590 --> 01:02:34.330 How many necessary coordinates are you 01:02:34.330 --> 01:02:36.870 going to need to completely describe the motion? 01:02:36.870 --> 01:02:38.220 I see some 1's going up. 01:02:38.220 --> 01:02:40.660 Everybody believe that? 01:02:40.660 --> 01:02:42.040 Lots of constraints. 01:02:42.040 --> 01:02:43.870 We're not going to let this-- we're 01:02:43.870 --> 01:02:46.895 constraining this to move in-- only allowing 01:02:46.895 --> 01:02:48.000 it to move up and down. 01:02:48.000 --> 01:02:52.060 These can't rotate, so there's one for each. 01:02:52.060 --> 01:02:57.870 These can't go in this direction, so two more. 01:02:57.870 --> 01:03:00.990 This can only rotate, no translations. 01:03:00.990 --> 01:03:03.660 So you end up with 1. 01:03:03.660 --> 01:03:05.800 And finally, there's this no slip condition, 01:03:05.800 --> 01:03:11.015 which means that R theta equals x, 01:03:11.015 --> 01:03:14.650 and that's one final one that you'd need to write down. 01:03:14.650 --> 01:03:17.780 So if you had that, gotten this far-- 01:03:17.780 --> 01:03:19.550 find the equation of motion. 01:03:19.550 --> 01:03:21.110 What method would you use? 01:03:21.110 --> 01:03:31.335 So, your choices are here-- direct method, Lagrange. 01:03:31.335 --> 01:03:31.960 Think about it. 01:03:31.960 --> 01:03:34.479 How would you-- what's the easiest way for you to 01:03:34.479 --> 01:03:35.145 do this problem? 01:03:42.400 --> 01:03:46.110 Maybe an ancillary question is, how many ways can you 01:03:46.110 --> 01:03:48.110 think of that you could do this problem? 01:03:48.110 --> 01:03:50.340 How many different approaches could you use? 01:04:15.510 --> 01:04:22.585 So, how many ways can you think of doing this problem, Kristen? 01:04:27.660 --> 01:04:29.010 I'm just picking on her, but-- 01:04:29.010 --> 01:04:30.093 AUDIENCE: I'm not Kristen. 01:04:32.911 --> 01:04:34.410 PROFESSOR: Give me an answer anyway. 01:04:34.410 --> 01:04:34.951 AUDIENCE: OK. 01:04:38.258 --> 01:04:40.522 I think you could do two, but I would choose Lagrange. 01:04:40.522 --> 01:04:42.480 PROFESSOR: You would want to do it by Lagrange. 01:04:42.480 --> 01:04:44.625 Helen, how else could you do it? 01:04:44.625 --> 01:04:48.270 If you did it by Lagrange, got the answer, and said, 01:04:48.270 --> 01:04:50.635 I want to check it, what would you do next? 01:04:50.635 --> 01:04:52.610 AUDIENCE: The direct method. 01:04:52.610 --> 01:04:53.360 PROFESSOR: Direct. 01:04:53.360 --> 01:04:55.990 But where would do apply? 01:04:55.990 --> 01:04:57.220 Would you use torques? 01:04:57.220 --> 01:04:58.710 Would you use forces? 01:04:58.710 --> 01:05:00.170 How would you go about doing this? 01:05:00.170 --> 01:05:03.800 AUDIENCE: I would use torques about the center of the pulley. 01:05:03.800 --> 01:05:05.300 PROFESSOR: Torques about the center. 01:05:05.300 --> 01:05:06.120 We did that. 01:05:06.120 --> 01:05:07.620 I mean, I worked that-- I actually 01:05:07.620 --> 01:05:11.830 did this problem by that method earlier in the term. 01:05:11.830 --> 01:05:13.980 Torques about A. Let's have the pivot. 01:05:13.980 --> 01:05:14.730 Works fine. 01:05:14.730 --> 01:05:17.950 Actually, it's pretty efficient. 01:05:17.950 --> 01:05:22.380 So, Lagrange or torques about A. You only 01:05:22.380 --> 01:05:24.260 need-- there's one degree of freedom, right? 01:05:24.260 --> 01:05:28.920 How many equations do you need to write to do this problem? 01:05:28.920 --> 01:05:29.420 Just one. 01:05:29.420 --> 01:05:32.870 So the sum of the torques about A will give you the answer. 01:05:32.870 --> 01:05:36.330 Lagrange equation will give you the answer. 01:05:36.330 --> 01:05:40.425 That's a good problem to practice on. 01:05:40.425 --> 01:05:46.815 Another problem-- this is not very big. 01:05:49.810 --> 01:05:51.420 It's basically this problem. 01:05:51.420 --> 01:05:52.891 This is just-- you've seen this. 01:05:52.891 --> 01:05:55.390 I'm sure you've been shown this in physics and stuff before. 01:05:55.390 --> 01:05:59.170 This is the falling stick problem. 01:05:59.170 --> 01:06:02.620 You can't set this up without having some friction. 01:06:02.620 --> 01:06:05.150 So there's definitely friction on the table. 01:06:05.150 --> 01:06:07.930 But until it hits, it's doing some things. 01:06:07.930 --> 01:06:10.020 And so, there's two problems that you 01:06:10.020 --> 01:06:12.170 could set up and try to do. 01:06:12.170 --> 01:06:19.070 One is the problem with no friction-- frictionless table. 01:06:19.070 --> 01:06:22.940 And then, you could allow friction. 01:06:22.940 --> 01:06:24.990 Tricky thing about allowing friction 01:06:24.990 --> 01:06:28.130 is-- you think there's a normal force. 01:06:28.130 --> 01:06:30.920 So let's say our usual friction model is-- the friction force 01:06:30.920 --> 01:06:33.940 is mu times the normal force. 01:06:33.940 --> 01:06:38.687 Does the normal force change with time in this problem? 01:06:43.360 --> 01:06:46.000 So, standing up, what's the normal force? 01:06:48.640 --> 01:06:51.060 Mg, right? 01:06:51.060 --> 01:06:54.010 And the sum of the forces in the vertical direction 01:06:54.010 --> 01:06:58.700 is mass times acceleration in the vertical direction. 01:06:58.700 --> 01:07:00.870 And what are the-- if you draw a free body 01:07:00.870 --> 01:07:03.495 diagram of this problem-- so let's draw it now. 01:07:03.495 --> 01:07:08.070 Here's my-- if there's no friction force, 01:07:08.070 --> 01:07:09.445 I still-- without friction force, 01:07:09.445 --> 01:07:12.260 I still have a normal force, and I still 01:07:12.260 --> 01:07:15.350 have Mg pulling down here. 01:07:15.350 --> 01:07:19.180 So the sum of the forces in the y 01:07:19.180 --> 01:07:26.160 certainly have a minus Mg plus N equals 01:07:26.160 --> 01:07:31.440 the mass times the acceleration in the y direction. 01:07:31.440 --> 01:07:44.020 And if you solve for N, do you-- then the issue-- 01:07:44.020 --> 01:07:47.110 my question is, does this remain constant? 01:07:47.110 --> 01:07:51.930 Depends on whether or not ay remains constant, right? 01:07:51.930 --> 01:07:55.310 Do you think the acceleration in the y direction of this thing 01:07:55.310 --> 01:08:00.670 will change as it goes more and more horizontal? 01:08:00.670 --> 01:08:02.970 There's some nods, up and down, and left and rights. 01:08:02.970 --> 01:08:04.178 I think it's going to change. 01:08:04.178 --> 01:08:07.840 For sure, you can't assume that it won't change. 01:08:07.840 --> 01:08:09.530 So you have to assume it'll change. 01:08:09.530 --> 01:08:11.820 And that means N-- this normal force becomes 01:08:11.820 --> 01:08:15.400 a function of time, which makes certain ways of doing 01:08:15.400 --> 01:08:18.410 this problem a little harder. 01:08:18.410 --> 01:08:22.080 So how many ways? 01:08:22.080 --> 01:08:38.310 Let's say the friction-- let's do a, no friction, and b, 01:08:38.310 --> 01:08:39.729 with friction. 01:08:39.729 --> 01:08:44.630 No friction-- first of all, is it a planar motion problem? 01:08:44.630 --> 01:08:47.029 Yeah, we can do that. 01:08:47.029 --> 01:08:48.729 So, at most three degrees of freedom. 01:08:48.729 --> 01:08:51.939 How many does this one have? 01:08:51.939 --> 01:08:53.229 Work this one out. 01:08:53.229 --> 01:08:55.970 Figure out how many degrees of freedom this problem has. 01:08:55.970 --> 01:08:58.712 How many separate equations do you need to come up with? 01:09:42.380 --> 01:09:44.600 So, you decide what your-- let's say 01:09:44.600 --> 01:09:46.120 we're going to use Lagrange. 01:09:46.120 --> 01:09:48.200 What would your generalized coordinates 01:09:48.200 --> 01:09:49.425 be to do this problem? 01:10:21.140 --> 01:10:24.930 So, problem a-- no friction. 01:10:24.930 --> 01:10:26.914 How many generalized coordinates? 01:10:26.914 --> 01:10:28.080 How many degrees of freedom? 01:10:28.080 --> 01:10:29.955 How many generalized coordinates do you need? 01:10:33.390 --> 01:10:34.275 I heard a three. 01:10:36.800 --> 01:10:38.388 I hear one. 01:10:38.388 --> 01:10:39.730 Somebody give me two. 01:10:39.730 --> 01:10:41.340 I got a two. 01:10:41.340 --> 01:10:43.250 All right. 01:10:43.250 --> 01:10:45.560 Obviously a good question. 01:10:45.560 --> 01:10:48.040 So, at most there can be three, because we've 01:10:48.040 --> 01:10:50.090 agreed that it's planar motion. 01:10:50.090 --> 01:10:51.830 Does this problem have any constraints? 01:10:55.730 --> 01:10:56.230 Where? 01:10:59.513 --> 01:11:03.130 AUDIENCE: The bottom of the line can't move the y. 01:11:03.130 --> 01:11:06.080 PROFESSOR: So it can't move in the y direction. 01:11:06.080 --> 01:11:10.860 So if any part of a body can't move in-- translate, 01:11:10.860 --> 01:11:13.840 then there's no translation in that direction, 01:11:13.840 --> 01:11:17.660 pure translation of the body in that direction. 01:11:17.660 --> 01:11:21.290 So there's a constraint in the y, 01:11:21.290 --> 01:11:23.550 if we draw a coordinate system here. 01:11:28.250 --> 01:11:29.440 Constraint in the y. 01:11:29.440 --> 01:11:30.040 True. 01:11:30.040 --> 01:11:32.260 So we're down to two. 01:11:32.260 --> 01:11:34.877 Are there any other constraints? 01:11:34.877 --> 01:11:38.162 AUDIENCE: I would say that, yes, because the upper line-- 01:11:38.162 --> 01:11:40.370 PROFESSOR: No, no, it's not leaning against the wall. 01:11:40.370 --> 01:11:42.585 This is just a skip. 01:11:42.585 --> 01:11:43.450 This is the problem. 01:11:50.974 --> 01:11:52.890 So it definitely can't move through the table. 01:11:52.890 --> 01:11:54.500 So it's constrained in y. 01:11:54.500 --> 01:11:57.215 We've got that so now that leaves us, at most, two. 01:12:00.737 --> 01:12:02.070 Are there any other constraints? 01:12:07.110 --> 01:12:09.500 How many say no? 01:12:09.500 --> 01:12:11.169 How many say yes? 01:12:11.169 --> 01:12:13.210 If you say yes, you've got to tell me what it is, 01:12:13.210 --> 01:12:14.740 but I don't see any others. 01:12:14.740 --> 01:12:15.910 So we're left with two. 01:12:15.910 --> 01:12:19.370 We need two coordinates. 01:12:19.370 --> 01:12:20.500 What might we pick here? 01:12:27.030 --> 01:12:28.500 What would you pick? 01:12:28.500 --> 01:12:30.620 You're now confronted with this problem. 01:12:30.620 --> 01:12:32.036 Do you have rotational-- if you're 01:12:32.036 --> 01:12:34.660 going to use Lagrange do you have rotational kinetic energy 01:12:34.660 --> 01:12:36.500 in this problem? 01:12:36.500 --> 01:12:39.509 You're probably going to need an angle. 01:12:39.509 --> 01:12:40.550 What else would you need? 01:12:40.550 --> 01:12:44.454 So we're going to need an angle for sure. 01:12:44.454 --> 01:12:44.995 Say it again? 01:12:44.995 --> 01:12:47.120 AUDIENCE: The height of the center of mass. 01:12:47.120 --> 01:12:48.150 PROFESSOR: You're going to need the height 01:12:48.150 --> 01:12:49.165 of the center of mass. 01:12:49.165 --> 01:12:51.540 Yeah, you're going to need a potential energy expression, 01:12:51.540 --> 01:12:55.590 but we've decided that y is constrained. 01:12:55.590 --> 01:12:58.410 So you can't have a theta and a y. 01:12:58.410 --> 01:12:59.642 Let's make this our theta. 01:13:03.370 --> 01:13:06.470 What else is there? 01:13:06.470 --> 01:13:06.970 AUDIENCE: x. 01:13:06.970 --> 01:13:07.750 PROFESSOR: x. 01:13:07.750 --> 01:13:09.480 You're going to need an x. 01:13:09.480 --> 01:13:11.320 So our generalized coordinates are 01:13:11.320 --> 01:13:16.740 going to be an x and a theta. 01:13:16.740 --> 01:13:19.370 But you do need to be able to write down a potential energy 01:13:19.370 --> 01:13:22.220 expression, and it involves motion in the y. 01:13:22.220 --> 01:13:23.220 So what do you do? 01:13:26.771 --> 01:13:27.270 Right. 01:13:27.270 --> 01:13:29.670 So what's the height of this thing 01:13:29.670 --> 01:13:37.755 is-- above the ground-- is sum L over 2 cosine theta equals y. 01:13:37.755 --> 01:13:38.880 Something like that, right? 01:13:38.880 --> 01:13:41.260 And I might have a-- depending on whether you make-- 01:13:41.260 --> 01:13:42.630 I haven't thought this through. 01:13:42.630 --> 01:13:45.180 Whether or not it's xy this way or xy that way, 01:13:45.180 --> 01:13:47.490 theta might be plus or minus. 01:13:47.490 --> 01:13:50.760 This could be-- there might be a sine in there. 01:13:50.760 --> 01:13:56.690 But basically, this is-- theta and y are constrained. 01:13:56.690 --> 01:13:58.554 If you know one, you know the other. 01:13:58.554 --> 01:14:00.720 So you do your potential energy expression this way. 01:14:05.680 --> 01:14:07.130 Now we've gotten that far. 01:14:07.130 --> 01:14:09.650 We know we need two coordinates. 01:14:09.650 --> 01:14:12.200 We're going to use x and theta. 01:14:12.200 --> 01:14:14.750 But now you have to decide what method to use. 01:14:14.750 --> 01:14:18.380 Direct method using one of these, or Lagrange? 01:14:18.380 --> 01:14:20.750 What would you do? 01:14:20.750 --> 01:14:21.833 How would you go about it? 01:14:21.833 --> 01:14:22.340 It's a quiz. 01:14:22.340 --> 01:14:26.580 You got 20 minutes to finish, and you 01:14:26.580 --> 01:14:30.153 want to do this in the safest, quickest possible way. 01:14:34.310 --> 01:14:36.270 What would you try? 01:14:36.270 --> 01:14:39.640 And this is the realistic situation, right? 01:14:39.640 --> 01:14:41.680 Next Tuesday. 01:14:41.680 --> 01:14:44.251 What do you trust yourself to do the most? 01:14:44.251 --> 01:14:46.542 It's probably what you ought to do in a quiz situation. 01:14:51.800 --> 01:14:53.800 Anne, how many ways can you think of doing this? 01:14:53.800 --> 01:14:55.657 How many ways could you do this, Rob? 01:14:55.657 --> 01:14:56.240 AUDIENCE: Two. 01:14:56.240 --> 01:14:57.010 PROFESSOR: Two. 01:14:57.010 --> 01:14:58.120 Any way. 01:14:58.120 --> 01:15:00.070 Direct, indirect, what would you use? 01:15:00.070 --> 01:15:02.210 Where would you choose to-- you're 01:15:02.210 --> 01:15:04.570 going to need a torque equation to do this problem, 01:15:04.570 --> 01:15:05.484 if you use direct. 01:15:05.484 --> 01:15:07.150 Where would you take your torques about? 01:15:12.790 --> 01:15:13.784 Think about that. 01:15:13.784 --> 01:15:15.950 For this problem, if you're doing the direct method, 01:15:15.950 --> 01:15:17.390 where are you going to compute the torques? 01:15:17.390 --> 01:15:18.170 About what point? 01:15:23.200 --> 01:15:24.700 So you can do it about that-- you 01:15:24.700 --> 01:15:30.630 could call this point A here. 01:15:30.630 --> 01:15:32.950 And what equation would you-- now 01:15:32.950 --> 01:15:34.790 you've got to use that equation that's got 01:15:34.790 --> 01:15:37.250 the problematic terms in it, right? 01:15:37.250 --> 01:15:38.380 But it's all right. 01:15:38.380 --> 01:15:40.060 You can punch those through and do it. 01:15:40.060 --> 01:15:41.490 That will work. 01:15:41.490 --> 01:15:43.692 And if you use Lagrange? 01:15:43.692 --> 01:15:45.150 You've got to be able to figure out 01:15:45.150 --> 01:15:47.630 the potential and kinetic energy, and so forth. 01:15:56.710 --> 01:16:01.109 So could all of you do this by Lagrange? 01:16:01.109 --> 01:16:02.650 This is a good one to go practice on. 01:16:02.650 --> 01:16:03.661 It's not that hard. 01:16:03.661 --> 01:16:04.160 Do it. 01:16:04.160 --> 01:16:07.860 It's a good practice problem. 01:16:07.860 --> 01:16:11.100 Now add friction. 01:16:11.100 --> 01:16:12.070 So, the b problem. 01:16:12.070 --> 01:16:13.105 It now has friction. 01:16:17.900 --> 01:16:26.334 Can you-- I don't think-- is Lagrange a good choice 01:16:26.334 --> 01:16:27.292 if it now has friction? 01:16:33.130 --> 01:16:37.620 I think it's got a problem, and I'd be careful using Lagrange 01:16:37.620 --> 01:16:39.540 if it had friction. 01:16:39.540 --> 01:16:42.550 Not that-- Lagrange is perfectly-- 01:16:42.550 --> 01:16:43.860 it is certainly valid. 01:16:43.860 --> 01:16:46.430 It's just hard. 01:16:46.430 --> 01:16:47.255 Why is it hard? 01:16:53.040 --> 01:16:55.870 Well, yeah, it might be hard to find. 01:16:55.870 --> 01:16:57.610 What do you need to know to-- so, 01:16:57.610 --> 01:17:00.850 are there any external non-conservative forces 01:17:00.850 --> 01:17:02.550 in the problem with friction? 01:17:02.550 --> 01:17:03.480 AUDIENCE: Friction. 01:17:03.480 --> 01:17:06.010 PROFESSOR: Friction. 01:17:06.010 --> 01:17:09.169 You're going to have to figure out the friction force. 01:17:09.169 --> 01:17:10.710 And to figure out the friction force, 01:17:10.710 --> 01:17:14.900 you're going to have to apply direct method. 01:17:14.900 --> 01:17:16.410 No other way. 01:17:16.410 --> 01:17:19.050 You are going to have to apply some direct method to do 01:17:19.050 --> 01:17:22.530 this problem, no matter what. 01:17:22.530 --> 01:17:26.750 So you can't just say-- Lagrange is not 01:17:26.750 --> 01:17:29.060 going to bail you out and not have 01:17:29.060 --> 01:17:32.400 to solve some of the F equals ma, 01:17:32.400 --> 01:17:33.920 and torque, and those sort of things 01:17:33.920 --> 01:17:36.820 to figure out what friction is. 01:17:36.820 --> 01:17:41.620 So there are problems that Lagrange-- isn't all as simple 01:17:41.620 --> 01:17:44.300 as we sometimes make it out to be. 01:17:44.300 --> 01:17:46.050 All right, I've run a couple minutes over. 01:17:46.050 --> 01:17:46.570 Thank you. 01:17:46.570 --> 01:17:48.820 We'll see you in class on Tuesday. 01:17:48.820 --> 01:17:51.030 We'll do some more review.