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 --> 00:00:06.070 Your support will help MIT OpenCourseWare 00:00:06.070 --> 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 --> 00:00:17.276 at ocw.mit.edu. 00:00:20.757 --> 00:00:21.340 PROFESSOR: OK. 00:00:21.340 --> 00:00:22.465 We're going to get started. 00:00:22.465 --> 00:00:26.016 The homework-- does everybody have a copy of the handout? 00:00:26.016 --> 00:00:27.900 If not, there's some on the steps there. 00:00:27.900 --> 00:00:29.610 There's Muddy Cards on the steps. 00:00:29.610 --> 00:00:32.049 And we're going to do three things today. 00:00:32.049 --> 00:00:33.500 That's this complex problem I want 00:00:33.500 --> 00:00:36.432 to talk about, a problem on center of percussion. 00:00:36.432 --> 00:00:37.890 And then as we have time, I'm going 00:00:37.890 --> 00:00:39.740 to summarize some summarising statements 00:00:39.740 --> 00:00:44.590 about imbalances, which we've talked a lot about off and on. 00:00:44.590 --> 00:00:49.310 So I want you restrain looking at the notes I've handed out. 00:00:49.310 --> 00:00:53.141 The notes I've handed out are this complex problem. 00:00:53.141 --> 00:00:54.890 And I'll let you look at them in a minute. 00:00:54.890 --> 00:00:57.348 But I want to get you to think about some things before you 00:00:57.348 --> 00:00:58.412 see what's on the notes. 00:00:58.412 --> 00:00:59.995 But the notes are intended so that you 00:00:59.995 --> 00:01:02.650 don't have to spend a lot of time writing down messy stuff. 00:01:02.650 --> 00:01:04.720 You can think and listen. 00:01:04.720 --> 00:01:07.650 And they'll also be sent out on the web, 00:01:07.650 --> 00:01:11.120 posted, so you don't have to grab a copy for your friends. 00:01:11.120 --> 00:01:14.690 Because I only made 100 copies and there are 122 of you. 00:01:14.690 --> 00:01:15.755 OK. 00:01:15.755 --> 00:01:17.130 So let's start with this problem. 00:01:17.130 --> 00:01:25.490 This is basically a complex system, a mass, a pendulum. 00:01:25.490 --> 00:01:29.940 And when you-- I've made up one for you. 00:01:29.940 --> 00:01:32.110 So here's the pendulum. 00:01:32.110 --> 00:01:33.930 It's on an axle stuck into my cart. 00:01:33.930 --> 00:01:36.470 The cart's got springs connecting to it. 00:01:36.470 --> 00:01:38.980 And it naturally has some damping. 00:01:38.980 --> 00:01:43.910 So this is a realization of what's been drawn there. 00:01:43.910 --> 00:01:45.850 So let's just give it a little bump. 00:01:49.670 --> 00:01:51.670 Cart moves back and forth. 00:01:51.670 --> 00:01:54.020 Pendulum swings back and forth. 00:01:54.020 --> 00:02:00.050 If I start it this way, it'll act a lot more crazy. 00:02:00.050 --> 00:02:02.210 It'll have a more chaotic looking kind of motion. 00:02:02.210 --> 00:02:04.150 The reason for that has to do with, this 00:02:04.150 --> 00:02:06.440 is a multiple degree of freedom system, has 00:02:06.440 --> 00:02:10.500 more than one natural frequency, has more than one frequency 00:02:10.500 --> 00:02:12.330 that responds, all mixed together. 00:02:12.330 --> 00:02:15.300 And that's why it'll do kind of crazy things. 00:02:15.300 --> 00:02:18.060 Like that'll stop, almost stop, and then start up again, 00:02:18.060 --> 00:02:19.150 stuff like that. 00:02:19.150 --> 00:02:22.620 All completely natural. 00:02:22.620 --> 00:02:24.490 But if I give it a nice, gentle start, 00:02:24.490 --> 00:02:26.290 it actually mostly vibrates in what 00:02:26.290 --> 00:02:29.701 I call one mode of vibration. 00:02:29.701 --> 00:02:30.200 OK? 00:02:30.200 --> 00:02:33.115 So we want to get the equations of motion of this system. 00:02:35.650 --> 00:02:40.290 So here, I've drawn it. 00:02:40.290 --> 00:02:42.350 And the first question about a system 00:02:42.350 --> 00:02:43.980 like this when you go to analyze it 00:02:43.980 --> 00:02:46.473 is, how many degrees of freedom does it have. 00:02:46.473 --> 00:02:47.800 All right? 00:02:47.800 --> 00:02:54.590 So I'm going to claim this to be a planar motion problem. 00:02:54.590 --> 00:02:58.980 It's confined to the board and confined to rotation 00:02:58.980 --> 00:03:01.010 perpendicular to the board. 00:03:01.010 --> 00:03:01.810 OK? 00:03:01.810 --> 00:03:08.660 Any time that happens, each object has, 00:03:08.660 --> 00:03:13.640 at most, three degrees of freedom, x y, and a rotation. 00:03:13.640 --> 00:03:18.150 So we said no rotations around x or around y. 00:03:18.150 --> 00:03:21.520 So three possible ones for this mass, three possible ones 00:03:21.520 --> 00:03:23.620 for this mass. 00:03:23.620 --> 00:03:30.490 And then we start looking for the constraints in addition. 00:03:30.490 --> 00:03:32.090 Well, this one is constrained. 00:03:32.090 --> 00:03:34.739 It can't move out of the-- let's see, 00:03:34.739 --> 00:03:36.030 we've already constrained that. 00:03:36.030 --> 00:03:36.613 Wait a second. 00:03:39.228 --> 00:03:40.990 How do I want to say this? 00:03:40.990 --> 00:03:44.930 This one is certainly constrained in the y because 00:03:44.930 --> 00:03:46.960 of the track here. 00:03:46.960 --> 00:03:48.390 So that's one constraint. 00:03:48.390 --> 00:03:50.390 We're starting off with six possible. 00:03:50.390 --> 00:03:55.000 We've got one because it's constrained in the y. 00:03:55.000 --> 00:03:57.120 But it can certainly move in the x. 00:03:57.120 --> 00:04:00.050 Can it rotate in the z, this big mass? 00:04:00.050 --> 00:04:01.190 So that's two. 00:04:01.190 --> 00:04:03.890 So out of the three possible, the top one only 00:04:03.890 --> 00:04:07.460 has one possible degree of freedom, x. 00:04:07.460 --> 00:04:11.890 This one also has three possible degrees of freedom. 00:04:11.890 --> 00:04:18.643 It's pinned at A. What does that do for constraint? 00:04:22.270 --> 00:04:24.850 So this one, take 30 seconds and talk to your neighbor. 00:04:24.850 --> 00:04:27.750 How many constraints are caused by the pin at A? 00:05:12.660 --> 00:05:14.740 All right. 00:05:14.740 --> 00:05:19.420 So at least-- how many believe that there's one constraint 00:05:19.420 --> 00:05:20.515 provided by the pin at A? 00:05:20.515 --> 00:05:22.390 Let's see the hands that believe we have one. 00:05:22.390 --> 00:05:24.560 Get them up high, up high now. 00:05:24.560 --> 00:05:26.311 How about two? 00:05:26.311 --> 00:05:26.810 OK. 00:05:26.810 --> 00:05:29.101 How about three? 00:05:29.101 --> 00:05:29.600 All right. 00:05:29.600 --> 00:05:31.460 A little uncertainty here. 00:05:31.460 --> 00:05:33.110 All right. 00:05:33.110 --> 00:05:36.590 Do you know the motion at A? 00:05:36.590 --> 00:05:40.260 Have you prescribed the motion at A in any way? 00:05:40.260 --> 00:05:40.940 In what way? 00:05:40.940 --> 00:05:42.566 I see you nodding your head. 00:05:42.566 --> 00:05:44.483 AUDIENCE: It has to move with the other block. 00:05:44.483 --> 00:05:46.190 PROFESSOR: It has to move with the block. 00:05:46.190 --> 00:05:48.550 But we've already given the other block a coordinate. 00:05:48.550 --> 00:05:49.050 What is it? 00:05:49.050 --> 00:05:50.260 AUDIENCE: In x. 00:05:50.260 --> 00:05:53.447 PROFESSOR: So if you know x, do you know the motion of A? 00:05:53.447 --> 00:05:53.946 OK. 00:05:53.946 --> 00:05:57.650 So you have established the motion of A. 00:05:57.650 --> 00:06:01.280 So if you know A-- you don't need any additional information 00:06:01.280 --> 00:06:02.400 about A do you? 00:06:02.400 --> 00:06:04.750 You've already chosen a coordinate for it. 00:06:04.750 --> 00:06:08.540 So if A is prescribed, that means 00:06:08.540 --> 00:06:14.700 x1 and y1, the motion-- the coordinates 00:06:14.700 --> 00:06:17.050 describing this thing-- and I've given 00:06:17.050 --> 00:06:19.443 started at a center of mass, a little x1 that way, 00:06:19.443 --> 00:06:20.630 and a little y1 that way. 00:06:20.630 --> 00:06:23.916 Those are my possible displacement coordinates. 00:06:23.916 --> 00:06:25.290 And it has a rotation coordinate. 00:06:25.290 --> 00:06:27.410 Three possible degrees of freedom, right? 00:06:27.410 --> 00:06:31.410 x and y are prescribed at this point. 00:06:31.410 --> 00:06:35.280 If you prescribe the motion-- fix the motion-- 00:06:35.280 --> 00:06:38.300 at any point on a rigid body, what 00:06:38.300 --> 00:06:41.810 does that say about translational motion 00:06:41.810 --> 00:06:45.715 any place else on the rigid body? 00:06:45.715 --> 00:06:48.510 Remember, this is back to that subtle definition 00:06:48.510 --> 00:06:51.890 of what we mean by translation, what we mean by rotation. 00:06:51.890 --> 00:06:53.830 AUDIENCE: It's a parallel motion. 00:06:53.830 --> 00:06:55.425 PROFESSOR: Parallel motion. 00:06:55.425 --> 00:06:59.510 The translation part of this-- every point on the object 00:06:59.510 --> 00:07:02.300 moves parallel to every other point. 00:07:02.300 --> 00:07:06.840 So if you've prescribed the translation of any one point 00:07:06.840 --> 00:07:09.370 you have prescribed the translation for all. 00:07:09.370 --> 00:07:12.180 So this second body, basically, its only translation 00:07:12.180 --> 00:07:15.820 that it can have is x of the main body. 00:07:15.820 --> 00:07:18.270 So you have confined it in x and y. 00:07:18.270 --> 00:07:19.890 It has no x and y possibilities. 00:07:19.890 --> 00:07:24.940 Those are two constraints leaving you with-- one. 00:07:24.940 --> 00:07:25.440 Right? 00:07:25.440 --> 00:07:26.890 One degree of freedom. 00:07:26.890 --> 00:07:28.530 And then we pick a coordinate for it. 00:07:28.530 --> 00:07:32.960 And the natural one to use for that is the angle here. 00:07:32.960 --> 00:07:34.350 So we need two. 00:07:34.350 --> 00:07:39.200 Two coordinates completely describe the motion. 00:07:39.200 --> 00:07:41.490 It will yield two equations of motion. 00:07:44.232 --> 00:07:45.150 OK? 00:07:45.150 --> 00:07:49.420 So on quizzes, you say, how many independent degrees of freedom 00:07:49.420 --> 00:07:50.050 are there. 00:07:50.050 --> 00:07:52.680 That is the same question as saying, 00:07:52.680 --> 00:07:57.730 how many equations of motion are required to completely describe 00:07:57.730 --> 00:07:59.100 this system. 00:07:59.100 --> 00:08:01.120 Or how many independent coordinates 00:08:01.120 --> 00:08:03.630 are required to completely describe the motion of system? 00:08:03.630 --> 00:08:05.254 They're, all three, the same questions. 00:08:05.254 --> 00:08:07.860 And that threw a couple people on the last quiz. 00:08:07.860 --> 00:08:08.470 OK. 00:08:08.470 --> 00:08:11.030 Let's move on to free body diagrams. 00:08:11.030 --> 00:08:14.124 So now we know we've got two coordinates, theta and x. 00:08:14.124 --> 00:08:15.665 And now you're free to look at your-- 00:08:15.665 --> 00:08:17.122 you can be free to look at things. 00:08:17.122 --> 00:08:18.830 And I want you to spend most of your time 00:08:18.830 --> 00:08:21.690 thinking and listening and not having to write. 00:08:21.690 --> 00:08:24.260 But make notes as you get insights about things. 00:08:24.260 --> 00:08:26.770 Free body diagram then, two of them. 00:08:26.770 --> 00:08:31.530 The pendulum piece is actually pretty simple. 00:08:31.530 --> 00:08:33.610 You've got Mg down. 00:08:33.610 --> 00:08:37.000 And you have two possible forces at this point of rotation. 00:08:37.000 --> 00:08:38.960 And I've just named them F1 and F2. 00:08:38.960 --> 00:08:41.299 And I've drawn them-- not arbitrarily, 00:08:41.299 --> 00:08:43.659 but I've picked the direction to draw them in. 00:08:43.659 --> 00:08:47.670 I don't know what direction they're in. 00:08:47.670 --> 00:08:50.040 And that's my complete free body diagram. 00:08:50.040 --> 00:08:52.110 If I've missed anything, tell me. 00:08:52.110 --> 00:08:55.501 Or if you have any questions about it, ask me. 00:08:55.501 --> 00:08:56.000 OK. 00:08:56.000 --> 00:08:59.530 The free body diagram for this guy-- this 00:08:59.530 --> 00:09:03.310 is capital M. This one's M1. 00:09:03.310 --> 00:09:06.500 This is mass M2 just to keep it straight. 00:09:06.500 --> 00:09:08.470 Lots of possible forces in this thing. 00:09:13.070 --> 00:09:14.730 Reaction forces through the wheels, 00:09:14.730 --> 00:09:16.390 there are only vertical. 00:09:16.390 --> 00:09:18.385 I've left out friction, ignored friction. 00:09:21.100 --> 00:09:23.760 But also, these F1 and F2 act at to pin. 00:09:23.760 --> 00:09:25.680 And notice I've drawn them exactly equal 00:09:25.680 --> 00:09:27.290 and opposite to these. 00:09:27.290 --> 00:09:28.610 Kind of a key thing to do. 00:09:28.610 --> 00:09:29.680 And why do we do that? 00:09:32.770 --> 00:09:35.655 There's a law. 00:09:35.655 --> 00:09:37.475 What's the law? 00:09:37.475 --> 00:09:38.820 Newton's third. 00:09:38.820 --> 00:09:41.491 You've bot to do that or else it won't work. 00:09:41.491 --> 00:09:41.990 OK? 00:09:41.990 --> 00:09:44.980 These are common unknowns. 00:09:44.980 --> 00:09:47.330 But they're equal and opposite at this point. 00:09:47.330 --> 00:09:49.930 And we have an M1g hanging down. 00:09:49.930 --> 00:09:51.560 The spring force resists. 00:09:51.560 --> 00:09:54.680 Any positive motion, a spring pulls back. 00:09:54.680 --> 00:09:58.690 Any positive velocity, the damper holds back. 00:09:58.690 --> 00:10:01.784 And that's all of the forces on this thing. 00:10:01.784 --> 00:10:03.700 It's going to be necessary to be able to break 00:10:03.700 --> 00:10:04.600 these things down. 00:10:04.600 --> 00:10:08.080 Because I'm going to sum things. 00:10:08.080 --> 00:10:12.580 Going to need to have bits and pieces of F2 and F1, 00:10:12.580 --> 00:10:14.410 so cosine thetas and sine thetas. 00:10:21.490 --> 00:10:26.160 Now resist looking at your paper for a second. 00:10:26.160 --> 00:10:29.740 Next concept question. 00:10:29.740 --> 00:10:31.220 We talked a lot in the last couple 00:10:31.220 --> 00:10:36.010 of the lectures about the best approach to do problems, 00:10:36.010 --> 00:10:37.965 right, especially using angular momentum. 00:10:37.965 --> 00:10:39.881 Do you think you're going to need some angular 00:10:39.881 --> 00:10:42.240 momentum to solve this problem? 00:10:42.240 --> 00:10:45.622 An approach using torques and angular momentum? 00:10:45.622 --> 00:10:46.330 More than likely. 00:10:46.330 --> 00:10:49.560 Anytime things are rotating, more than likely. 00:10:49.560 --> 00:10:53.550 So what's the best approach here? 00:10:53.550 --> 00:10:56.490 Are we going to compute angular momentum with respect 00:10:56.490 --> 00:10:58.890 to A, that pivot point? 00:10:58.890 --> 00:11:01.295 With respect to g, the center of mass? 00:11:03.900 --> 00:11:05.940 And it's not too obvious. 00:11:05.940 --> 00:11:09.430 So if you were starting out this problem, 00:11:09.430 --> 00:11:10.540 how would you begin it? 00:11:10.540 --> 00:11:15.070 Would you decide, I'm going to sum my torques around point A 00:11:15.070 --> 00:11:17.810 or, I'm going to sum my torques around g? 00:11:17.810 --> 00:11:18.700 Did I mark g? 00:11:18.700 --> 00:11:21.790 Yeah, it's right at the center. 00:11:21.790 --> 00:11:22.850 All right. 00:11:22.850 --> 00:11:24.580 Think about that for a second. 00:11:24.580 --> 00:11:26.352 Got a question? 00:11:26.352 --> 00:11:27.234 All right. 00:11:27.234 --> 00:11:29.450 I'm going to ask you this. 00:11:29.450 --> 00:11:30.970 Any questions about the question? 00:11:30.970 --> 00:11:33.030 I want a real show of hands here. 00:11:33.030 --> 00:11:35.505 I want you to just-- what would you do to start with? 00:11:35.505 --> 00:11:39.710 So how many would sum torques, compute angular momentum 00:11:39.710 --> 00:11:43.490 with respect to point A? 00:11:43.490 --> 00:11:44.080 OK. 00:11:44.080 --> 00:11:47.390 How about g? 00:11:47.390 --> 00:11:48.510 Hmm, interesting. 00:11:48.510 --> 00:11:50.782 And another way? 00:11:50.782 --> 00:11:52.240 Some of you are holding back on me. 00:11:52.240 --> 00:11:54.130 Not everybody raised their hands here. 00:11:54.130 --> 00:11:55.040 OK. 00:11:55.040 --> 00:11:55.540 All right. 00:11:55.540 --> 00:11:57.920 Most people would do it around A. 00:11:57.920 --> 00:11:58.700 A would work. 00:12:02.070 --> 00:12:03.470 g would work too. 00:12:03.470 --> 00:12:06.030 And in fact, when I sat down to do this problem, 00:12:06.030 --> 00:12:08.310 I did it with A first. 00:12:08.310 --> 00:12:09.850 And then I went and did it with g. 00:12:09.850 --> 00:12:11.945 And it turns out that doing it with respect to g 00:12:11.945 --> 00:12:13.230 is just slightly easier. 00:12:18.910 --> 00:12:19.410 OK. 00:12:19.410 --> 00:12:23.720 So the approach we need to find two equations of motion. 00:12:23.720 --> 00:12:25.090 We have two bodies. 00:12:25.090 --> 00:12:28.560 We're going to use Newton's and Euler's laws to go after them. 00:12:28.560 --> 00:12:32.840 So starting out, first one, sum of the forces 00:12:32.840 --> 00:12:35.625 in the x direction on this body. 00:12:38.470 --> 00:12:39.850 There's a lot of unknowns. 00:12:39.850 --> 00:12:41.610 So I'm going to end up-- or start off-- 00:12:41.610 --> 00:12:43.640 with more than two equations. 00:12:43.640 --> 00:12:45.350 Because I've got how many unknowns? 00:12:45.350 --> 00:12:53.850 One, two, three, four, x, the motion x, five, 00:12:53.850 --> 00:12:56.170 and the angle theta, six. 00:12:56.170 --> 00:13:01.070 I could need as many as six equations to start with. 00:13:01.070 --> 00:13:03.670 If I sum forces in the vertical direction I can-- 00:13:03.670 --> 00:13:06.700 it turns out that N1 and N2 two here, I never actually 00:13:06.700 --> 00:13:09.990 have to deal with. 00:13:09.990 --> 00:13:11.250 So don't start there. 00:13:11.250 --> 00:13:12.810 If you're thinking you might not have to mess with them, 00:13:12.810 --> 00:13:13.580 don't start there. 00:13:13.580 --> 00:13:14.940 You're going to waste a lot of time. 00:13:14.940 --> 00:13:16.350 I don't think I'm going to have to deal with them. 00:13:16.350 --> 00:13:18.058 And in fact, I'm only going to need four. 00:13:18.058 --> 00:13:23.516 I'm going to find four equations involving F1, F2, x, and theta. 00:13:23.516 --> 00:13:24.016 OK. 00:13:30.266 --> 00:13:32.140 So here's how I'm saving a little time today. 00:13:32.140 --> 00:13:36.150 I've written it down for you and I've written it out. 00:13:36.150 --> 00:13:37.670 Same thing on your paper. 00:13:37.670 --> 00:13:39.800 Let's talk about the first one. 00:13:39.800 --> 00:13:44.990 This is a sum of forces in this capital X direction, which 00:13:44.990 --> 00:13:51.650 is our inertial frame, X, on the main cart on mass 1. 00:13:51.650 --> 00:13:54.340 So it's got to be the mass times the acceleration. 00:13:54.340 --> 00:13:55.700 And this is not [INAUDIBLE]. 00:13:55.700 --> 00:13:59.590 I don't how that sneaked in there. 00:13:59.590 --> 00:14:02.400 And I've called it-- in capital, I had direction. 00:14:02.400 --> 00:14:03.020 OK? 00:14:03.020 --> 00:14:04.210 Mass times acceleration. 00:14:04.210 --> 00:14:06.540 I just sum up all the forces on the cart. 00:14:06.540 --> 00:14:07.910 Spring force holds back. 00:14:07.910 --> 00:14:12.780 The direction of the arrow on my free body diagram 00:14:12.780 --> 00:14:14.450 tells you the sense of it. 00:14:14.450 --> 00:14:19.290 Minus KX minus bx dot plus F2 cosine theta 00:14:19.290 --> 00:14:23.620 plus F1 sine theta, all in the X I hat direction. 00:14:23.620 --> 00:14:24.550 OK? 00:14:24.550 --> 00:14:29.415 Some of the forces on my little bar here in the x1 direction. 00:14:29.415 --> 00:14:34.180 Now x1 is down here. 00:14:34.180 --> 00:14:36.040 And y1 is off that way. 00:14:38.930 --> 00:14:42.440 So sum of the forces in x1, which is now on mass 2, 00:14:42.440 --> 00:14:45.935 must its mass times its acceleration. 00:14:45.935 --> 00:14:50.834 And I've just called it-- it doesn't matter where I do my-- 00:14:50.834 --> 00:14:52.250 what point I'm going to do angular 00:14:52.250 --> 00:14:53.900 momentum with respect to. 00:14:53.900 --> 00:14:55.330 This is Newton's law. 00:14:55.330 --> 00:14:57.480 And it's the acceleration of the center 00:14:57.480 --> 00:15:00.420 of the gravity with respect to an inertial frame. 00:15:00.420 --> 00:15:02.600 But this is in the little i hat direction. 00:15:02.600 --> 00:15:04.710 So it's the complement in that direction. 00:15:04.710 --> 00:15:08.850 I have a minus F1i in that direction. 00:15:08.850 --> 00:15:15.460 And I have a plus M2g cosine theta i in the i hat direction. 00:15:15.460 --> 00:15:17.970 Pretty straightforward Newton's second law. 00:15:17.970 --> 00:15:20.030 Again, the other component for Newton's 00:15:20.030 --> 00:15:25.180 in the Y1 direction, mass times the acceleration in the j hat 00:15:25.180 --> 00:15:28.900 direction, minus F2j-- because my F2 happens 00:15:28.900 --> 00:15:31.570 to be in the minus direction in the free body 00:15:31.570 --> 00:15:38.020 diagram-- minus M2g sine theta in the j hat. 00:15:38.020 --> 00:15:41.510 And finally, I'm going to sum my moments 00:15:41.510 --> 00:15:45.680 about the center of mass. 00:15:45.680 --> 00:15:49.760 It would also work-- if I did it around A, 00:15:49.760 --> 00:15:54.380 I would-- the reason you normally do it around A, which 00:15:54.380 --> 00:15:56.870 is why I started there, is why? 00:15:56.870 --> 00:15:58.960 What's the advantage of doing it around A? 00:15:58.960 --> 00:16:01.824 The potential advantage of going to A? 00:16:01.824 --> 00:16:04.534 AUDIENCE: So you can eliminate forces at the pin. 00:16:04.534 --> 00:16:05.200 PROFESSOR: Yeah. 00:16:05.200 --> 00:16:08.474 F1 and F2 don't generate moments at the pin. 00:16:08.474 --> 00:16:09.890 And there's a hope that, then, you 00:16:09.890 --> 00:16:11.920 can get what you need to know without ever 00:16:11.920 --> 00:16:13.730 having to evaluate F1 and F2. 00:16:13.730 --> 00:16:14.690 Right? 00:16:14.690 --> 00:16:20.884 And in this case, they're going to pop up in these equations. 00:16:20.884 --> 00:16:23.550 And you're going to have to deal with them anyway, it turns out. 00:16:23.550 --> 00:16:26.120 So it doesn't actually give you much of an advantage. 00:16:26.120 --> 00:16:29.390 So it's easier because the torque equation's 00:16:29.390 --> 00:16:33.610 easier in terms of the number of terms it has. 00:16:33.610 --> 00:16:35.970 That's what makes it slightly easier in this problem. 00:16:35.970 --> 00:16:38.260 You've got to deal with F1 and F2 anyway. 00:16:38.260 --> 00:16:40.670 So it's just the time rate of change of the angular 00:16:40.670 --> 00:16:43.580 momentum with respect to g. 00:16:43.580 --> 00:16:45.010 And that's external torques. 00:16:45.010 --> 00:16:49.480 And the only external torque is caused by F2. 00:16:49.480 --> 00:16:52.200 And I left out a minus sign. 00:16:52.200 --> 00:16:54.945 You have it on your paper, I think, here. 00:16:58.600 --> 00:16:59.610 No, it's correct. 00:16:59.610 --> 00:17:00.790 It's positive. 00:17:00.790 --> 00:17:05.707 So you have, F2 acts on a moment arm about g, L2. 00:17:05.707 --> 00:17:07.540 And it's going to be in the k hat direction. 00:17:07.540 --> 00:17:10.200 That's the external torque on the system. 00:17:10.200 --> 00:17:16.050 Its gravity causes no torque because it's acting at g. 00:17:16.050 --> 00:17:17.069 All right? 00:17:17.069 --> 00:17:19.364 And this has got to be the time rate of change of h. 00:17:24.280 --> 00:17:28.349 I claim in this problem-- you don't 00:17:28.349 --> 00:17:31.290 have to do this-- but in this problem, 00:17:31.290 --> 00:17:34.540 I claim I can write h as a mass moment of inertia 00:17:34.540 --> 00:17:42.760 matrix times a vector that tells you what the rotation rate is. 00:17:42.760 --> 00:17:45.740 You, in fact, can always do this about g. 00:17:45.740 --> 00:17:48.925 This is I with respect to g. 00:17:48.925 --> 00:17:53.210 If you know what it is, this I-- I 00:17:53.210 --> 00:17:55.900 have chosen a set of coordinates that pass 00:17:55.900 --> 00:17:59.090 through the center of mass. 00:17:59.090 --> 00:18:03.150 x1 that way, y1 perpendicular to it, are they principal coord-- 00:18:03.150 --> 00:18:06.100 and z coming out of the board-- are they principal coordinates? 00:18:06.100 --> 00:18:07.130 Sure, you know that. 00:18:07.130 --> 00:18:09.080 This is just a uniform rod. 00:18:09.080 --> 00:18:12.230 But just symmetry, immediately, should tell you that they are. 00:18:12.230 --> 00:18:13.092 Yes. 00:18:13.092 --> 00:18:16.290 AUDIENCE: I thought you said that IMA could only work 00:18:16.290 --> 00:18:18.996 for a stationary rotation axis. 00:18:18.996 --> 00:18:21.702 I thought IMA could only work for a stationary axis 00:18:21.702 --> 00:18:22.450 of rotation. 00:18:22.450 --> 00:18:26.290 PROFESSOR: And rotation about the center of mass. 00:18:26.290 --> 00:18:29.361 You can always do things with the center of mass. 00:18:29.361 --> 00:18:29.860 OK? 00:18:29.860 --> 00:18:35.580 But you could just work this out by the basic definition 00:18:35.580 --> 00:18:37.280 of angular momentum. 00:18:37.280 --> 00:18:40.731 Don't do it the hard way, r cross p's and those kind 00:18:40.731 --> 00:18:41.230 of things. 00:18:41.230 --> 00:18:43.870 And you end up in the same place. 00:18:43.870 --> 00:18:47.940 This one, because it's 0 0 omega z, 00:18:47.940 --> 00:18:50.230 when you do the multiplication, this is a diagonal. 00:18:50.230 --> 00:18:52.877 The only term that matters is this one. 00:18:52.877 --> 00:18:54.100 Right? 00:18:54.100 --> 00:18:57.980 So that's going to give us an IzzG theta double dot, which 00:18:57.980 --> 00:19:01.211 is omega z, k hat direction. 00:19:01.211 --> 00:19:02.460 And it's not theta double dot. 00:19:02.460 --> 00:19:04.270 The h gives you theta dot. 00:19:04.270 --> 00:19:05.790 Omega z is theta dot. 00:19:05.790 --> 00:19:07.330 And we've taken the time derivative, 00:19:07.330 --> 00:19:10.230 which gets us theta double dot. 00:19:10.230 --> 00:19:13.330 And we know what the mass moment of inertia 00:19:13.330 --> 00:19:16.440 about a uniform stick is with respect 00:19:16.440 --> 00:19:19.600 to G, ML squared over 12. 00:19:19.600 --> 00:19:20.700 Theta dot-- OK. 00:19:20.700 --> 00:19:23.600 So there's our four equations that we have to work with. 00:19:23.600 --> 00:19:26.980 And do they involve N1 or N2? 00:19:26.980 --> 00:19:27.589 Not at all. 00:19:27.589 --> 00:19:29.380 That's why I said we had, potentially, six. 00:19:29.380 --> 00:19:31.850 We really only have four unknowns. 00:19:31.850 --> 00:19:39.962 And now you use these two to solve for F1 and F2. 00:19:39.962 --> 00:19:43.210 And once you get expressions for F1 and F2 00:19:43.210 --> 00:19:47.230 you can eliminate them from here and here. 00:19:47.230 --> 00:19:49.060 And you're done. 00:19:49.060 --> 00:19:52.360 But there's a bit of work left to do that. 00:19:52.360 --> 00:19:54.092 But that's the approach. 00:19:54.092 --> 00:19:57.860 You use these two to isolate F1 and F2 00:19:57.860 --> 00:19:59.500 and plug them into these two to get 00:19:59.500 --> 00:20:00.890 your final equations of motion. 00:20:04.600 --> 00:20:08.140 But we have a couple of things we don't know yet 00:20:08.140 --> 00:20:11.000 that we need in here, the acceleration of that center 00:20:11.000 --> 00:20:14.870 of mass in the i hat-- and break into two 00:20:14.870 --> 00:20:17.310 components, i hat and j hat. 00:20:17.310 --> 00:20:20.700 But that's what we've been studying kinematics for. 00:20:34.290 --> 00:20:38.310 So we need to know the velocity of G with respect to O. 00:20:38.310 --> 00:20:40.410 And we need to know the acceleration of G 00:20:40.410 --> 00:20:42.380 with respect to O. OK? 00:20:46.700 --> 00:20:49.580 So the velocity is pretty straightforward. 00:20:49.580 --> 00:20:50.890 We've done this many times. 00:20:50.890 --> 00:20:53.680 So the velocity of A with respect-- 00:20:53.680 --> 00:20:58.890 Remember, you pick things you know to work from. 00:20:58.890 --> 00:21:02.970 And you try to make as few as possible things you don't know. 00:21:02.970 --> 00:21:04.740 Or put them in forms that we know 00:21:04.740 --> 00:21:06.250 how to go about getting it. 00:21:06.250 --> 00:21:08.720 These are vectors. 00:21:08.720 --> 00:21:10.812 Do we know the velocity of A with respect to O? 00:21:15.740 --> 00:21:16.410 What's A? 00:21:16.410 --> 00:21:20.170 A is the place where the pin is. 00:21:20.170 --> 00:21:23.124 What's its velocity at that point? 00:21:23.124 --> 00:21:23.980 AUDIENCE: x dot. 00:21:23.980 --> 00:21:24.830 PROFESSOR: x dot. 00:21:24.830 --> 00:21:26.682 And in what direction? 00:21:26.682 --> 00:21:27.807 AUDIENCE: Capital I hat. 00:21:27.807 --> 00:21:28.390 PROFESSOR: OK. 00:21:28.390 --> 00:21:33.700 So this is x dot capital I hat. 00:21:33.700 --> 00:21:36.970 And now this term, the velocity of G with respect to A, 00:21:36.970 --> 00:21:44.320 this is a rigid body which is rotating and translating. 00:21:44.320 --> 00:21:46.480 And we've run into this before, right? 00:21:46.480 --> 00:21:50.630 And basically, the equation for such things 00:21:50.630 --> 00:21:55.150 is the motion of the translational velocity 00:21:55.150 --> 00:22:01.430 of the object plus the velocities within the object, 00:22:01.430 --> 00:22:03.870 including anything contributed by rotation. 00:22:03.870 --> 00:22:10.192 So this term is the velocity of the point G, with respect to A, 00:22:10.192 --> 00:22:14.590 to the center-- velocity of the center G with respect 00:22:14.590 --> 00:22:15.780 to point A. 00:22:15.780 --> 00:22:20.650 And that is-- I'll write it consistently here. 00:22:20.650 --> 00:22:26.500 This is the velocity of G with respect to A evaluated 00:22:26.500 --> 00:22:30.640 if you had no rotation. 00:22:30.640 --> 00:22:32.300 If you're sitting-- the other way 00:22:32.300 --> 00:22:34.380 the books often say it is this is 00:22:34.380 --> 00:22:37.370 what you would see if you were on the object sitting at A 00:22:37.370 --> 00:22:39.150 looking and G. Is it moving? 00:22:39.150 --> 00:22:40.660 You'd say no. 00:22:40.660 --> 00:22:42.760 Well, another way of saying that, that's 00:22:42.760 --> 00:22:46.288 the velocity of this thing if there were no rotation. 00:22:46.288 --> 00:22:47.180 All right? 00:22:47.180 --> 00:22:49.750 So this term happens to be 0. 00:22:49.750 --> 00:22:53.890 Plus omega-- now we're kind of doing this on purpose. 00:22:53.890 --> 00:22:58.300 We know this is omega z in the k direction. 00:22:58.300 --> 00:23:00.685 But reminding you, make it always 00:23:00.685 --> 00:23:05.060 with respect to the inertial frame-- cross 00:23:05.060 --> 00:23:12.460 RGA, the position vector from between the two points. 00:23:12.460 --> 00:23:15.810 And so this case, we end up with an x 00:23:15.810 --> 00:23:31.860 dot I hat plus-- this is going to be-- omega zk cross L/2 00:23:31.860 --> 00:23:34.010 little i hat. 00:23:34.010 --> 00:23:36.264 k cross i is j. 00:23:36.264 --> 00:23:42.380 So x dot I hat. 00:23:42.380 --> 00:23:47.530 This is in the little j moving coordinate system direction. 00:23:47.530 --> 00:23:50.650 Omega z is theta dot. 00:23:50.650 --> 00:23:57.520 Theta dot L over 2 j hat. 00:24:04.220 --> 00:24:06.520 I didn't even leave enough room here. 00:24:06.520 --> 00:24:08.680 Let me see this. 00:24:08.680 --> 00:24:13.104 x dot I plus. 00:24:13.104 --> 00:24:13.980 OK? 00:24:13.980 --> 00:24:16.620 Familiar, our omega term. 00:24:16.620 --> 00:24:20.260 So that's your velocity of G with respect to O. 00:24:20.260 --> 00:24:27.920 We need to find an acceleration of that same point. 00:24:27.920 --> 00:24:32.540 And we're going to take the derivative of this to get it. 00:24:32.540 --> 00:24:37.800 A point to-- we haven't talked about this in a while. 00:24:37.800 --> 00:24:41.110 This has mixed unit vectors. 00:24:41.110 --> 00:24:41.610 Right? 00:24:41.610 --> 00:24:44.670 It's got unit vectors in the inertial frame and the unit 00:24:44.670 --> 00:24:45.950 vectors in the rotating frame. 00:24:45.950 --> 00:24:49.080 Is that allowed? 00:24:49.080 --> 00:24:50.680 No problem. 00:24:50.680 --> 00:24:53.060 It's perfectly legitimate, right? 00:24:53.060 --> 00:24:54.665 You have to reduce it, eventually, 00:24:54.665 --> 00:24:56.740 to get a workable equation. 00:24:56.740 --> 00:25:00.270 But that's just fine at this intermediate stage. 00:25:00.270 --> 00:25:01.200 OK. 00:25:01.200 --> 00:25:05.900 The next thing you want to do is find this a with respect 00:25:05.900 --> 00:25:10.760 to G-- aG with respect to O, the acceleration. 00:25:10.760 --> 00:25:17.790 And we know that's just a derivative, remember, 00:25:17.790 --> 00:25:20.735 with respect to the inertial frame-- that's 00:25:20.735 --> 00:25:24.590 why we have to deal with this rotation business-- of velocity 00:25:24.590 --> 00:25:28.660 of G with respect to O. But the derivative of this-- does I 00:25:28.660 --> 00:25:29.960 change direction? 00:25:29.960 --> 00:25:30.460 Nope. 00:25:30.460 --> 00:25:32.020 So this is just x double dot. 00:25:35.710 --> 00:25:38.170 And we have a single term here that we 00:25:38.170 --> 00:25:39.390 have to take a derivative of. 00:25:39.390 --> 00:25:41.935 So we get a theta double dot L/2 j. 00:25:48.730 --> 00:25:51.460 This is a little lowercase j. 00:25:51.460 --> 00:25:53.250 But now, does j rotate? 00:25:56.010 --> 00:25:56.775 Yeah? 00:25:56.775 --> 00:25:59.910 And so this is the derivative of a rotating vector. 00:25:59.910 --> 00:26:06.810 The derivative of j is minus theta dot I. 00:26:06.810 --> 00:26:21.690 So minus theta dot squared L over 2 i hat. 00:26:25.298 --> 00:26:26.200 All right? 00:26:26.200 --> 00:26:28.870 And that's all there is to getting G. 00:26:28.870 --> 00:26:35.185 Now we could have done-- oops, I want that one. 00:26:41.990 --> 00:26:51.780 I don't want you to be afraid of using the big, kind of hairy 00:26:51.780 --> 00:26:58.450 looking 3D vector equation for acceleration. 00:26:58.450 --> 00:27:01.230 Acceleration and this gets called-- this is A with respect 00:27:01.230 --> 00:27:03.870 to O, putting it in terms of this problem, 00:27:03.870 --> 00:27:07.490 plus the acceleration of G with respect 00:27:07.490 --> 00:27:23.230 to A evaluated with no rotation plus omega dot cross RGA 00:27:23.230 --> 00:27:39.310 plus omega cross omega cross RGA plus 2 omega cross 00:27:39.310 --> 00:27:46.610 velocity GA omega equals 0. 00:27:46.610 --> 00:27:50.870 And I should have one, two, three, four, five terms. 00:27:50.870 --> 00:27:52.980 There's always five potential terms 00:27:52.980 --> 00:27:58.280 when you're evaluating the acceleration of a point 00:27:58.280 --> 00:28:02.640 on a rigid body which is translating and rotating. 00:28:02.640 --> 00:28:07.950 This is translating and rotating reference frames, 00:28:07.950 --> 00:28:10.090 attached to the rigid body. 00:28:10.090 --> 00:28:13.470 And then you just go in and fill it in. 00:28:13.470 --> 00:28:15.570 This is the acceleration of the rigid body, 00:28:15.570 --> 00:28:17.050 the translational acceleration. 00:28:17.050 --> 00:28:20.550 In this problem, what is that? 00:28:20.550 --> 00:28:21.542 Louder. 00:28:21.542 --> 00:28:22.500 AUDIENCE: X double dot. 00:28:22.500 --> 00:28:23.083 PROFESSOR: OK. 00:28:23.083 --> 00:28:26.286 This guy is capital X double dot I hat. 00:28:26.286 --> 00:28:28.780 All right? 00:28:28.780 --> 00:28:32.000 This is the acceleration of point G relative to A 00:28:32.000 --> 00:28:33.900 if you were in the object, on the object. 00:28:33.900 --> 00:28:34.933 It is? 00:28:34.933 --> 00:28:35.432 AUDIENCE: 0. 00:28:35.432 --> 00:28:36.307 PROFESSOR: All right. 00:28:36.307 --> 00:28:37.430 0. 00:28:37.430 --> 00:28:41.560 This is the Eulerian term. 00:28:41.560 --> 00:28:46.310 This is theta double dot crossed with the distance 00:28:46.310 --> 00:28:47.720 from between the two points. 00:28:47.720 --> 00:28:49.240 Is this 0? 00:28:49.240 --> 00:28:50.290 No, not necessarily. 00:28:50.290 --> 00:28:53.524 So this is theta double dot. 00:28:56.842 --> 00:28:59.740 This is the L/2. 00:28:59.740 --> 00:29:01.180 That's this. 00:29:01.180 --> 00:29:05.130 And when you do the cross product of k and I, you get j. 00:29:05.130 --> 00:29:07.280 That's that term. 00:29:07.280 --> 00:29:11.380 Omega cross omega cross RGA, this is the centripetal term. 00:29:11.380 --> 00:29:13.180 Would you think it's going to be 0? 00:29:13.180 --> 00:29:14.120 Nope. 00:29:14.120 --> 00:29:18.402 So this is k cross I. That's j. k cross j is 00:29:18.402 --> 00:29:22.070 I minus-- this is the term that gives you minus-- theta 00:29:22.070 --> 00:29:27.880 dot squared L/2 i. 00:29:27.880 --> 00:29:30.440 And this is our Coriolis term. 00:29:30.440 --> 00:29:33.735 It requires motion of that point, G, relative to A. 00:29:33.735 --> 00:29:34.920 Is that moving? 00:29:34.920 --> 00:29:35.420 Nope. 00:29:35.420 --> 00:29:37.250 So this term just goes to 0. 00:29:37.250 --> 00:29:39.197 Get the same answer? 00:29:39.197 --> 00:29:39.696 All right. 00:29:39.696 --> 00:29:40.990 So don't be afraid of this. 00:29:40.990 --> 00:29:42.350 This thing, just lay it down. 00:29:42.350 --> 00:29:46.850 Just plug the things in and the right things will fall out. 00:29:46.850 --> 00:29:47.350 All right. 00:29:47.350 --> 00:29:54.160 On the paper, I break down the acceleration of G with respect 00:29:54.160 --> 00:29:58.680 to O into its little i and little j components. 00:30:03.000 --> 00:30:07.200 And I end up with acceleration of G with respect 00:30:07.200 --> 00:30:09.940 to O. I group the terms. 00:30:09.940 --> 00:30:15.871 The I capital X double dot sine theta-- 00:30:15.871 --> 00:30:17.370 I actually have to break this-- see, 00:30:17.370 --> 00:30:20.570 this is not in the direction of little i or little j. 00:30:20.570 --> 00:30:28.030 So I know that I can express capital I as a lowercase i sine 00:30:28.030 --> 00:30:30.090 theta plus j cosine theta. 00:30:30.090 --> 00:30:31.110 And I use that. 00:30:31.110 --> 00:30:35.220 So it's X double dot sine theta minus L/2 00:30:35.220 --> 00:30:40.090 theta dot squared-- it is in the little i hat direction. 00:30:40.090 --> 00:30:48.610 So this is the i hat term-- plus X double dot cosine theta 00:30:48.610 --> 00:30:51.610 plus L/2 theta double dot. 00:30:51.610 --> 00:30:53.900 And this is the j hat term. 00:30:53.900 --> 00:30:54.820 OK? 00:30:54.820 --> 00:30:55.590 Yeah. 00:30:55.590 --> 00:30:58.085 AUDIENCE: Why would you choose to put everything in terms 00:30:58.085 --> 00:31:02.004 of the rotating inertial frame? 00:31:02.004 --> 00:31:03.420 PROFESSOR: You have to always have 00:31:03.420 --> 00:31:04.760 these same decisions to make. 00:31:04.760 --> 00:31:06.960 And it's what you're comfortable with, 00:31:06.960 --> 00:31:09.160 what you think is going to lead to the least work. 00:31:09.160 --> 00:31:11.980 You have no idea how much time I spent working on this problem 00:31:11.980 --> 00:31:13.520 to put it in a form that I thought 00:31:13.520 --> 00:31:14.650 I could teach it to you. 00:31:14.650 --> 00:31:19.220 I spent a lot of time on it, the point being, occasionally, you 00:31:19.220 --> 00:31:22.820 have to spend a lot of time working them out, going down 00:31:22.820 --> 00:31:26.090 a path that doesn't pay off, backing up, going down 00:31:26.090 --> 00:31:27.240 the next one. 00:31:27.240 --> 00:31:31.410 On quizzes, on homework, most of what 00:31:31.410 --> 00:31:34.927 stops you from getting to the final right answer 00:31:34.927 --> 00:31:35.510 is confidence. 00:31:38.080 --> 00:31:39.490 You've got to believe that you've 00:31:39.490 --> 00:31:41.510 learned these things well enough that you know 00:31:41.510 --> 00:31:43.710 this is the right thing to do. 00:31:43.710 --> 00:31:46.740 And it'll get you there if you just do the arithmetic right. 00:31:46.740 --> 00:31:47.647 OK? 00:31:47.647 --> 00:31:48.980 And we all make little mistakes. 00:31:48.980 --> 00:31:51.120 And you'll all have to back up and do it again. 00:31:51.120 --> 00:31:53.161 But that's why the fundamentals and understanding 00:31:53.161 --> 00:31:54.980 the basic concepts are so important. 00:31:54.980 --> 00:31:57.140 If you've got the concepts down cold, 00:31:57.140 --> 00:32:00.730 you'll have confidence that your method's going to work. 00:32:00.730 --> 00:32:01.630 OK. 00:32:01.630 --> 00:32:06.810 So this is my i direction term. 00:32:06.810 --> 00:32:08.740 This is my j direction term. 00:32:08.740 --> 00:32:17.290 I can take those and take this bit and plug it in here. 00:32:17.290 --> 00:32:18.650 Let's just give it a name. 00:32:18.650 --> 00:32:25.694 Let's call this A and this piece here B. 00:32:25.694 --> 00:32:27.240 And your A goes right here. 00:32:30.730 --> 00:32:35.899 That's A. And this is B. OK? 00:32:35.899 --> 00:32:37.690 And not that you've made that substitution, 00:32:37.690 --> 00:32:40.310 you can solve for F1. 00:32:40.310 --> 00:32:42.880 And they're all in one direction now. 00:32:42.880 --> 00:32:45.170 They're all in the little i hat direction. 00:32:45.170 --> 00:32:48.100 You make the other substitution, everything's in little j hat. 00:32:48.100 --> 00:32:49.450 And you can drop them. 00:32:49.450 --> 00:32:51.530 You no longer have to carry it along. 00:32:51.530 --> 00:32:57.910 You now have scalar equations you can solve for F1 and F2. 00:32:57.910 --> 00:33:00.870 So that's done. 00:33:00.870 --> 00:33:04.020 And you take those two expressions-- 00:33:04.020 --> 00:33:13.440 so this implies F1 equals-- and this one implies F2 equals-- 00:33:13.440 --> 00:33:14.740 and you take those. 00:33:14.740 --> 00:33:23.150 And you put F1 and F2-- you need F1 and F2. 00:33:23.150 --> 00:33:25.690 And you plug it in here. 00:33:25.690 --> 00:33:29.125 And in this equation, you only need an F2. 00:33:37.739 --> 00:33:42.210 And when you do that, you get your two equations of motion. 00:33:42.210 --> 00:33:43.360 Now a couple things happen. 00:33:43.360 --> 00:33:49.920 It turns out, when you do this-- when 00:33:49.920 --> 00:33:54.970 you make the substitution in here-- you end up with an MX 00:33:54.970 --> 00:34:00.420 double dot cosine squared theta and an M2 X double dot sine 00:34:00.420 --> 00:34:01.920 squared theta. 00:34:01.920 --> 00:34:04.740 And sine squared plus cosine squared 00:34:04.740 --> 00:34:07.110 conveniently equals-- 1. 00:34:07.110 --> 00:34:09.159 And that collapses and goes away. 00:34:09.159 --> 00:34:19.300 So you end up with the two final expressions 00:34:19.300 --> 00:34:21.260 here after you've made those combinations. 00:34:21.260 --> 00:34:23.464 So from four over here-- I'll call it four prime-- 00:34:23.464 --> 00:34:25.130 you get one of your equations of motion. 00:35:01.760 --> 00:35:03.010 That's one equation of motion. 00:35:03.010 --> 00:35:07.500 It's primarily about the translation of-- I 00:35:07.500 --> 00:35:09.260 mean rotation-- of the system. 00:35:09.260 --> 00:35:13.320 It derives from all those substitutions in this equation. 00:35:13.320 --> 00:35:15.910 Curiously, these two bits go together, 00:35:15.910 --> 00:35:18.790 which, if you had worked around A and you used, 00:35:18.790 --> 00:35:20.980 dangerously, perhaps, a parallel axis theorem, 00:35:20.980 --> 00:35:23.160 you would have ended up with ML squared over 3. 00:35:29.842 --> 00:35:31.300 But that's your first equa-- that's 00:35:31.300 --> 00:35:32.640 one of your equations of motion. 00:35:32.640 --> 00:35:37.150 And from the first one, it comes from the sum of the forces 00:35:37.150 --> 00:35:43.990 on the main mass, M1, plus M2 X double dot 00:35:43.990 --> 00:35:48.640 plus b X dot plus KX. 00:35:48.640 --> 00:35:51.490 All the usual, just things for mass spring oscillator. 00:35:51.490 --> 00:35:54.540 But now you've got these additional forces 00:35:54.540 --> 00:35:57.180 that are exerted on that because it's got this pendulum 00:35:57.180 --> 00:35:58.420 flagging back and forth. 00:36:18.670 --> 00:36:22.548 And if I made a mistake on the board, believe the paper. 00:36:22.548 --> 00:36:25.320 I think I've got it all transcribed right. 00:36:25.320 --> 00:36:28.810 This is your force equation on the main mass. 00:36:28.810 --> 00:36:31.520 If you didn't have the pendulum hanging there, 00:36:31.520 --> 00:36:35.560 there's your equation of motion for a cart going 00:36:35.560 --> 00:36:37.460 back and forth with springs. 00:36:37.460 --> 00:36:40.570 Then you have this pendulum hanging on it, which 00:36:40.570 --> 00:36:42.740 puts additional forces on it. 00:36:42.740 --> 00:36:45.690 It comes through those F1 and F2 terms. 00:36:45.690 --> 00:36:48.410 And you can see they have to do with having to accelerate 00:36:48.410 --> 00:36:50.770 things down there. 00:36:50.770 --> 00:36:51.270 OK? 00:36:55.270 --> 00:36:56.900 Now these two equations are actually 00:36:56.900 --> 00:36:59.350 quite easy to linearize. 00:36:59.350 --> 00:37:01.910 So linearize means you can always 00:37:01.910 --> 00:37:04.630 linearize around the equilibrium position. 00:37:04.630 --> 00:37:06.560 This is its equilibrium position. 00:37:06.560 --> 00:37:07.465 Small motions. 00:37:10.010 --> 00:37:12.390 That can be described, what you see there, 00:37:12.390 --> 00:37:14.790 by linearized equations of motion. 00:37:14.790 --> 00:37:25.590 And to linearize-- so for theta small sine theta 00:37:25.590 --> 00:37:28.240 is approximately equal to theta. 00:37:28.240 --> 00:37:30.560 Cosine theta is approximately equal to 1. 00:37:30.560 --> 00:37:32.880 And you just have sines and cosines in here. 00:37:32.880 --> 00:37:34.380 You let this guy go to 1. 00:37:34.380 --> 00:37:36.710 You let this go to theta. 00:37:36.710 --> 00:37:39.090 Here, 1. 00:37:39.090 --> 00:37:40.880 Here, theta. 00:37:40.880 --> 00:37:42.740 And you have linearized equations of motion. 00:37:45.360 --> 00:37:48.340 Now other problems can be harder to linearize. 00:37:48.340 --> 00:37:51.200 But this is particularly simple. 00:37:51.200 --> 00:37:53.992 Two linear equations, solvable. 00:37:53.992 --> 00:37:55.450 If you solve them, they'll give you 00:37:55.450 --> 00:38:03.970 two natural frequencies for the system and two vibration modes. 00:38:03.970 --> 00:38:04.470 All right. 00:38:04.470 --> 00:38:07.910 That's the end of this problem. 00:38:07.910 --> 00:38:10.170 Any last questions about it? 00:38:10.170 --> 00:38:12.800 I'm going to move on to this topic of center percussion. 00:38:28.470 --> 00:38:28.970 OK. 00:38:28.970 --> 00:38:30.178 This next one is kind of fun. 00:38:30.178 --> 00:38:32.220 It has a real practical purpose in life. 00:38:32.220 --> 00:38:35.870 I didn't bring my tennis racket or my baseball bat. 00:38:35.870 --> 00:38:38.300 But if any of you play sports that 00:38:38.300 --> 00:38:42.640 use things that hit things, you know, when hit the baseball 00:38:42.640 --> 00:38:44.660 and you hit it down on the handle, 00:38:44.660 --> 00:38:45.900 it really stings your hands. 00:38:45.900 --> 00:38:47.840 Right? 00:38:47.840 --> 00:38:49.930 Or the tennis racket, if you don't hit it right, 00:38:49.930 --> 00:38:51.850 you feel a lot of forces in your hands. 00:38:51.850 --> 00:38:55.425 And you hit it really sweet, you feel almost no force at all. 00:38:55.425 --> 00:38:58.010 How many have had that experience? 00:38:58.010 --> 00:38:58.650 Really common. 00:38:58.650 --> 00:38:59.149 OK. 00:38:59.149 --> 00:39:01.665 So is there a right place to hit it? 00:39:01.665 --> 00:39:02.820 The answer's yes. 00:39:02.820 --> 00:39:08.020 And we're going to go through that right now. 00:39:08.020 --> 00:39:10.760 So here's our-- I'm looking down. 00:39:10.760 --> 00:39:12.250 Here's my baseball bat. 00:39:12.250 --> 00:39:14.170 z is in this direction. 00:39:14.170 --> 00:39:16.750 This is kind of the top view. 00:39:16.750 --> 00:39:18.490 So here's my handle of my bat. 00:39:21.920 --> 00:39:22.890 Ball's coming in. 00:39:27.710 --> 00:39:29.200 Put some force on it. 00:39:29.200 --> 00:39:33.400 I think I put it lowercase in the notes. 00:39:33.400 --> 00:39:35.060 OK? 00:39:35.060 --> 00:39:38.060 And I want to know-- and I want to minimize that force 00:39:38.060 --> 00:39:39.124 at my hands. 00:39:39.124 --> 00:39:40.540 I'm going to call this place where 00:39:40.540 --> 00:39:47.685 it hits P. I'm going to say here's its center of mass at G. 00:39:47.685 --> 00:39:49.730 The point about which you're holding it 00:39:49.730 --> 00:39:57.100 and it's rotating is A. And my coordinate system attached 00:39:57.100 --> 00:40:01.240 to the bat-- we always have this coordinate system attached 00:40:01.240 --> 00:40:03.520 to the bad in these rotational problems 00:40:03.520 --> 00:40:06.670 so that we can divine things like moments of inertia. 00:40:06.670 --> 00:40:09.540 So it's a coordinate system attached to the bat. 00:40:09.540 --> 00:40:11.770 And here's my y direction. 00:40:11.770 --> 00:40:13.110 And z's coming out of the board. 00:40:16.490 --> 00:40:21.940 And this distance is going to be important to me, RGA. 00:40:25.770 --> 00:40:30.990 And this distance here, I'm going to call q. 00:40:30.990 --> 00:40:31.950 It's my unknown. 00:40:31.950 --> 00:40:35.490 And I want to know, where's the sweet spot. 00:40:35.490 --> 00:40:40.180 What do I want q to be so that I minimize the force at A, 00:40:40.180 --> 00:40:41.380 where I'm hanging onto it. 00:40:45.970 --> 00:40:47.440 So how many degrees of freedom? 00:40:47.440 --> 00:40:51.210 How many independent coordinates do I need to do this problem? 00:40:51.210 --> 00:40:54.160 This problem is complicated, because it's a-- you know, 00:40:54.160 --> 00:40:56.270 a real life situation is probably complicated. 00:40:56.270 --> 00:40:59.346 And I'm going to do a lot of simplification and claim 00:40:59.346 --> 00:41:01.220 that the answer-- and then look at the answer 00:41:01.220 --> 00:41:02.480 and say, does it makes sense. 00:41:02.480 --> 00:41:05.590 Did my simplifications make sense? 00:41:05.590 --> 00:41:08.850 So I'm going to do a number of simplifications. 00:41:08.850 --> 00:41:12.210 I'm going to argue that I think I can get at most of the answer 00:41:12.210 --> 00:41:16.240 by essentially saying, right here where I'm hitting it, 00:41:16.240 --> 00:41:18.430 my wrists are like a hinge point. 00:41:18.430 --> 00:41:21.970 That's point A. I'm going to assume it rotates about a fixed 00:41:21.970 --> 00:41:25.522 point at that moment. 00:41:25.522 --> 00:41:26.880 OK? 00:41:26.880 --> 00:41:29.600 Pretty gross simplification of the real thing going on. 00:41:29.600 --> 00:41:32.090 You've really got muscles, you're putting moments on it, 00:41:32.090 --> 00:41:34.280 your wrists actually are moving some. 00:41:34.280 --> 00:41:37.450 But I'm just going to say rotating only about that point. 00:41:37.450 --> 00:41:39.010 It stays in this plane. 00:41:39.010 --> 00:41:40.360 It's a planar motion problem. 00:41:40.360 --> 00:41:43.010 At most, three possible degrees of freedom, 00:41:43.010 --> 00:41:45.620 x, y, and some theta. 00:41:45.620 --> 00:41:46.904 Right? 00:41:46.904 --> 00:41:48.570 And I'm going to make the simplification 00:41:48.570 --> 00:41:50.150 that it's pinned at A. 00:41:50.150 --> 00:41:53.665 So if it's pinned at A, how many constraints is that? 00:41:53.665 --> 00:41:54.455 AUDIENCE: Two. 00:41:54.455 --> 00:41:55.080 PROFESSOR: Two. 00:41:55.080 --> 00:41:56.160 One left. 00:41:56.160 --> 00:42:00.660 So I only need-- I'm going to pick one coordinate. 00:42:00.660 --> 00:42:03.220 That's going to be my theta. 00:42:03.220 --> 00:42:06.210 So even though I've made these gross simplifications, 00:42:06.210 --> 00:42:09.470 will the answer actually turn out to be meaningful? 00:42:09.470 --> 00:42:12.940 And this G is my center of mass. 00:42:12.940 --> 00:42:22.850 And let's assume that we know Izz with respect 00:42:22.850 --> 00:42:29.020 to G, that is, mass moment of inertia for rotation in z. 00:42:29.020 --> 00:42:29.950 OK? 00:42:29.950 --> 00:42:31.085 Assume that we know that. 00:42:31.085 --> 00:42:31.650 It's given. 00:42:35.110 --> 00:42:41.480 So for this problem, when you have fixed 00:42:41.480 --> 00:42:44.430 rotation points-- this problem here, 00:42:44.430 --> 00:42:46.330 that A point wasn't fixed. 00:42:46.330 --> 00:42:48.230 It was moving, accelerating. 00:42:48.230 --> 00:42:50.980 It could have had lots of complicated terms 00:42:50.980 --> 00:42:54.910 in its expression for torque if you'd done it that way. 00:42:54.910 --> 00:42:57.070 This is fixed at A. It makes a lot of sense 00:42:57.070 --> 00:43:01.790 to compute moments about A. All right? 00:43:01.790 --> 00:43:04.440 So my first equation here, that I 00:43:04.440 --> 00:43:11.350 want, is the sum of the moments, torques with respect to A. 00:43:11.350 --> 00:43:13.790 And these are-- well, they will all turn out 00:43:13.790 --> 00:43:17.460 to be in the z direction here. 00:43:17.460 --> 00:43:27.280 And they are time rated change of the angular momentum 00:43:27.280 --> 00:43:31.180 with respect to G, which, in these planar motion problems, 00:43:31.180 --> 00:43:35.600 always then boils down to the mass moment of inertia 00:43:35.600 --> 00:43:40.250 for the axis you're rotating it about times the angular 00:43:40.250 --> 00:43:41.250 acceleration. 00:43:41.250 --> 00:43:43.890 That's that term. 00:43:43.890 --> 00:43:50.450 Oops, not G. A. 00:43:50.450 --> 00:43:50.950 Excuse me. 00:43:50.950 --> 00:43:51.783 We picked our point. 00:43:51.783 --> 00:43:53.170 We've got to stick with it here. 00:43:53.170 --> 00:43:53.669 OK. 00:43:53.669 --> 00:43:56.680 I need that. 00:43:56.680 --> 00:44:03.690 And this is in the k hat direction. 00:44:03.690 --> 00:44:07.420 This is the only term in our torque equation. 00:44:10.390 --> 00:44:13.560 What are the external torques? 00:44:13.560 --> 00:44:17.830 Well now, z is out of the board like this. 00:44:17.830 --> 00:44:21.920 I've got a force acting on a moment arm 00:44:21.920 --> 00:44:24.780 giving me a torque in the other direction. 00:44:24.780 --> 00:44:30.448 So this is minus fq-- also k hat. 00:44:39.074 --> 00:44:40.490 Now I might have done this sooner, 00:44:40.490 --> 00:44:43.530 but we should look at a free body diagram. 00:44:43.530 --> 00:44:47.830 So here's a little stick figure of my bat. 00:44:47.830 --> 00:44:50.850 And I potentially have an unknown force 00:44:50.850 --> 00:44:56.260 in the y direction, Ry, and another unknown force 00:44:56.260 --> 00:45:00.040 in the x direction here at A, that pin point 00:45:00.040 --> 00:45:02.552 where it's rotating it about. 00:45:02.552 --> 00:45:05.920 Out here is G. 00:45:05.920 --> 00:45:11.100 And at G, is there an mG term? 00:45:11.100 --> 00:45:12.610 Well, it's gravity, right? 00:45:12.610 --> 00:45:15.300 But it's acting-- it's in the minus z direction. 00:45:15.300 --> 00:45:17.420 And I'm looking down on it. 00:45:17.420 --> 00:45:18.480 This is a top view. 00:45:21.340 --> 00:45:25.040 So gravity would create a moment in the y direction. 00:45:25.040 --> 00:45:27.360 And I don't have to deal with that. 00:45:27.360 --> 00:45:30.710 My only moment-- this is my torque equation-- 00:45:30.710 --> 00:45:33.310 only has k hat terms in it. 00:45:33.310 --> 00:45:36.490 Gravity creates a moment in the y, 00:45:36.490 --> 00:45:39.841 which, in fact, you have to-- that's a static equilibrium 00:45:39.841 --> 00:45:41.590 problem-- you do have to provide that when 00:45:41.590 --> 00:45:43.440 you're holding the bat or else the bat would droop. 00:45:43.440 --> 00:45:43.939 Right? 00:45:43.939 --> 00:45:46.610 It doesn't have anything to do with the dynamics, actually. 00:45:46.610 --> 00:45:47.110 OK. 00:45:47.110 --> 00:45:50.100 Here's G. There's no gravity term that you can see. 00:45:50.100 --> 00:45:52.400 It's pointing into the board. 00:45:52.400 --> 00:45:57.750 And out here is my point P. And here's f. 00:45:57.750 --> 00:46:00.510 OK. 00:46:00.510 --> 00:46:05.460 So I need an equation-- I have an equation for moment 00:46:05.460 --> 00:46:06.590 equilibrium. 00:46:06.590 --> 00:46:08.960 I need some dynamic equation. 00:46:08.960 --> 00:46:13.320 I have two possible equations for force equilibrium. 00:46:13.320 --> 00:46:14.701 How many unknowns do I have? 00:46:20.960 --> 00:46:22.270 But no others. 00:46:22.270 --> 00:46:23.270 Three possible unknowns. 00:46:23.270 --> 00:46:24.645 I'm going to need three equations 00:46:24.645 --> 00:46:26.840 to get rid of these two terms. 00:46:26.840 --> 00:46:27.900 Right? 00:46:27.900 --> 00:46:30.040 And I am looking ahead. 00:46:30.040 --> 00:46:32.929 My objective is to make that term go to 0. 00:46:32.929 --> 00:46:34.720 So I both have to find an expression for it 00:46:34.720 --> 00:46:36.511 and then figure out how to make it go to 0. 00:46:36.511 --> 00:46:38.630 That's how you do this problem. 00:46:41.140 --> 00:46:41.640 OK. 00:46:41.640 --> 00:46:46.510 So let's do sum of the forces in the y. 00:46:51.220 --> 00:47:02.106 So you have an Ry j hat minus f, also in the j hat direction. 00:47:02.106 --> 00:47:03.855 And that's basically all there is to that. 00:47:03.855 --> 00:47:07.140 But that must be the mass of the bat 00:47:07.140 --> 00:47:09.760 times the acceleration of G with respect 00:47:09.760 --> 00:47:13.460 to O in the y direction. 00:47:13.460 --> 00:47:15.800 So that's our j hat component. 00:47:15.800 --> 00:47:19.000 This is just the component of acceleration in the j hat 00:47:19.000 --> 00:47:19.870 direction. 00:47:19.870 --> 00:47:25.770 And the sum of the forces in x, we look at those. 00:47:25.770 --> 00:47:28.850 Well, we have an Rx in the i. 00:47:28.850 --> 00:47:31.630 And there are no others. 00:47:31.630 --> 00:47:34.180 And that must be, then, the mass of the bat 00:47:34.180 --> 00:47:38.480 times the acceleration, G, with respect to O, of the bit 00:47:38.480 --> 00:47:40.160 in the x direction. 00:47:40.160 --> 00:47:44.230 And that'll be in the i hat. 00:47:44.230 --> 00:47:46.640 So I've got two unknown accelerations, now, 00:47:46.640 --> 00:47:48.410 that I have to deal with. 00:47:48.410 --> 00:47:51.260 But we know lots of kinematics now. 00:48:03.100 --> 00:48:06.330 So let's do it this way just to give you the practice. 00:48:06.330 --> 00:48:10.060 Acceleration of G with respect to O. Acceleration of A 00:48:10.060 --> 00:48:15.890 with respect to O plus the acceleration of G with respect 00:48:15.890 --> 00:48:32.940 A omega equals 0 plus omega dot cross RGA plus 2 omega 00:48:32.940 --> 00:48:46.160 cross VGA omega equals 0, and finally, plus omega 00:48:46.160 --> 00:48:51.215 cross omega cross RGA. 00:48:54.350 --> 00:48:59.150 One, two, three, four, five terms. 00:48:59.150 --> 00:49:02.230 Our vector 3D equation is figured out quickly. 00:49:02.230 --> 00:49:02.970 What's this term? 00:49:06.218 --> 00:49:08.080 AUDIENCE: 0. 00:49:08.080 --> 00:49:08.900 PROFESSOR: 0? 00:49:08.900 --> 00:49:09.716 Why? 00:49:09.716 --> 00:49:10.590 AUDIENCE: It's fixed. 00:49:10.590 --> 00:49:12.048 PROFESSOR: Fixed point of rotation, 00:49:12.048 --> 00:49:13.090 that's our assumption. 00:49:13.090 --> 00:49:14.270 This guy goes to 0. 00:49:14.270 --> 00:49:15.100 OK. 00:49:15.100 --> 00:49:18.662 The velocity of G with respect to A? 00:49:18.662 --> 00:49:19.162 AUDIENCE: 0. 00:49:19.162 --> 00:49:19.745 PROFESSOR: OK. 00:49:19.745 --> 00:49:21.580 Another 0. 00:49:21.580 --> 00:49:22.890 Omega dot cross RGA? 00:49:27.920 --> 00:49:29.861 So that could-- you know, if there's 00:49:29.861 --> 00:49:31.980 an angular acceleration, which there might be, 00:49:31.980 --> 00:49:35.130 cross R, that's a perfectly legitimate term. 00:49:35.130 --> 00:49:38.600 And that'll give you a-- this will give you 00:49:38.600 --> 00:49:42.260 a theta double dot, just before. 00:49:42.260 --> 00:49:49.590 It's in the k direction, cross RGA in the i. 00:49:49.590 --> 00:49:52.020 That's this term right here. 00:49:52.020 --> 00:49:52.520 OK. 00:49:52.520 --> 00:49:54.820 This velocity of G with respect to A 00:49:54.820 --> 00:49:56.140 on the bat, that went to 0. 00:49:56.140 --> 00:49:58.140 The Coriolis, that goes away. 00:49:58.140 --> 00:50:01.200 So at this point, I've only got one term from there. 00:50:01.200 --> 00:50:02.910 And this term is our centripetal term. 00:50:02.910 --> 00:50:04.993 You think there's going to be a centripetal force? 00:50:07.694 --> 00:50:14.580 Common sense should be any time you have a rotational moment, 00:50:14.580 --> 00:50:18.570 something rotating about a point, if the center of mass 00:50:18.570 --> 00:50:24.700 is not the point of rotation, you 00:50:24.700 --> 00:50:27.310 will be forcing a mass to move in a circle. 00:50:27.310 --> 00:50:31.270 And that always requires-- that centripetal acceleration-- 00:50:31.270 --> 00:50:32.520 always requires a force. 00:50:32.520 --> 00:50:35.040 This, for sure, will give you a term 00:50:35.040 --> 00:50:43.057 which is a minus omega z squared type term times the distance 00:50:43.057 --> 00:50:43.556 RGA. 00:50:46.180 --> 00:50:47.660 That's its radius. 00:50:47.660 --> 00:50:48.840 And it's inward. 00:50:48.840 --> 00:50:52.010 So in this direction, it will be i hat. 00:50:52.010 --> 00:50:53.710 Three terms, same as before. 00:50:53.710 --> 00:50:54.960 We now have our accelerations. 00:51:05.190 --> 00:51:05.750 OK. 00:51:05.750 --> 00:51:07.790 So from our top equation up there 00:51:07.790 --> 00:51:11.650 from the summation of forces in the y, 00:51:11.650 --> 00:51:13.770 we'll now substitute these in. 00:51:13.770 --> 00:51:21.940 We have M RGA j hat. 00:51:21.940 --> 00:51:28.750 The acceleration term, we ended up with the-- the k cross i 00:51:28.750 --> 00:51:29.800 term gives you j. 00:51:29.800 --> 00:51:33.110 So the Eulerian term is going to show up here. 00:51:33.110 --> 00:51:38.490 So that, we need a theta double dot. 00:51:38.490 --> 00:51:39.810 And this is in the j direction. 00:51:39.810 --> 00:51:44.680 That comes from the acceleration we figured out. 00:51:44.680 --> 00:51:50.390 And that's equal to minus f plus Ry. 00:51:50.390 --> 00:51:52.881 And this is all little j hat. 00:52:03.690 --> 00:52:05.570 Let me write the other one first here. 00:52:05.570 --> 00:52:10.330 Here's the summation in the x direction. 00:52:10.330 --> 00:52:16.550 It's mass times our acceleration in the i hat direction. 00:52:16.550 --> 00:52:19.790 And we only ended up with one term doing that, 00:52:19.790 --> 00:52:21.870 and that's our centripetal term. 00:52:21.870 --> 00:52:36.460 So we get minus RGA theta dot squared i hat. 00:52:36.460 --> 00:52:39.430 That's the centripetal acceleration times the mass. 00:52:39.430 --> 00:52:42.950 And the only term that it is equated to 00:52:42.950 --> 00:52:45.788 is the other reaction force. 00:52:45.788 --> 00:52:47.150 OK? 00:52:47.150 --> 00:52:51.290 So here's the key step in this problem. 00:52:51.290 --> 00:52:56.910 We want what to be 0 so that we're 00:52:56.910 --> 00:53:00.032 hitting it at the sweet spot? 00:53:00.032 --> 00:53:01.282 What's the original objective? 00:53:05.220 --> 00:53:07.730 We want Ry to go away. 00:53:07.730 --> 00:53:08.710 All right? 00:53:13.610 --> 00:53:15.127 Just make it go away. 00:53:18.330 --> 00:53:37.010 So f, then-- I'll do it as an intermediate step. 00:53:37.010 --> 00:53:39.240 Just a second. f is then-- I'm going to leave Ry 00:53:39.240 --> 00:53:40.560 for just a second longer. 00:53:40.560 --> 00:53:48.270 Ry minus M RGA theta double dot. 00:53:48.270 --> 00:53:50.910 And since this is my-- I can drop the j hats. 00:53:50.910 --> 00:53:53.380 This is now a scalar equation. 00:53:53.380 --> 00:53:55.200 I can find an expression for this force 00:53:55.200 --> 00:53:56.950 that-- I don't even know what the force is 00:53:56.950 --> 00:53:59.190 that the ball exerts on the bat. 00:53:59.190 --> 00:54:01.060 But it exists. 00:54:01.060 --> 00:54:02.990 I can say this must be true. 00:54:02.990 --> 00:54:07.000 And my objective is that this should be 0. 00:54:07.000 --> 00:54:10.330 So that says, when that's true, f 00:54:10.330 --> 00:54:14.010 is minus M RGA theta double dot. 00:54:14.010 --> 00:54:17.056 And I'm going to use that in just a moment. 00:54:17.056 --> 00:54:22.660 So I'm going to substitute this into our first equation 00:54:22.660 --> 00:54:26.640 up there for moment, equation number one. 00:54:26.640 --> 00:54:35.870 So this implies that f is minus M RG with respect 00:54:35.870 --> 00:54:38.570 to A theta double dot. 00:54:38.570 --> 00:54:46.590 And this goes into 1, which I have in fq. 00:54:46.590 --> 00:54:49.891 So I do that. 00:54:53.249 --> 00:54:54.040 Where are my notes? 00:55:17.640 --> 00:55:19.610 This is now also a scalar equation. 00:55:19.610 --> 00:55:21.190 I can drop the k hats. 00:55:21.190 --> 00:55:28.520 So Izz with respect to A theta double dot equals minus fq. 00:55:28.520 --> 00:55:32.680 But f is minus, so the minuses cancel. 00:55:32.680 --> 00:55:42.916 So I get M RGA theta double dot times q. 00:55:48.580 --> 00:56:10.370 Now Izz with respect to A can be expressed as M times some thing 00:56:10.370 --> 00:56:14.240 we call the radius of gyration squared with respect-- 00:56:14.240 --> 00:56:16.250 and this has got to be-- with respect to A. 00:56:16.250 --> 00:56:20.676 The radius of gyration for rotation about A, 00:56:20.676 --> 00:56:23.440 I can find an expression like this 00:56:23.440 --> 00:56:26.310 so that it's equal to Izz with respect to A. 00:56:26.310 --> 00:56:28.670 And all the radius of gyration means 00:56:28.670 --> 00:56:33.210 is that if I took all the mass and put it that distance away, 00:56:33.210 --> 00:56:36.450 the mass moment of inertia of that concentrated point 00:56:36.450 --> 00:56:38.240 mass with respect to the point of rotation 00:56:38.240 --> 00:56:41.410 is the same as in the real object. 00:56:41.410 --> 00:56:44.640 So this is just a convenience that we use. 00:56:44.640 --> 00:56:47.370 So that's M Kappa squared. 00:56:47.370 --> 00:57:00.840 And that says, then, that M Kappa squared A equals M RGA 00:57:00.840 --> 00:57:03.450 theta double dot-- I have a theta double dot here too. 00:57:03.450 --> 00:57:06.310 Sorry-- times q. 00:57:06.310 --> 00:57:08.380 And I can solve for q. 00:57:08.380 --> 00:57:11.000 The theta double dots go away. 00:57:11.000 --> 00:57:13.540 The M's go away. 00:57:13.540 --> 00:57:21.797 And q is just Kappa squared A divided by RGA. 00:57:33.870 --> 00:57:42.610 And that's the answer to this center of percussion problem. 00:57:42.610 --> 00:57:43.370 Here's your bat. 00:57:47.900 --> 00:57:59.686 Here's A. Here's G. Here's P. The q is about here. 00:58:02.560 --> 00:58:04.280 Whoops, excuse me. 00:58:04.280 --> 00:58:12.530 q is the distance to P. It's always 00:58:12.530 --> 00:58:16.811 outside of G. It will always be outside of G. 00:58:16.811 --> 00:58:19.060 We'll have to think about, maybe, the reason for that. 00:58:19.060 --> 00:58:21.290 So here's the center of mass. 00:58:21.290 --> 00:58:21.790 Here's RGA. 00:58:26.440 --> 00:58:27.870 and here's this point at which you 00:58:27.870 --> 00:58:29.840 want to hit at the sweet spot. 00:58:29.840 --> 00:58:35.390 And it's always Kappa squared A over RGA. 00:58:35.390 --> 00:58:38.540 Now how do you get to Kappa squared A? 00:58:38.540 --> 00:58:41.250 Well, you need to evaluate Izz with respect to A. 00:58:41.250 --> 00:58:45.770 But Izz with respect to A, you can do parallel axis theorem, 00:58:45.770 --> 00:58:51.883 Izz with respect with respect to G plus M RGA squared. 00:58:51.883 --> 00:58:52.630 Right? 00:58:52.630 --> 00:58:53.860 And then you have that. 00:58:53.860 --> 00:58:56.400 So if we were to do this-- I'll give you a quick example. 00:58:56.400 --> 00:59:13.980 If this were a uniform rod, Izz G equals ML squared over 12. 00:59:13.980 --> 00:59:27.960 Izz with respect to A ML squared over 12 plus MRG 00:59:27.960 --> 00:59:31.840 with respect to A squared. 00:59:31.840 --> 00:59:39.410 That is ML squared over 12 plus ML squared-- 00:59:39.410 --> 00:59:41.830 this is L/2 to RGA, half the length, 00:59:41.830 --> 00:59:44.130 if I put A right at the end. 00:59:44.130 --> 00:59:46.090 So I'm doing the simple rod. 00:59:46.090 --> 00:59:50.040 I'm putting A right here, G right in the middle. 00:59:50.040 --> 00:59:56.830 So this is L/2-- ML squared over 2 squared, 4. 00:59:56.830 --> 01:00:03.440 You add those two together, you get ML squared over 3 01:00:03.440 --> 01:00:08.150 equals M Kappa A squared. 01:00:08.150 --> 01:00:10.566 So Kappa squared A is L squared over 3. 01:00:15.880 --> 01:00:24.940 And q equals L squared over 3 divided by L/2. 01:00:24.940 --> 01:00:33.080 So you get 2/3 L. 2/3 of the length puts you out here. 01:00:33.080 --> 01:00:35.650 Or at 1/2 the length-- so here's L/2. 01:00:35.650 --> 01:00:39.640 Here's P at 2L/3. 01:00:45.970 --> 01:00:51.650 So anytime you run into these center of percussion problems-- 01:00:51.650 --> 01:00:56.810 tennis racquet, baseball bats, whatever-- the right place 01:00:56.810 --> 01:01:01.671 to hit it is away from the point of rotation. 01:01:01.671 --> 01:01:03.920 Now do you think this is-- now how good is this model? 01:01:12.100 --> 01:01:14.220 You know, you start thinking about, well, 01:01:14.220 --> 01:01:18.000 what are the things-- if the bat handles really moving. 01:01:18.000 --> 01:01:21.605 It's still actually remarkably good. 01:01:21.605 --> 01:01:23.390 It's remarkably good. 01:01:23.390 --> 01:01:25.504 In fact, the fact these other things 01:01:25.504 --> 01:01:26.920 are happening, like you're putting 01:01:26.920 --> 01:01:30.160 some moment on with your wrists, the fact that you still 01:01:30.160 --> 01:01:32.610 have some speed down here, all you're doing 01:01:32.610 --> 01:01:36.680 is kind of maybe changing the point it's rotating around. 01:01:36.680 --> 01:01:40.237 But you're still having to exert forces with your hands 01:01:40.237 --> 01:01:41.820 at that point where you're holding it. 01:01:41.820 --> 01:01:44.070 And you just want those forces to go to 0. 01:01:44.070 --> 01:01:45.842 All of these other complications, 01:01:45.842 --> 01:01:48.300 you're still going to find out, to make that force go to 0, 01:01:48.300 --> 01:01:51.350 it's approximately that. 01:01:51.350 --> 01:01:53.660 Really a pretty good answer. 01:01:53.660 --> 01:01:54.180 All right. 01:01:58.440 --> 01:02:03.710 Ah, let's go back. 01:02:03.710 --> 01:02:05.799 And we have another piece of information 01:02:05.799 --> 01:02:08.090 that we developed in this problem that we haven't used. 01:02:11.960 --> 01:02:13.430 We had another equation. 01:02:13.430 --> 01:02:17.080 We had the acceleration in the i direction. 01:02:17.080 --> 01:02:19.763 And it's just equal to RX. 01:02:19.763 --> 01:02:24.015 The RX force is M times the acceleration in the i 01:02:24.015 --> 01:02:24.515 direction. 01:02:28.810 --> 01:02:42.510 And that's minus RG with respect to A theta dot squared. 01:02:42.510 --> 01:02:46.770 That's our old centripetal force term, right? 01:02:46.770 --> 01:02:48.655 Centrifugal force, centripetal acceleration. 01:02:48.655 --> 01:02:50.910 Here's the centripetal acceleration times 01:02:50.910 --> 01:02:56.070 the mass is the force to make the bat go in the circle. 01:02:56.070 --> 01:02:58.763 You have to provide that force. 01:02:58.763 --> 01:02:59.263 OK? 01:03:06.010 --> 01:03:11.420 So I want to finish up by a loose end 01:03:11.420 --> 01:03:15.524 that many people have-- continue to have-- 01:03:15.524 --> 01:03:16.440 a little trouble with. 01:03:16.440 --> 01:03:20.250 Because I get questions on Muddy cards about this. 01:03:20.250 --> 01:03:21.800 And it was on this last homework. 01:03:21.800 --> 01:03:26.340 I don't know how that wheel with the little bit missing-- 01:03:26.340 --> 01:03:28.240 if you all got that sorted out. 01:03:28.240 --> 01:03:30.684 If you're all perfectly clear in your mind about that then 01:03:30.684 --> 01:03:32.100 I don't need to say what I'm going 01:03:32.100 --> 01:03:33.350 to talk about for a second. 01:03:33.350 --> 01:03:36.040 But I thought I would just tie that 01:03:36.040 --> 01:03:37.920 up, the last little bit here. 01:03:37.920 --> 01:03:40.280 This problem is a nice lead to it. 01:03:46.090 --> 01:03:54.380 This last force that I computed is 01:03:54.380 --> 01:03:58.760 the force required to swing this thing in a circle. 01:03:58.760 --> 01:04:00.400 And I'm putting that force right here. 01:04:00.400 --> 01:04:03.280 And I'm saying it's-- actually the point of rotation is right 01:04:03.280 --> 01:04:04.780 here. 01:04:04.780 --> 01:04:09.680 And that force is the mass times the acceleration of that point. 01:04:09.680 --> 01:04:14.350 And that point is a distance from my point of rotation 01:04:14.350 --> 01:04:16.000 to the center of mass. 01:04:16.000 --> 01:04:18.230 No accident. 01:04:18.230 --> 01:04:21.180 You're accelerating the center of mass. 01:04:21.180 --> 01:04:25.440 That times the rotation rate squared 01:04:25.440 --> 01:04:28.680 is the acceleration, the centripetal acceleration. 01:04:28.680 --> 01:04:30.220 This is the force. 01:04:30.220 --> 01:04:46.250 So any time you have an object-- maybe it's a wheel, 01:04:46.250 --> 01:04:51.690 and maybe it has, stuck on it, a little extra mass 01:04:51.690 --> 01:04:53.800 you don't know about, a rock stuck in your tire. 01:04:57.760 --> 01:04:59.600 This is the center of rotation. 01:04:59.600 --> 01:05:01.520 Your axle on your wheel doesn't move. 01:05:01.520 --> 01:05:03.470 Tire doesn't move relative to the axle. 01:05:03.470 --> 01:05:05.980 You've got this little bit of mass missing 01:05:05.980 --> 01:05:08.880 or added, doesn't much matter. 01:05:08.880 --> 01:05:12.370 The center of mass of this system 01:05:12.370 --> 01:05:15.230 is no longer at the center of rotation. 01:05:15.230 --> 01:05:17.180 So I'll call the center of rotation here A. 01:05:17.180 --> 01:05:19.560 That's what it's rotating about. 01:05:19.560 --> 01:05:23.280 This addition or loss of a piece of mass 01:05:23.280 --> 01:05:26.940 makes the center of mass of the system move a little bit. 01:05:26.940 --> 01:05:30.100 So I have an extra little bit of mass out here. 01:05:30.100 --> 01:05:34.150 The new actual center gravity of this system 01:05:34.150 --> 01:05:36.430 is a little distance away. 01:05:36.430 --> 01:05:41.093 And I'll call that distance e, eccentricity. 01:05:43.840 --> 01:05:46.990 So now I have a system whose center of mass 01:05:46.990 --> 01:05:49.560 is not at the center of rotation. 01:05:49.560 --> 01:06:01.260 And its distance, RGA, I'm calling the eccentricity. 01:06:01.260 --> 01:06:04.960 And as this rotates-- and I'll put a coordinate system 01:06:04.960 --> 01:06:05.620 on here. 01:06:05.620 --> 01:06:11.330 i, x, y, rotating with the system. 01:06:11.330 --> 01:06:18.820 The force, it's going to appear here on this axle. 01:06:18.820 --> 01:06:26.390 It's going to have a force that I'll call Rx. 01:06:26.390 --> 01:06:36.310 Rx equals the mass times the acceleration of the center 01:06:36.310 --> 01:06:39.040 of gravity of the system. 01:06:39.040 --> 01:06:41.000 And the acceleration of the center 01:06:41.000 --> 01:06:52.090 of gravity in this system is minus e theta dot squared. 01:06:55.480 --> 01:06:58.790 And it's inward in the i hat direction. 01:06:58.790 --> 01:07:01.180 Therefore, the minus sign. 01:07:01.180 --> 01:07:05.040 So the force-- this is the centripetal acceleration 01:07:05.040 --> 01:07:07.940 times mass-- the force required to make it 01:07:07.940 --> 01:07:14.750 do that is an inward force minus M e theta dot squared i hat. 01:07:17.280 --> 01:07:18.080 OK? 01:07:18.080 --> 01:07:20.370 I drew Rx in a positive direction. 01:07:20.370 --> 01:07:21.530 The answer comes out minus. 01:07:21.530 --> 01:07:23.054 It says it's going the other way. 01:07:28.780 --> 01:07:31.760 So this is always the case. 01:07:31.760 --> 01:07:35.620 If you have a system that, for some reason, 01:07:35.620 --> 01:07:40.450 does not rotate about its true center of mass, 01:07:40.450 --> 01:07:44.440 but in fact the center of mass is off a little bit or a lot, 01:07:44.440 --> 01:07:47.700 you will have to provide a force that's inwardly 01:07:47.700 --> 01:07:50.554 directed as that thing spins. 01:07:50.554 --> 01:07:51.850 OK? 01:07:51.850 --> 01:07:57.650 Now a subtlety I want you to know, to really go away with, 01:07:57.650 --> 01:08:02.900 is that we've talked quite a lot about unbalance 01:08:02.900 --> 01:08:06.640 due to these masses that aren't concentrated 01:08:06.640 --> 01:08:10.690 at the center, these unbalanced masses. 01:08:10.690 --> 01:08:16.960 There are two kinds of unbalance the engineers have 01:08:16.960 --> 01:08:20.430 chosen to describe the world with, two kinds. 01:08:20.430 --> 01:08:26.410 One is static imbalance and the other is dynamic imbalance. 01:08:26.410 --> 01:08:34.180 A statically imbalanced system is one in which you-- 01:08:34.180 --> 01:08:45.609 I'm looking at a side view of-- an edge view of this thing 01:08:45.609 --> 01:08:51.680 as it's spinning, maybe like this. 01:08:51.680 --> 01:08:54.510 And the center of rotation is here. 01:08:54.510 --> 01:08:57.189 And the G is here. 01:08:57.189 --> 01:09:00.029 That force is inward on it. 01:09:00.029 --> 01:09:04.520 But really, it's perfectly symmetric. 01:09:04.520 --> 01:09:08.270 So this force that I'm having to provide 01:09:08.270 --> 01:09:12.029 to keep this thing from flying off 01:09:12.029 --> 01:09:13.779 is perfectly inward directed. 01:09:13.779 --> 01:09:16.979 Does it generate any moments about this point? 01:09:16.979 --> 01:09:17.710 None. 01:09:17.710 --> 01:09:19.359 This is called static imbalance. 01:09:22.290 --> 01:09:26.729 And here's an example of static imbalance. 01:09:26.729 --> 01:09:28.340 This is my rotor. 01:09:28.340 --> 01:09:30.370 I'm going to put it on its side. 01:09:30.370 --> 01:09:32.597 Gravity says, I want this to hang down. 01:09:32.597 --> 01:09:34.680 Because where's the center of mass of this system? 01:09:34.680 --> 01:09:36.350 Up here? 01:09:36.350 --> 01:09:38.280 Along here? 01:09:38.280 --> 01:09:41.270 It's down here somewhere, right? 01:09:41.270 --> 01:09:44.540 That center of mass, you do a statics calculation, MG sine 01:09:44.540 --> 01:09:48.017 theta, it's going to hang down. 01:09:48.017 --> 01:09:49.850 So that's why they call it static imbalance. 01:09:49.850 --> 01:09:53.050 And the way you can correct a statically imbalanced system 01:09:53.050 --> 01:09:54.570 is just do the test. 01:09:54.570 --> 01:09:57.630 Put the thing-- hang the thing on an axle. 01:09:57.630 --> 01:09:59.810 And do I have a handy axle here today? 01:10:05.630 --> 01:10:08.770 Put it on an axle and see if it rotates. 01:10:08.770 --> 01:10:12.292 And if it always rotates to some point hanging down, 01:10:12.292 --> 01:10:14.125 you know this thing's statically imbalanced. 01:10:19.668 --> 01:10:21.910 Look in my kit of parts here. 01:10:21.910 --> 01:10:23.718 Just maybe do this. 01:10:28.558 --> 01:10:30.910 I don't know if this is heavy enough to do the job. 01:10:30.910 --> 01:10:34.330 But now I've put a little bit of mass on here. 01:10:34.330 --> 01:10:41.270 And I presume that if I do this with it, OK, it goes down. 01:10:41.270 --> 01:10:43.140 This system is statically imbalanced. 01:10:43.140 --> 01:10:46.450 This system, now, is also dynamically imbalanced. 01:10:46.450 --> 01:10:46.950 Why? 01:10:51.620 --> 01:10:54.180 It kind of depends on where I do the calculation. 01:10:54.180 --> 01:10:59.020 But if I say that I want to know-- this 01:10:59.020 --> 01:11:03.520 is when put a force, a mass, off-center here. 01:11:03.520 --> 01:11:10.950 And if I compute moments about a point that's perfectly lined up 01:11:10.950 --> 01:11:15.590 with it, this system will create no moments about that point. 01:11:15.590 --> 01:11:17.860 This system is perfectly balanced now. 01:11:17.860 --> 01:11:21.170 The mass that I've stuck on here is there. 01:11:21.170 --> 01:11:22.940 This system is perfectly balanced. 01:11:22.940 --> 01:11:24.610 It generates no moments as it stands, 01:11:24.610 --> 01:11:26.520 no matter where you calculate this point. 01:11:26.520 --> 01:11:33.270 But this extra little bit-- this is 01:11:33.270 --> 01:11:36.320 A. If I compute H with respect to A 01:11:36.320 --> 01:11:38.700 here and do the derivatives of it, 01:11:38.700 --> 01:11:42.240 I get no torques except in the z direction. 01:11:42.240 --> 01:11:44.720 But as soon as I move away from there, 01:11:44.720 --> 01:11:48.500 and there is a distance here, and I compute moments 01:11:48.500 --> 01:11:51.280 about that, do I get some static moments 01:11:51.280 --> 01:11:53.310 not in the direction of spin? 01:11:53.310 --> 01:11:54.170 Yes. 01:11:54.170 --> 01:11:58.360 So when you do that, this system-- if this is my point A, 01:11:58.360 --> 01:12:01.250 let's call it, over here-- this system, 01:12:01.250 --> 01:12:03.690 is it statically imbalanced? 01:12:03.690 --> 01:12:04.190 Yep. 01:12:04.190 --> 01:12:06.700 Is it dynamically imbalanced? 01:12:06.700 --> 01:12:08.010 Yeah. 01:12:08.010 --> 01:12:11.320 So how would you balance it? 01:12:11.320 --> 01:12:14.340 Final question of the term for me. 01:12:17.250 --> 01:12:20.696 How would you put this system into balance? 01:12:20.696 --> 01:12:25.088 AUDIENCE: Put another mass on the other side. 01:12:25.088 --> 01:12:26.552 PROFESSOR: All right. 01:12:26.552 --> 01:12:29.720 So if I put another mass here, will that statically 01:12:29.720 --> 01:12:30.350 balance it? 01:12:33.098 --> 01:12:36.550 Statically balance, meaning if I let go, 01:12:36.550 --> 01:12:39.520 will it have to rotate around and find a low point. 01:12:39.520 --> 01:12:41.570 Is there any low point-- if I put 01:12:41.570 --> 01:12:46.720 one equal and opposite, equal distance away, opposite side, 01:12:46.720 --> 01:12:48.303 would it be statically balanced? 01:12:48.303 --> 01:12:48.886 AUDIENCE: Yes. 01:12:48.886 --> 01:12:51.219 PROFESSOR: Yeah, because gravity pulls down on this one, 01:12:51.219 --> 01:12:53.240 equal and opposite down on this one. 01:12:53.240 --> 01:12:54.740 So it would be statically balanced. 01:12:54.740 --> 01:12:57.480 Would it be dynamically balanced? 01:12:57.480 --> 01:13:00.345 The system, as you spin, will try to twist like this. 01:13:00.345 --> 01:13:01.720 So it's not dynamically balanced. 01:13:01.720 --> 01:13:02.886 What if I move it over here? 01:13:06.670 --> 01:13:08.042 Is it statically balanced? 01:13:08.042 --> 01:13:08.625 AUDIENCE: Yes. 01:13:08.625 --> 01:13:10.221 PROFESSOR: Is it dynamically balanced? 01:13:10.221 --> 01:13:10.804 AUDIENCE: Yes. 01:13:13.444 --> 01:13:14.069 PROFESSOR: Hmm. 01:13:17.450 --> 01:13:18.740 Compute the torques. 01:13:18.740 --> 01:13:19.820 You know? 01:13:19.820 --> 01:13:24.050 Let's put A right in the center, compute the angular momentum, 01:13:24.050 --> 01:13:25.500 take its time derivative, and you 01:13:25.500 --> 01:13:28.350 will find that this one, as it spins, 01:13:28.350 --> 01:13:30.940 tries to create a moment like that. 01:13:30.940 --> 01:13:34.170 And this one, as it spins tries to create a moment like that. 01:13:34.170 --> 01:13:37.360 And they're exactly equal and opposite. 01:13:37.360 --> 01:13:41.143 This is dynamically balanced and statically balanced. 01:13:45.080 --> 01:13:45.580 Hmm. 01:13:45.580 --> 01:13:47.204 AUDIENCE: When there was just one mass, 01:13:47.204 --> 01:13:50.855 was it dynamically balanced about A, the axis of rotation. 01:13:50.855 --> 01:13:52.480 PROFESSOR: This is not dynamic-- if you 01:13:52.480 --> 01:13:55.740 put A right in the usual center-- the original center 01:13:55.740 --> 01:13:59.610 of gravity, we'll call A, at this wheel-- this system is not 01:13:59.610 --> 01:14:02.380 dynamically balanced with respect to an A that's 01:14:02.380 --> 01:14:03.990 right on the axle and on the center. 01:14:03.990 --> 01:14:06.770 Because it has this little offset. 01:14:06.770 --> 01:14:08.320 OK?