WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.810 Commons license. 00:00:03.810 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.258 at ocw.mit.edu. 00:00:25.830 --> 00:00:29.820 PROFESSOR: So today we're going to do two things in particular. 00:00:29.820 --> 00:00:33.540 One is finish off with the discussion 00:00:33.540 --> 00:00:36.830 of this device, a shaker. 00:00:36.830 --> 00:00:39.640 This, by the way, this is a commercial thing. 00:00:39.640 --> 00:00:44.120 And out of the catalog, this is the littlest one. 00:00:44.120 --> 00:00:45.770 This is a 50 pound shaker. 00:00:45.770 --> 00:00:48.900 At full speed it actually puts out 50 pounds. 00:00:48.900 --> 00:00:51.902 All it is is masses inside going around and around. 00:00:51.902 --> 00:00:53.636 AUDIENCE: What is its commercial purpose? 00:00:53.636 --> 00:00:55.510 PROFESSOR: Ah, what's its commercial purpose? 00:00:55.510 --> 00:01:03.410 Well, the big ones that are maybe 100 pounds of moving mass 00:01:03.410 --> 00:01:07.650 are they bolt them to the floor in nuclear power plants 00:01:07.650 --> 00:01:09.390 and test them. 00:01:09.390 --> 00:01:13.745 Shake the buildings to represent earthquake kind 00:01:13.745 --> 00:01:15.770 of loads and things like that. 00:01:15.770 --> 00:01:18.690 And the smaller ones, you can buy these for. 00:01:18.690 --> 00:01:20.860 This kind is actually if you're running 00:01:20.860 --> 00:01:23.330 an operation like in a flour mill 00:01:23.330 --> 00:01:26.600 and you've got particulate stuff trying to get 00:01:26.600 --> 00:01:28.310 it to slide through chutes. 00:01:28.310 --> 00:01:32.376 Does stuff slide down chutes easier if the things 00:01:32.376 --> 00:01:33.500 are vibrating a little bit? 00:01:33.500 --> 00:01:34.600 Have you ever banged on something 00:01:34.600 --> 00:01:35.870 to get stuff to come loose? 00:01:35.870 --> 00:01:38.660 You can just stick one of these on the side and let it run. 00:01:38.660 --> 00:01:39.620 Nothing sticks. 00:01:39.620 --> 00:01:43.150 So there's lots and lots of purposes for shakers like this. 00:01:43.150 --> 00:01:48.340 So we were in the process of analyzing how one of these 00:01:48.340 --> 00:01:48.840 works. 00:01:55.020 --> 00:01:57.300 And I want to finish that. 00:01:57.300 --> 00:02:00.670 And then part two today is we're going to-- we've only 00:02:00.670 --> 00:02:03.760 really talked about angular momentum with respect 00:02:03.760 --> 00:02:06.010 to particles, individual particles. 00:02:06.010 --> 00:02:09.240 And even in your physics classes you 00:02:09.240 --> 00:02:11.352 did things with mass moment of inertia. 00:02:11.352 --> 00:02:13.060 And so we're going to make the connection 00:02:13.060 --> 00:02:17.260 today between particles, mass moment of inertia, 00:02:17.260 --> 00:02:18.520 unbalanced shakers. 00:02:18.520 --> 00:02:24.180 It all comes together in the second part of today's lecture. 00:02:24.180 --> 00:02:29.230 So the problem we are analyzing, literally that little shaker, 00:02:29.230 --> 00:02:31.100 can be modeled. 00:02:31.100 --> 00:02:35.480 Well, in that one particular application 00:02:35.480 --> 00:02:37.365 we can find that thing on rollers. 00:02:40.850 --> 00:02:43.280 This is the problem we are discussing. 00:02:43.280 --> 00:02:48.920 It has inside of it an unbalanced rotating mass 00:02:48.920 --> 00:02:51.110 with an arm that's E long. 00:02:51.110 --> 00:02:54.380 It's called the eccentricity in the trade. 00:02:54.380 --> 00:02:57.580 And it has some mass m. 00:02:57.580 --> 00:03:00.440 And this body that it's in, we'll call it mass of the body, 00:03:00.440 --> 00:03:02.870 mb. 00:03:02.870 --> 00:03:05.270 And this thing's going around and round. 00:03:05.270 --> 00:03:10.050 So this is some angle theta which is described as omega t. 00:03:10.050 --> 00:03:13.500 And they're constant rotation rate devices. 00:03:13.500 --> 00:03:17.535 So theta dot equals omega and that's a constant. 00:03:20.697 --> 00:03:22.280 That's how they're basically designed. 00:03:24.790 --> 00:03:27.660 And we label this point A. Over here we 00:03:27.660 --> 00:03:34.820 have a inertial coordinate system, xy. 00:03:34.820 --> 00:03:38.493 This point a, this point we've called b in our analysis. 00:03:42.580 --> 00:03:45.740 And we set out to find the equation 00:03:45.740 --> 00:03:48.180 of motion of this thing in the x direction. 00:03:48.180 --> 00:03:49.905 Has no movement in the y. 00:03:49.905 --> 00:03:51.860 It's confined in the y. 00:03:51.860 --> 00:03:54.090 It puts out lots of force in the y direction. 00:03:54.090 --> 00:03:56.940 You really have to restrain it to keep it from moving. 00:03:56.940 --> 00:03:59.200 But it doesn't move in the y direction, 00:03:59.200 --> 00:04:03.426 but it will move in the x. 00:04:03.426 --> 00:04:03.925 OK. 00:04:09.390 --> 00:04:11.140 So we came to the conclusion that we 00:04:11.140 --> 00:04:14.690 could write for the main body the summation 00:04:14.690 --> 00:04:20.269 of the external forces on mb. 00:04:20.269 --> 00:04:25.062 It's mb times its acceleration. 00:04:25.062 --> 00:04:26.520 And it's acceleration is completely 00:04:26.520 --> 00:04:27.985 defined by this coordinate. 00:04:35.502 --> 00:04:48.660 And if we draw a free body diagram of this mass, 00:04:48.660 --> 00:04:55.630 you're going to have a normal force I'll call n 00:04:55.630 --> 00:04:58.280 in the y direction upwards. 00:04:58.280 --> 00:05:04.275 You're going to have its weight downwards. 00:05:07.020 --> 00:05:11.540 And you're going to have some force exerted 00:05:11.540 --> 00:05:16.720 on it through this shaft that comes from the little mass. 00:05:16.720 --> 00:05:19.750 So we're accounting for everything the little mass, 00:05:19.750 --> 00:05:22.700 all its influence on this big block 00:05:22.700 --> 00:05:25.670 by the forces that are passed through that rod, which 00:05:25.670 --> 00:05:27.574 is hinged at the center. 00:05:27.574 --> 00:05:28.870 OK. 00:05:28.870 --> 00:05:34.750 And I'm going to call that f mb. 00:05:34.750 --> 00:05:37.640 OK, the force from that little link. 00:05:37.640 --> 00:05:44.190 Now, that happens to be equal to minus the forces on the mass 00:05:44.190 --> 00:05:45.465 that this rod exerts. 00:05:45.465 --> 00:05:46.965 It must exert some force on the mass 00:05:46.965 --> 00:05:49.630 to make it go around and around. 00:05:49.630 --> 00:05:51.960 And because of Newton's third law, 00:05:51.960 --> 00:05:54.280 those two forces have to be equal and opposite 00:05:54.280 --> 00:05:56.730 because they're operating on the same massless 00:05:56.730 --> 00:06:01.345 shaft for the purpose of this example. 00:06:04.670 --> 00:06:11.030 OK, in order find that equation of motion, 00:06:11.030 --> 00:06:12.810 the sum of the external forces, these 00:06:12.810 --> 00:06:14.420 are both in the y direction. 00:06:14.420 --> 00:06:17.970 So we just need to find the horizontal component 00:06:17.970 --> 00:06:21.240 of this force and we'll be able to complete that equation. 00:06:21.240 --> 00:06:22.665 So the point of the exercise here 00:06:22.665 --> 00:06:26.030 is just is to find this horizontal component. 00:06:26.030 --> 00:06:33.030 So to do that, let's move on to thinking about the little mass, 00:06:33.030 --> 00:06:37.070 small mass, and what its free body diagram looks like. 00:06:41.690 --> 00:06:46.380 So viewed from the here's our rod. 00:06:46.380 --> 00:06:47.540 Here's the small mass. 00:06:47.540 --> 00:06:48.690 This is a side view. 00:06:48.690 --> 00:06:51.350 So there's the point it's rotating about A. 00:06:51.350 --> 00:06:56.140 But I want to draw a free body diagram of this rod. 00:06:56.140 --> 00:06:58.950 The rod puts a force on this mass, 00:06:58.950 --> 00:07:03.090 which will have a vertical component, fm. 00:07:03.090 --> 00:07:05.190 And I'll just call it y. 00:07:05.190 --> 00:07:11.680 And it'll put a force that's in horizontal component fm x. 00:07:11.680 --> 00:07:15.760 And I'm drawing them both positive, 00:07:15.760 --> 00:07:17.709 because I don't know which direction they act. 00:07:17.709 --> 00:07:19.500 And if the answer turns out to be positive, 00:07:19.500 --> 00:07:20.360 then I guessed right. 00:07:20.360 --> 00:07:22.526 If it's negative, it means it's going the other way. 00:07:24.870 --> 00:07:27.440 And what other forces that are acting on this? 00:07:27.440 --> 00:07:31.720 Well, there is certainly is an mg downwards 00:07:31.720 --> 00:07:32.750 acting on that mass. 00:07:36.690 --> 00:07:37.190 OK. 00:07:37.190 --> 00:07:42.090 And there's no forces in and out of the page on it. 00:07:42.090 --> 00:07:43.720 And this is operating in the plane. 00:07:43.720 --> 00:07:47.900 So this is a planar motion problem. 00:07:47.900 --> 00:07:56.549 And we note that in here r dot equals r double dot equals 0. 00:07:56.549 --> 00:07:58.340 This thing doesn't change in length at all. 00:07:58.340 --> 00:08:02.350 It's just going round and round fixed length. 00:08:02.350 --> 00:08:06.290 So we can write, then, that the summation 00:08:06.290 --> 00:08:10.572 of the external forces on this little mass 00:08:10.572 --> 00:08:17.800 had better equal its mass times the acceleration of point A 00:08:17.800 --> 00:08:20.854 with respect to the inertial frame. 00:08:20.854 --> 00:08:22.440 And whoops. 00:08:22.440 --> 00:08:25.020 Not A, but what? 00:08:30.660 --> 00:08:34.370 B. The acceleration of this point. 00:08:34.370 --> 00:08:35.280 This is B. 00:08:35.280 --> 00:08:38.360 We need to figure out what the acceleration of that point 00:08:38.360 --> 00:08:40.900 is in the inertial frame. 00:08:40.900 --> 00:08:42.730 But we've done enough of these problems, 00:08:42.730 --> 00:08:44.290 so this should be pretty easy. 00:08:44.290 --> 00:08:48.430 This is the mass times the acceleration 00:08:48.430 --> 00:08:51.930 of point A with respect to O plus the mass 00:08:51.930 --> 00:08:59.020 times the acceleration of B with respect to A. B and A. 00:08:59.020 --> 00:09:05.006 These are all vectors until I break them down 00:09:05.006 --> 00:09:06.255 into their x and y components. 00:09:15.200 --> 00:09:17.140 So what's the acceleration of A with respect 00:09:17.140 --> 00:09:22.190 to O in the coordinate systems that we have written here? 00:09:29.569 --> 00:09:31.610 So that's just kind of our generic representation 00:09:31.610 --> 00:09:34.140 of acceleration, right? 00:09:34.140 --> 00:09:38.592 But we've chosen some coordinates here. 00:09:38.592 --> 00:09:40.050 Specifically have a coordinate that 00:09:40.050 --> 00:09:42.300 describes the motion of the main mass, right? 00:09:42.300 --> 00:09:43.950 What is that? 00:09:43.950 --> 00:09:46.620 So what's the acceleration of point A? 00:09:46.620 --> 00:09:47.770 x double dot. 00:09:47.770 --> 00:09:57.060 So we know that this then is m x double dot and plus. 00:09:57.060 --> 00:10:01.190 Now, it's easiest to describe this in terms 00:10:01.190 --> 00:10:04.030 of cylindrical coordinates. 00:10:04.030 --> 00:10:06.600 And we can then write that, well, then this 00:10:06.600 --> 00:10:18.600 must be a mass times the terms in the r hat direction. 00:10:26.105 --> 00:10:28.030 r hat. 00:10:28.030 --> 00:10:34.840 And then terms over here in the theta hat direction. 00:10:34.840 --> 00:10:39.830 Theta double dot plus 2r dot theta dot. 00:10:42.800 --> 00:10:45.440 Now, which of these are 0? 00:10:45.440 --> 00:10:48.645 Does that arm change length? 00:10:48.645 --> 00:10:49.145 No. 00:10:49.145 --> 00:10:52.710 So this is 0. 00:10:52.710 --> 00:10:56.700 Is the angular acceleration constant? 00:10:56.700 --> 00:10:58.710 So this is 0. 00:10:58.710 --> 00:11:00.310 The arm doesn't change length. 00:11:00.310 --> 00:11:01.300 The Coriolis is 0. 00:11:01.300 --> 00:11:04.370 So there's no Coriolis force, no [INAUDIBLE] force, 00:11:04.370 --> 00:11:07.480 no radial acceleration, only a single term. 00:11:07.480 --> 00:11:09.550 Just the centrifugal. 00:11:09.550 --> 00:11:15.592 So this becomes a pretty simple expression. 00:11:22.770 --> 00:11:27.250 So the summation of the external forces on our little mass, 00:11:27.250 --> 00:11:35.092 then, we can write as mx double dot in the i hat direction. 00:11:35.092 --> 00:11:37.710 I'm going to break it into its vector components here. 00:11:40.810 --> 00:11:43.570 Minus m. 00:11:43.570 --> 00:11:47.210 And I know that r here equals e. 00:11:47.210 --> 00:11:48.630 That's the eccentricity. 00:11:48.630 --> 00:11:50.630 I'm going to start using these terms. 00:11:50.630 --> 00:11:54.380 Minus me omega squared. 00:11:59.140 --> 00:12:02.460 r hat. 00:12:02.460 --> 00:12:06.450 But I'm going to break that r. 00:12:06.450 --> 00:12:07.680 Goes round and round. 00:12:07.680 --> 00:12:09.970 I need to break it into x and y components, 00:12:09.970 --> 00:12:13.060 but we've done that many times before. 00:12:13.060 --> 00:12:18.340 That looks like a cosine omega t in the i hat 00:12:18.340 --> 00:12:25.484 direction plus a sine omega t in the j hat direction. 00:12:25.484 --> 00:12:27.150 So as this thing goes around and around, 00:12:27.150 --> 00:12:29.699 it has a cosine term and a sine term. 00:12:29.699 --> 00:12:31.740 And this is in the x direction, this is in the y. 00:12:36.840 --> 00:12:42.150 So we're really interested in the equation of motion 00:12:42.150 --> 00:12:47.110 on small mass m in the x direction. 00:12:47.110 --> 00:12:49.900 So we just need to pull out the x components from this. 00:12:49.900 --> 00:13:03.420 So we have an m x double dot i hat minus me 00:13:03.420 --> 00:13:08.390 omega squared cosine omega t i hat. 00:13:08.390 --> 00:13:10.990 And we can drop the i hats now because we just have 00:13:10.990 --> 00:13:12.400 one single component equation. 00:13:15.780 --> 00:13:19.540 And this is this quantity I called fm. 00:13:24.080 --> 00:13:29.090 And fm x then is the x term in my little free body diagram. 00:13:29.090 --> 00:13:31.810 And the force that it exerts on the main mass 00:13:31.810 --> 00:13:33.640 is in the x direction. 00:13:39.020 --> 00:13:40.950 So this is the force that the rod 00:13:40.950 --> 00:13:43.680 places on that little mass in the x direction. 00:13:43.680 --> 00:13:45.480 What's the force that the rod places 00:13:45.480 --> 00:13:48.420 on the big mass in the x direction? 00:13:48.420 --> 00:13:49.580 Minus that. 00:13:49.580 --> 00:13:57.750 So this is minus f mb in the x direction. 00:13:57.750 --> 00:13:59.060 That's what we're after. 00:13:59.060 --> 00:14:03.890 We need that force so we can go back now and we'll finish out 00:14:03.890 --> 00:14:05.310 are the equation of motion that we 00:14:05.310 --> 00:14:09.080 were after for the main mass. 00:14:09.080 --> 00:14:09.670 This up here. 00:14:09.670 --> 00:14:12.990 We need to sum the external forces to get that, 00:14:12.990 --> 00:14:14.820 to fill out that expression. 00:14:14.820 --> 00:14:17.470 But while I'm here, just to have it, 00:14:17.470 --> 00:14:22.446 the summation of the forces on the small mass in the y 00:14:22.446 --> 00:14:22.945 direction. 00:14:25.470 --> 00:14:27.250 Look at our free body diagram. 00:14:27.250 --> 00:14:29.780 It has a minus mg. 00:14:32.510 --> 00:14:43.520 And it then has this term, minus me omega squared sine omega t. 00:14:43.520 --> 00:14:46.320 So just for completeness, we have also the y component 00:14:46.320 --> 00:14:50.020 of the force that the rod places on the small mass. 00:14:50.020 --> 00:14:53.422 And minus this amount is what it places on the main mass 00:14:53.422 --> 00:14:54.380 that it's connected to. 00:15:00.790 --> 00:15:03.540 So now let's go back to our equation up here, 00:15:03.540 --> 00:15:08.650 the summation of the forces on the main body. 00:15:13.410 --> 00:15:16.425 In the x direction. 00:15:16.425 --> 00:15:19.170 This is going to be the main body. 00:15:19.170 --> 00:15:21.080 x double dot. 00:15:21.080 --> 00:15:28.440 And it's now the x direction forces. 00:15:28.440 --> 00:15:32.830 There's only x component of this force. 00:15:32.830 --> 00:15:34.700 And that's what we have right here. 00:15:34.700 --> 00:15:35.835 It's minus that. 00:16:02.260 --> 00:16:06.010 And that's our equation of motion. 00:16:06.010 --> 00:16:09.520 We can rearrange it a little bit and it remarkably simplifies, 00:16:09.520 --> 00:16:10.540 actually. 00:16:10.540 --> 00:16:13.750 You end up, if you collect the motion terms involving 00:16:13.750 --> 00:16:32.510 x on the left hand side equals an external excitation 00:16:32.510 --> 00:16:34.160 on the right hand side. 00:16:34.160 --> 00:16:38.560 And I've been kind of following the commentaries in mb. 00:16:38.560 --> 00:16:40.290 Little confusion about some questions. 00:16:40.290 --> 00:16:43.400 When you're asked to find an equation of motion, 00:16:43.400 --> 00:16:46.680 is that the same thing as meaning 00:16:46.680 --> 00:16:47.965 solve the equation of motion? 00:16:47.965 --> 00:16:51.374 No, asking find the equation of motion means get this far. 00:16:51.374 --> 00:16:53.540 Now, if I wanted to know a solution for this, pretty 00:16:53.540 --> 00:16:56.540 trivial in this case, it's going to look like cosine omega t, 00:16:56.540 --> 00:17:01.210 but then I'd say solve that equation of motion. 00:17:01.210 --> 00:17:04.630 OK, now let's see. 00:17:04.630 --> 00:17:06.599 We also know-- let's just finish this-- 00:17:06.599 --> 00:17:11.869 that the summation of the forces on this main body in the y 00:17:11.869 --> 00:17:15.770 direction must be 0 because it can't move. 00:17:15.770 --> 00:17:16.479 No acceleration. 00:17:19.329 --> 00:17:22.660 And from the free body diagram for that, 00:17:22.660 --> 00:17:40.350 we can write that this is n and y minus mmbg 00:17:40.350 --> 00:17:55.400 minus mg from the little mass plus me 00:17:55.400 --> 00:18:00.302 omega squared sine omega t. 00:18:00.302 --> 00:18:02.700 That's the other phase of this. 00:18:02.700 --> 00:18:10.860 And the interesting thing here, then, is to solve for the force 00:18:10.860 --> 00:18:14.870 that it takes to hold this thing in place. 00:18:14.870 --> 00:18:25.288 So you get mb plus m times g plus or minus. 00:18:25.288 --> 00:18:26.252 All right, yeah. 00:18:36.297 --> 00:18:37.380 All right, so what's that? 00:18:37.380 --> 00:18:39.810 So just kind of step back and look at these things 00:18:39.810 --> 00:18:41.800 and say what's it telling us. 00:18:41.800 --> 00:18:43.830 So first of all, just to keep this thing 00:18:43.830 --> 00:18:49.040 from moving up and down, there's a force on it 00:18:49.040 --> 00:18:50.740 that has to support its weight. 00:18:50.740 --> 00:18:52.540 And it's the combined weight of whatever's 00:18:52.540 --> 00:18:54.350 inside that container. 00:18:54.350 --> 00:18:57.340 The weight of the rotating mass and the weight of the object. 00:18:57.340 --> 00:19:00.320 They have to be supported by a normal force, which 00:19:00.320 --> 00:19:03.240 this is a constant term. 00:19:03.240 --> 00:19:05.220 Weight down, normal force up. 00:19:05.220 --> 00:19:10.700 And around that constant force is an oscillating force. 00:19:10.700 --> 00:19:13.222 me omega squared sine omega t. 00:19:13.222 --> 00:19:15.520 e omega squared you should recognize 00:19:15.520 --> 00:19:18.530 as a centripetal acceleration. 00:19:18.530 --> 00:19:21.180 Mass times acceleration to force. 00:19:21.180 --> 00:19:26.370 And because it goes round and round, when 00:19:26.370 --> 00:19:28.240 it's like this it's pulling up and when 00:19:28.240 --> 00:19:30.740 it's like this it's pulling down and when it's like this is, 00:19:30.740 --> 00:19:32.150 it's only going to the sides. 00:19:32.150 --> 00:19:35.850 So sine omega t for the vertical parts, cosine omega 00:19:35.850 --> 00:19:36.935 t for the horizontal. 00:19:41.680 --> 00:19:44.550 And that's actually all there is to the shake. 00:19:44.550 --> 00:19:47.610 That's all there is to the shakers. 00:19:47.610 --> 00:19:50.740 The rotating mass inside. 00:19:50.740 --> 00:19:55.330 Now, in the homework, from the second homework where 00:19:55.330 --> 00:20:00.090 you had this thing, this ball running around inside, 00:20:00.090 --> 00:20:01.640 where I posed the question in a way 00:20:01.640 --> 00:20:03.075 I didn't really quite intend. 00:20:07.510 --> 00:20:19.280 But I asked here's the track. 00:20:19.280 --> 00:20:23.790 And you had this roller going around inside. 00:20:23.790 --> 00:20:29.900 And I asked to find the normal force that the track exerts 00:20:29.900 --> 00:20:32.420 on the roller. 00:20:32.420 --> 00:20:34.830 So it's an unknown. 00:20:34.830 --> 00:20:46.380 And there must also be a tangential force on this thing. 00:20:46.380 --> 00:20:51.600 And there's also going to be this thing certainly has 00:20:51.600 --> 00:20:52.915 weight mg. 00:21:02.140 --> 00:21:05.460 And so that's the complete free body diagram. 00:21:05.460 --> 00:21:08.370 Now, let's if this is frictionless, which it won't 00:21:08.370 --> 00:21:11.190 be in reality, but for the purposes of analysis, 00:21:11.190 --> 00:21:14.100 let's say it's frictionless, it's only a normal force. 00:21:14.100 --> 00:21:16.800 Where does this tangential force come from? 00:21:16.800 --> 00:21:20.890 Why's it there in this problem? 00:21:20.890 --> 00:21:23.010 There's a key piece of information you're told, 00:21:23.010 --> 00:21:25.400 and that is that the angular acceleration of this thing 00:21:25.400 --> 00:21:26.090 is constant. 00:21:26.090 --> 00:21:28.005 It's constant speed going around. 00:21:28.005 --> 00:21:31.770 If you had a ball rolling around there at constant speed, 00:21:31.770 --> 00:21:33.610 would it go constant if you just pushed 00:21:33.610 --> 00:21:35.679 it and it started rolling? 00:21:35.679 --> 00:21:38.220 It would slowdown going up and it would speed up coming down. 00:21:38.220 --> 00:21:39.920 Why? 00:21:39.920 --> 00:21:41.330 AUDIENCE: [INAUDIBLE]. 00:21:41.330 --> 00:21:42.420 PROFESSOR: Gravity, right? 00:21:42.420 --> 00:21:45.490 So there must be something that has to overcome gravity 00:21:45.490 --> 00:21:48.140 going uphill and holding it back coming downhill. 00:21:48.140 --> 00:21:50.010 So the way these things actually work 00:21:50.010 --> 00:21:52.999 is they've got ports pushing compressed air in here. 00:21:52.999 --> 00:21:54.790 And this is driven around by compressed air 00:21:54.790 --> 00:21:56.844 and there's a pressure difference 00:21:56.844 --> 00:21:58.510 between this side and that side and that 00:21:58.510 --> 00:22:01.120 generates the necessary tangential force to make 00:22:01.120 --> 00:22:02.960 the thing go around and around. 00:22:02.960 --> 00:22:04.300 But they're really easy to make. 00:22:04.300 --> 00:22:05.980 You can imagine very few moving parts. 00:22:05.980 --> 00:22:08.130 Just hook up a compressed air hose to that 00:22:08.130 --> 00:22:10.139 and it's just pushing the ball around inside. 00:22:10.139 --> 00:22:11.180 You get the same outcome. 00:22:17.130 --> 00:22:20.280 On this ball, on this roller, if there 00:22:20.280 --> 00:22:29.190 is a-- the problem we just solved is we found fm 00:22:29.190 --> 00:22:34.980 in the y and fm in the x. 00:22:34.980 --> 00:22:38.400 And this problem said yeah, but why can't we get the same thing 00:22:38.400 --> 00:22:46.240 but have those coordinates be f normal and f tangential? 00:22:46.240 --> 00:22:49.270 And sure, that's just a coordinate rotation. 00:22:49.270 --> 00:22:51.960 So what can you say about these forces? 00:22:51.960 --> 00:22:59.000 Well, one thing you could say is fn squared plus ft squared 00:22:59.000 --> 00:23:06.230 had better be equal to fmx squared plus fmy squared, 00:23:06.230 --> 00:23:08.040 right? 00:23:08.040 --> 00:23:10.410 And then just like converting from polar 00:23:10.410 --> 00:23:15.000 to Cartesian coordinates, you can do these conversions. 00:23:15.000 --> 00:23:17.100 And you could find out, for example, 00:23:17.100 --> 00:23:28.600 that fn is-- keep my notation consistent 00:23:28.600 --> 00:23:42.820 here. fn will be fm in the x cosine omega t plus fm in the y 00:23:42.820 --> 00:23:45.320 sine omega t. 00:23:45.320 --> 00:23:46.620 And so there's the answer. 00:23:46.620 --> 00:23:49.056 This is what you're asked for in that problem set. 00:23:59.968 --> 00:24:02.470 OK. 00:24:02.470 --> 00:24:07.170 So all you need to know about shakers. 00:24:07.170 --> 00:24:10.170 If you're ever confronted with something like this, 00:24:10.170 --> 00:24:14.410 what's the magnitude of the force that the shaker puts out? 00:24:22.825 --> 00:24:25.049 AUDIENCE: [INAUDIBLE]. 00:24:25.049 --> 00:24:26.090 PROFESSOR: Little louder? 00:24:26.090 --> 00:24:27.780 AUDIENCE: Is it mr omega squared? 00:24:27.780 --> 00:24:30.540 PROFESSOR: mr omega squared, but substitute 00:24:30.540 --> 00:24:35.920 for r the actual eccentricity. 00:24:35.920 --> 00:24:38.960 It's whatever that mass in length 00:24:38.960 --> 00:24:40.960 out there that's spinning around. 00:24:40.960 --> 00:24:44.300 me omega squared is the magnitude of the force 00:24:44.300 --> 00:24:46.710 and it's going to oscillate up and down 00:24:46.710 --> 00:24:49.070 and it's going to have gravity that it adds to. 00:24:49.070 --> 00:24:51.910 But the important part is me omega squared 00:24:51.910 --> 00:24:54.480 is the magnitude of the force. 00:24:54.480 --> 00:24:58.730 OK, now we're going to move on to the next topic. 00:24:58.730 --> 00:25:03.210 The next topic is mass moments of inertia. 00:25:03.210 --> 00:25:05.360 And it has a strong connection to these. 00:25:05.360 --> 00:25:07.640 And I'm going to use this kind of analysis 00:25:07.640 --> 00:25:11.470 as the transition to talking about moments of inertia. 00:25:11.470 --> 00:25:14.170 Moments of inertia and products of inertia. 00:25:14.170 --> 00:25:17.830 So any final questions about this before we go on? 00:25:17.830 --> 00:25:18.901 Yeah? 00:25:18.901 --> 00:25:21.727 AUDIENCE: Your summation of [INAUDIBLE], 00:25:21.727 --> 00:25:24.082 why did you not include mg? 00:25:24.082 --> 00:25:25.530 PROFESSOR: Why didn't I include? 00:25:25.530 --> 00:25:26.480 AUDIENCE: Mg. 00:25:26.480 --> 00:25:28.600 PROFESSOR: Mg. 00:25:28.600 --> 00:25:29.890 in the. 00:25:29.890 --> 00:25:31.875 AUDIENCE: First summation. 00:25:31.875 --> 00:25:32.374 [INAUDIBLE] 00:25:35.080 --> 00:25:37.380 PROFESSOR: Oh. 00:25:37.380 --> 00:25:39.000 Yeah, you're right. 00:25:42.900 --> 00:25:44.335 And where's my free body diagram? 00:25:48.650 --> 00:25:50.150 Has it on it, right? 00:25:50.150 --> 00:25:52.660 Just didn't get it down into the-- and what direction's 00:25:52.660 --> 00:25:53.160 it in? 00:25:57.570 --> 00:25:59.265 Because then we did get it back in. 00:26:02.134 --> 00:26:04.370 AUDIENCE: [INAUDIBLE]. 00:26:04.370 --> 00:26:08.540 PROFESSOR: Back in the last line for the y component. 00:26:08.540 --> 00:26:11.334 Now, does it appear in this one, this equation at all? 00:26:11.334 --> 00:26:13.000 It has absolutely nothing to do with it. 00:26:13.000 --> 00:26:15.470 Gravity is in the j hat direction. 00:26:15.470 --> 00:26:18.460 This is a force equation in i hat. 00:26:18.460 --> 00:26:21.390 But it does appear in that normal tangential expression 00:26:21.390 --> 00:26:24.550 when you go look at the solution for that problem, 00:26:24.550 --> 00:26:29.390 because it has compounds in both of the i and j directions. 00:26:29.390 --> 00:26:30.960 And so it'll show up. 00:26:30.960 --> 00:26:34.730 Gravity will show up in this expression. 00:26:34.730 --> 00:26:37.030 Right through this term. 00:26:37.030 --> 00:26:38.925 Yeah? 00:26:38.925 --> 00:26:42.725 AUDIENCE: In the bottom equation on that middle board, 00:26:42.725 --> 00:26:48.340 you have my minus mbg minus mg plus. 00:26:48.340 --> 00:26:51.274 I don't understand where that last plus came from. 00:26:51.274 --> 00:26:53.719 Because in your equation on the left 00:26:53.719 --> 00:26:57.142 you're using the force of little f, correct? 00:26:57.142 --> 00:26:59.910 And you have two negatives there. 00:26:59.910 --> 00:27:01.510 PROFESSOR: This is 0. 00:27:01.510 --> 00:27:03.720 I left the n where it was and moved everything 00:27:03.720 --> 00:27:05.840 to the other side. 00:27:05.840 --> 00:27:08.330 So that plus becomes a minus. 00:27:08.330 --> 00:27:13.070 AUDIENCE: No, from the board to the left to the middle board. 00:27:13.070 --> 00:27:13.960 PROFESSOR: OK. 00:27:13.960 --> 00:27:15.421 AUDIENCE: So down. 00:27:15.421 --> 00:27:18.343 So you have the summation of the force 00:27:18.343 --> 00:27:25.585 on the little mass as negative mg minus m-- yep, that one. 00:27:25.585 --> 00:27:28.495 And from what I can understand, you just 00:27:28.495 --> 00:27:32.375 moved that force over to the large force, 00:27:32.375 --> 00:27:34.315 but you [INAUDIBLE], correct? 00:27:43.510 --> 00:27:47.000 PROFESSOR: It should be minus this thing. 00:27:47.000 --> 00:27:48.180 The summation here. 00:27:48.180 --> 00:27:52.340 This force is minus the little mass force. 00:27:52.340 --> 00:27:55.440 So that ought to become a plus and a plus, right? 00:27:55.440 --> 00:27:58.085 And so if I do that carefully. 00:28:19.155 --> 00:28:25.660 To this one is OK. 00:28:25.660 --> 00:28:28.815 But this one appears to have a sign problem, right? 00:28:32.110 --> 00:28:36.470 But these two terms have got to be the same. 00:28:36.470 --> 00:28:38.080 And so I've got a mistake somewhere. 00:28:38.080 --> 00:28:42.790 And rather than spend 10 minutes fixing it on the fly, 00:28:42.790 --> 00:28:44.060 I'll take note of that. 00:28:50.520 --> 00:28:51.770 This should be OK. 00:28:51.770 --> 00:28:53.736 AUDIENCE: Yeah, intuitively makes sense too, 00:28:53.736 --> 00:28:55.235 I just don't understand [INAUDIBLE]. 00:29:03.170 --> 00:29:05.680 PROFESSOR: Ah, wait a second. 00:29:08.540 --> 00:29:11.630 No, I'm not going to try to fix it right now. 00:29:11.630 --> 00:29:14.120 I made a slip in my notes somewhere. 00:29:14.120 --> 00:29:15.490 But I will repair that. 00:29:15.490 --> 00:29:16.466 Yeah? 00:29:16.466 --> 00:29:18.450 AUDIENCE: Why do we need mg at all? 00:29:18.450 --> 00:29:22.418 Because doesn't this force the angular acceleration 00:29:22.418 --> 00:29:23.410 is constant? 00:29:23.410 --> 00:29:26.386 Or the angular velocity is constant, right? 00:29:26.386 --> 00:29:28.122 So the centrifugal acceleration is 00:29:28.122 --> 00:29:32.834 going to be constant, which means that the part that's 00:29:32.834 --> 00:29:34.818 driven by the motor is going to be changing 00:29:34.818 --> 00:29:36.802 to account for gravity. 00:29:36.802 --> 00:29:39.778 So isn't gravity taking into account 00:29:39.778 --> 00:29:44.738 that we have a constant force or [INAUDIBLE]? 00:29:50.000 --> 00:29:53.410 PROFESSOR: Yeah, you're asking if gravity is not 00:29:53.410 --> 00:29:58.520 taken into account somehow by that rotating. 00:29:58.520 --> 00:30:05.530 The gravitational force that is on the main mass that 00:30:05.530 --> 00:30:07.900 comes from the little mass certainly 00:30:07.900 --> 00:30:11.690 has to pass through the rod. 00:30:11.690 --> 00:30:17.535 It's got to be contained in the forces in the connecting rod. 00:30:17.535 --> 00:30:19.960 So it's definitely there. 00:30:19.960 --> 00:30:24.510 But the force that causes the centripetal acceleration 00:30:24.510 --> 00:30:29.670 of that rotating mass is completely 00:30:29.670 --> 00:30:31.870 independent of gravity. 00:30:31.870 --> 00:30:34.700 With or without gravity, it takes a particular force 00:30:34.700 --> 00:30:37.640 to make that thing travel in a circular path. 00:30:37.640 --> 00:30:42.340 And that's m minus mr theta dot squared always. 00:30:42.340 --> 00:30:42.840 Yep? 00:30:42.840 --> 00:30:45.877 AUDIENCE: So doesn't that mean that on your first expression 00:30:45.877 --> 00:30:49.919 on that board, there should be no mb? 00:30:49.919 --> 00:30:51.210 PROFESSOR: On which expression? 00:30:51.210 --> 00:30:52.127 AUDIENCE: That one. 00:30:52.127 --> 00:30:52.960 PROFESSOR: This one. 00:30:52.960 --> 00:30:55.593 OK, this is the total forces on the little mass. 00:30:55.593 --> 00:30:56.509 AUDIENCE: [INAUDIBLE]. 00:31:05.990 --> 00:31:08.520 PROFESSOR: We need to back up to here. 00:31:08.520 --> 00:31:13.160 The total forces are mass times the acceleration 00:31:13.160 --> 00:31:17.180 of the main body it's connected to plus the mass 00:31:17.180 --> 00:31:19.940 times the acceleration of B with respect to A. 00:31:19.940 --> 00:31:22.950 So we have to have that term. 00:31:22.950 --> 00:31:25.960 And we then go into our four terms 00:31:25.960 --> 00:31:29.190 here and find there's only one left. 00:31:29.190 --> 00:31:34.580 So that's the force exerted on the small mass by the rod. 00:31:34.580 --> 00:31:41.980 And that is positive mx double dot minus mr theta dot squared. 00:31:41.980 --> 00:31:46.040 So we sum the forces on that little mass. 00:31:46.040 --> 00:31:50.529 It has got to be equal to-- ah, I 00:31:50.529 --> 00:31:51.820 know where we made the mistake. 00:31:51.820 --> 00:31:54.020 So we've just discovered our mistake. 00:31:54.020 --> 00:31:56.640 This has got to be able to mass times acceleration. 00:31:56.640 --> 00:32:01.450 And what are the forces? 00:32:01.450 --> 00:32:04.150 The summation the forces is mass times acceleration. 00:32:04.150 --> 00:32:08.220 So the acceleration is this plus this. 00:32:08.220 --> 00:32:09.555 But the sum of the forces. 00:32:50.040 --> 00:32:56.260 The problem here is I've used a notation where 00:32:56.260 --> 00:33:00.820 this is very similar looking to the forces 00:33:00.820 --> 00:33:02.440 that I've noted here. 00:33:02.440 --> 00:33:09.260 So this is the actual force in the y direction, j hat, 00:33:09.260 --> 00:33:16.302 plus the actual force in the x direction, i hat, minus mgi. 00:33:20.860 --> 00:33:23.670 So when I solve for the i component, 00:33:23.670 --> 00:33:28.340 I'm going to get the i pieces of that plus mgi. 00:33:32.019 --> 00:33:33.310 I mean, excuse me, j component. 00:33:39.128 --> 00:33:40.535 Should it be like that? 00:33:44.580 --> 00:33:48.990 The j component will have this piece times sine omega 00:33:48.990 --> 00:33:51.710 t with a minus. 00:33:51.710 --> 00:34:01.250 And you move the mgj to that side and it becomes a plus. 00:34:01.250 --> 00:34:02.200 All right. 00:34:02.200 --> 00:34:02.965 That makes sense. 00:34:05.970 --> 00:34:10.710 The rod has to hold up the weight of that little mass, 00:34:10.710 --> 00:34:12.560 right? 00:34:12.560 --> 00:34:13.760 The weight's down. 00:34:13.760 --> 00:34:19.340 But the rod has to push up on it in the y direction. 00:34:19.340 --> 00:34:23.060 So the force the rod puts on the little mass 00:34:23.060 --> 00:34:26.880 has got to be equal to the weight of the small mass 00:34:26.880 --> 00:34:29.350 minus this me omega squared term, which 00:34:29.350 --> 00:34:37.530 is the force necessary to create the centripetal acceleration. 00:34:37.530 --> 00:34:40.120 OK, so we've got this now fixed. 00:34:45.530 --> 00:34:48.450 This term is OK. 00:34:56.449 --> 00:35:01.730 And the minus that force is then the force on the main body. 00:35:01.730 --> 00:35:02.545 So minus. 00:35:08.721 --> 00:35:09.220 Plus. 00:35:11.950 --> 00:35:15.020 Now I've got to figure out what I did wrong here. 00:35:15.020 --> 00:35:17.620 You're doing what I said I wasn't going to do. 00:35:17.620 --> 00:35:21.200 We're on the fly trying to figure out where the. 00:35:21.200 --> 00:35:24.200 AUDIENCE: [INAUDIBLE]. 00:35:24.200 --> 00:35:25.200 PROFESSOR: OK. 00:35:25.200 --> 00:35:26.250 You're happy now. 00:35:26.250 --> 00:35:28.250 Good. 00:35:28.250 --> 00:35:29.294 OK. 00:35:29.294 --> 00:35:30.710 All right, we're going to move on. 00:35:39.474 --> 00:35:40.390 AUDIENCE: [INAUDIBLE]. 00:35:43.960 --> 00:35:45.460 PROFESSOR: What about the summation? 00:35:45.460 --> 00:35:47.043 AUDIENCE: It's not really a summation. 00:35:47.043 --> 00:35:51.340 It's just the force that arm is [INAUDIBLE]. 00:35:55.760 --> 00:35:57.230 PROFESSOR: Right. 00:35:57.230 --> 00:35:58.614 Fair enough. 00:35:58.614 --> 00:35:59.114 Yep. 00:36:03.600 --> 00:36:06.850 This is just minus f. 00:36:06.850 --> 00:36:08.570 This is on the little mass. 00:36:08.570 --> 00:36:10.950 This is the force on the little mass from the free body 00:36:10.950 --> 00:36:11.670 diagram. 00:36:11.670 --> 00:36:12.400 Right. 00:36:12.400 --> 00:36:16.660 And that helps. 00:36:16.660 --> 00:36:22.050 And that's different from the summation one here. 00:36:22.050 --> 00:36:23.110 OK. 00:36:23.110 --> 00:36:23.610 All right. 00:36:23.610 --> 00:36:26.310 I think we've got it sorted out. 00:36:26.310 --> 00:36:34.170 Now, I've put on the Stellar website 00:36:34.170 --> 00:36:36.370 under readings a little one page thing called 00:36:36.370 --> 00:36:37.750 "Moments of Inertia." 00:36:37.750 --> 00:36:42.360 It's two pages of information taken from the Williams 00:36:42.360 --> 00:36:44.120 textbook on dynamics. 00:36:44.120 --> 00:36:49.186 And it's going to show some of what 00:36:49.186 --> 00:36:51.060 I'm going to put on the board, and especially 00:36:51.060 --> 00:36:52.910 the detailed stuff you don't have to copy. 00:36:52.910 --> 00:36:55.410 OK, we're going to come up with some expressions for angular 00:36:55.410 --> 00:36:59.050 momentum in terms of particles and their positions. 00:36:59.050 --> 00:37:03.060 And this is now the subject of mass and moments 00:37:03.060 --> 00:37:04.945 of inertia and products of inertia. 00:37:04.945 --> 00:37:07.320 And I'm going to put some of these equations on the board 00:37:07.320 --> 00:37:10.068 and you don't have to copy them all. 00:37:10.068 --> 00:37:13.730 All these expressions become the definitions 00:37:13.730 --> 00:37:16.760 of mass moments of inertia and products of inertia. 00:37:16.760 --> 00:37:21.130 And if you just drop down one last little bit, 00:37:21.130 --> 00:37:24.050 we come up with an expression for angular momentum. 00:37:24.050 --> 00:37:30.240 Three vector components look like ixx omega x plus ixy omega 00:37:30.240 --> 00:37:33.540 y and so forth. 00:37:33.540 --> 00:37:38.400 These compounds in terms of particle masses and positions 00:37:38.400 --> 00:37:41.630 are defined in these final equations. 00:37:41.630 --> 00:37:44.370 So I'm going to tell you what I'm going to tell you. 00:37:44.370 --> 00:37:45.990 We're going to make the transition 00:37:45.990 --> 00:37:55.000 from dealing with particles and angular momentum of particles 00:37:55.000 --> 00:38:00.200 to angular momentum of rigid bodies. 00:38:00.200 --> 00:38:01.200 OK? 00:38:01.200 --> 00:38:07.450 And in my own experience this is something 00:38:07.450 --> 00:38:09.860 that is generally done badly. 00:38:09.860 --> 00:38:11.710 And I'm going to try to do it well. 00:38:11.710 --> 00:38:17.240 I'm going to try to give you an intuitive understanding of why 00:38:17.240 --> 00:38:23.020 we have these diagonal terms called the moments of inertia 00:38:23.020 --> 00:38:25.420 and what they're useful for and why 00:38:25.420 --> 00:38:28.850 these off diagonal terms call products of inertia turn up 00:38:28.850 --> 00:38:30.470 and what they actually mean. 00:38:30.470 --> 00:38:31.970 When I was taught the stuff, I never 00:38:31.970 --> 00:38:34.370 got a gut feeling for why or what 00:38:34.370 --> 00:38:36.180 the off diagonal terms meant. 00:38:36.180 --> 00:38:38.494 You don't know it, but we've been using them. 00:38:38.494 --> 00:38:39.910 And then I'll tell you the answer. 00:38:39.910 --> 00:38:50.060 The answer is that when we have a problem like well, 00:38:50.060 --> 00:38:56.000 the motorcycle problem were talking about is this. 00:38:56.000 --> 00:39:00.140 Basically here's the motorcycle wheel spinning around 00:39:00.140 --> 00:39:00.740 and round. 00:39:00.740 --> 00:39:02.920 And it has these two masses. 00:39:02.920 --> 00:39:05.650 This is set up B. So one little mass 00:39:05.650 --> 00:39:08.920 was off to the side of the rim a bit. 00:39:08.920 --> 00:39:11.390 And the other mass was off to the side. 00:39:11.390 --> 00:39:13.030 On the picture it looked like this. 00:39:13.030 --> 00:39:14.680 Here's the axle, motorcycle, and forks 00:39:14.680 --> 00:39:16.350 would be coming down like this. 00:39:16.350 --> 00:39:19.090 And these two little masses. 00:39:19.090 --> 00:39:21.610 Equal distance but opposite sides from one another. 00:39:21.610 --> 00:39:24.880 And if this spins, it puts a heck of a wobble 00:39:24.880 --> 00:39:25.560 into this thing. 00:39:28.930 --> 00:39:34.140 And this puts a moment about this point. 00:39:34.140 --> 00:39:36.342 It tries to make this thing rock back and forth 00:39:36.342 --> 00:39:37.050 as it's spinning. 00:39:37.050 --> 00:39:39.560 It's really hard to hold. 00:39:39.560 --> 00:39:41.996 You hold the axle there. 00:39:41.996 --> 00:39:43.370 And you got to do it so you don't 00:39:43.370 --> 00:39:45.870 get hit by the-- there you go. 00:39:45.870 --> 00:39:47.940 Now tell me if you feel a moment. 00:39:47.940 --> 00:39:50.281 It's really hard to keep that thing straight, right? 00:39:50.281 --> 00:39:52.780 Well that's what it's trying to do to that motorcycle wheel. 00:39:56.460 --> 00:39:57.600 OK. 00:39:57.600 --> 00:40:02.870 For this problem, those off diagonal terms, 00:40:02.870 --> 00:40:07.110 those products of inertia are not 0. 00:40:07.110 --> 00:40:10.440 The product of inertia terms cause these things 00:40:10.440 --> 00:40:13.380 called dynamic imbalances. 00:40:13.380 --> 00:40:17.940 It causes there to be angular-- makes the angular momentum 00:40:17.940 --> 00:40:20.280 terms instead of the angular momentum 00:40:20.280 --> 00:40:25.672 being aligned with the axis of rotation, the rotation vector, 00:40:25.672 --> 00:40:27.130 it's pointed off in this direction. 00:40:30.270 --> 00:40:34.160 Anytime the angular momentum vector and the rotation vector 00:40:34.160 --> 00:40:37.280 are not aligned, you have off diagonal terms 00:40:37.280 --> 00:40:40.060 and you will have dynamic imbalance. 00:40:40.060 --> 00:40:42.100 So there's a physical consequence 00:40:42.100 --> 00:40:44.000 of those off diagonal terms. 00:40:44.000 --> 00:40:46.580 And they explain the dynamic imbalance. 00:40:46.580 --> 00:40:52.904 So let's see if we can't make some headway on that. 00:40:52.904 --> 00:40:53.945 So you've seen the rotor. 00:40:58.360 --> 00:40:59.540 Let's look at two cases. 00:41:06.170 --> 00:41:08.830 One that looks like that, which I just had set up a second ago. 00:41:20.910 --> 00:41:22.180 One that looks like that. 00:41:22.180 --> 00:41:29.710 Call this A, B. And in both cases, 00:41:29.710 --> 00:41:32.990 the rotation is around the vertical axis 00:41:32.990 --> 00:41:36.455 and it's constant at omega. 00:41:39.590 --> 00:41:44.500 And I just mean these to be two different cases. 00:41:44.500 --> 00:41:46.750 I'll make it lowercase so I don't confuse it 00:41:46.750 --> 00:41:50.520 with my coordinate system notation. 00:41:50.520 --> 00:41:55.320 This is going to be point A in both of these problems. 00:41:55.320 --> 00:41:58.560 And it's going to be the origin of a coordinate system. 00:42:09.830 --> 00:42:14.144 So if you cause this to spin, these both have-- 00:42:14.144 --> 00:42:16.615 did I write these masses as m over 2? 00:42:22.850 --> 00:42:26.220 For a moment, let's just think of these as being equal masses. 00:42:26.220 --> 00:42:30.860 If you do this problem, do you think this one will wobble? 00:42:30.860 --> 00:42:34.860 No, it's perfectly balanced. 00:42:34.860 --> 00:42:38.010 And it'll just spin nice and smoothly. 00:42:38.010 --> 00:42:43.010 It has angular momentum around the z-axis, the omega axis. 00:42:43.010 --> 00:42:46.490 It has angular momentum in that direction, 00:42:46.490 --> 00:42:48.570 certainly, when calculated. 00:42:48.570 --> 00:42:54.240 This one has same mass, same distance away from this axis, 00:42:54.240 --> 00:42:55.830 but now one up and one down. 00:42:55.830 --> 00:42:57.870 This one wobbles. 00:42:57.870 --> 00:43:00.450 But this one has a component of angular momentum 00:43:00.450 --> 00:43:03.380 in this direction, which is exactly equal to this one. 00:43:03.380 --> 00:43:08.340 But this also has a component that's in this direction. 00:43:08.340 --> 00:43:10.570 And we're going take a look and see what that is. 00:43:15.100 --> 00:43:18.678 So we're going to do this problem here. 00:43:18.678 --> 00:43:29.450 We're going to analyze B. This case B. And here's the goal. 00:43:29.450 --> 00:43:35.540 The goal is to show you that the angular 00:43:35.540 --> 00:43:42.850 momentum of this system with respect to this point 00:43:42.850 --> 00:43:47.500 can be written as a matrix with constants in it, 00:43:47.500 --> 00:43:51.260 which you can call the mass moment of inertia matrix. 00:43:51.260 --> 00:44:01.390 Times the vector components of the rotation rate. 00:44:01.390 --> 00:44:06.060 Now, this problem, the z-axis will be 00:44:06.060 --> 00:44:10.440 upwards and will only have one component one non 0 component. 00:44:10.440 --> 00:44:15.380 But in general, we want to be able to express the angular 00:44:15.380 --> 00:44:19.462 momentum as a product of this inertia matrix. 00:44:19.462 --> 00:44:21.170 And these are the inertias we'll find out 00:44:21.170 --> 00:44:28.360 with respect to A. Times the vector of angular velocities. 00:44:28.360 --> 00:44:30.860 We've got to be very careful about some definitions. 00:44:42.720 --> 00:44:45.770 So we're going to do this specific problem, 00:44:45.770 --> 00:44:50.380 but we're going to use methods that are completely general. 00:44:50.380 --> 00:44:53.410 So I want to describe the general problem. 00:44:53.410 --> 00:45:07.420 Here is a inertial coordinate system fixed. 00:45:07.420 --> 00:45:13.430 Here's a body out here in space. 00:45:13.430 --> 00:45:26.450 And it is rotating about some point A. So point one 00:45:26.450 --> 00:45:30.230 and the rotation vector, the angular 00:45:30.230 --> 00:45:32.260 rotation and some omega. 00:45:32.260 --> 00:45:36.100 And it's just in some direction. 00:45:36.100 --> 00:45:38.880 And that omega is with respect. 00:45:38.880 --> 00:45:41.670 We always in these angular momentum problems 00:45:41.670 --> 00:45:45.750 define rotation rate with respect to an inertia 00:45:45.750 --> 00:45:46.540 coordinate system. 00:45:49.440 --> 00:45:54.910 Now, this point A. So first carefully 00:45:54.910 --> 00:46:00.865 define A is a fixed point. 00:46:04.867 --> 00:46:06.075 So is that an inertial point? 00:46:09.710 --> 00:46:10.940 Yeah. 00:46:10.940 --> 00:46:12.780 You can do Newton's laws from this point 00:46:12.780 --> 00:46:15.370 just as well as you could any fixed point 00:46:15.370 --> 00:46:18.420 in this inertial reference frame is an inertial point 00:46:18.420 --> 00:46:20.040 and you can use Newton's laws. 00:46:20.040 --> 00:46:21.230 So this is a fixed point. 00:46:21.230 --> 00:46:22.970 I'm defining it that way. 00:46:22.970 --> 00:46:30.540 This body is rotating about that point with this angular rate. 00:46:33.630 --> 00:46:45.220 But attached to the body is a coordinate system 00:46:45.220 --> 00:46:46.630 that rotates with the body. 00:46:46.630 --> 00:46:50.480 So this would be some a xyz coordinate 00:46:50.480 --> 00:46:52.640 system attached to the body. 00:46:52.640 --> 00:46:56.500 So it's like this problem where I've 00:46:56.500 --> 00:46:58.870 got a coordinate system attached to my wheel. 00:46:58.870 --> 00:47:01.780 There's x, here's y, z coming out of it. 00:47:01.780 --> 00:47:03.160 And in a really simple case, it's 00:47:03.160 --> 00:47:05.130 rotating around the z-axis. 00:47:05.130 --> 00:47:08.920 But I can make it rotate around some other axis. 00:47:08.920 --> 00:47:10.997 I pushed a nail through here and I'm 00:47:10.997 --> 00:47:12.330 trying to hold it constant here. 00:47:12.330 --> 00:47:16.520 And now it's rotating about a different axis, right? 00:47:19.440 --> 00:47:23.070 Same rotation rate, but it doesn't 00:47:23.070 --> 00:47:25.130 have to be lined up in any pretty way. 00:47:25.130 --> 00:47:29.200 If I make that thing rotate around that other axis, 00:47:29.200 --> 00:47:31.530 it looks weird, but we can define it. 00:47:31.530 --> 00:47:33.390 And that's what we're talking about here. 00:47:33.390 --> 00:47:37.870 So this body is rotating around, has some rotation 00:47:37.870 --> 00:47:42.800 rate with respect to a reference frame attached to the body. 00:47:46.710 --> 00:47:58.260 So A xyz is a frame that can't-- going to make this go up. 00:48:01.810 --> 00:48:02.500 Come on. 00:48:05.210 --> 00:48:11.770 This is attached to the body. 00:48:16.860 --> 00:48:19.530 And I've drawn them at kind of funny angles here, 00:48:19.530 --> 00:48:23.620 just to emphasize that they're not necessarily 00:48:23.620 --> 00:48:26.240 lined up with these. 00:48:26.240 --> 00:48:28.024 And it's going to rotate. 00:48:28.024 --> 00:48:28.524 OK. 00:48:46.885 --> 00:48:47.385 Omega. 00:49:13.900 --> 00:49:14.890 Just to emphasize. 00:49:14.890 --> 00:49:16.500 It's always in the inertial frame. 00:49:21.020 --> 00:49:26.220 The last point may be confusing to start with. 00:49:26.220 --> 00:49:37.770 Omega measured with respect to O can 00:49:37.770 --> 00:49:49.445 be expressed in terms of the axyz unit vectors. 00:49:56.870 --> 00:49:57.870 We're going to do that. 00:49:57.870 --> 00:50:00.440 It turns out it vastly simplifies the problem 00:50:00.440 --> 00:50:04.100 to express the rotation in the unit 00:50:04.100 --> 00:50:07.600 vectors of the frame attached to the body. 00:50:07.600 --> 00:50:10.190 Remember, that frame is still fit. 00:50:10.190 --> 00:50:14.520 Its origin A is at a fixed point in the inertial frame. 00:50:14.520 --> 00:50:18.360 So it's just the system's going around and inside 00:50:18.360 --> 00:50:22.090 of that system you have a rotation 00:50:22.090 --> 00:50:24.990 and you can break it down into xyz components. 00:50:24.990 --> 00:50:27.520 Just a vector and you can express it in those components. 00:50:27.520 --> 00:50:28.728 That's all we're saying here. 00:50:35.970 --> 00:50:39.920 Now I want to do the motorcycle problem. 00:50:39.920 --> 00:50:43.470 I'm going to just turn it on its side. 00:50:43.470 --> 00:50:45.730 And the reason I'm going to do this specific example, 00:50:45.730 --> 00:50:47.580 the hope here is to actually now give you 00:50:47.580 --> 00:50:50.160 a physical feeling for what's going on. 00:50:50.160 --> 00:50:53.250 We've done a lot of illustrations of it. 00:50:53.250 --> 00:50:55.890 And you know that it produces imbalances. 00:50:55.890 --> 00:51:01.530 So here's my z-axis and my rotation rate. 00:51:04.280 --> 00:51:10.610 Omega with respect to O is some omega in the k hat 00:51:10.610 --> 00:51:14.370 direction in the fixed frame. 00:51:14.370 --> 00:51:19.540 And in this case, it's going to be simpler 00:51:19.540 --> 00:51:23.760 than the general case, so that we can do it 00:51:23.760 --> 00:51:26.660 in a reasonable length of time. 00:51:26.660 --> 00:51:28.195 So actually here's my rod. 00:51:30.820 --> 00:51:40.190 Here's my point A. This is my coordinate system axyz. 00:51:43.440 --> 00:51:47.970 So this is now attached to the body. 00:51:47.970 --> 00:51:53.102 My rigid body is a massless rod with two masses on it. 00:51:56.060 --> 00:52:00.035 And this distance, this is the x. 00:52:02.860 --> 00:52:06.310 Going that way will be a y, which we have little use of. 00:52:06.310 --> 00:52:09.540 There's nothing happening in that direction. 00:52:09.540 --> 00:52:16.770 So this distance here I'll call x1. 00:52:16.770 --> 00:52:20.700 This distance here is z1. 00:52:20.700 --> 00:52:28.769 Over here, this is z2 and x2. 00:52:28.769 --> 00:52:30.310 Now we're going to make this problem. 00:52:30.310 --> 00:52:31.680 We'll substitute a number. 00:52:31.680 --> 00:52:33.380 So this is symmetric. 00:52:33.380 --> 00:52:37.150 So x2 is going to be minus x1 and so forth. 00:52:37.150 --> 00:52:40.400 But we want to keep them separate for the moment 00:52:40.400 --> 00:52:43.620 so you see what happens to different terms. 00:52:43.620 --> 00:52:45.055 OK, so that defines a problem. 00:52:52.710 --> 00:52:55.080 So the coordinates. 00:52:55.080 --> 00:52:57.975 And we'll call this mass m1. 00:52:57.975 --> 00:52:59.950 I'll keep this a little general for a moment. 00:52:59.950 --> 00:53:01.750 And this is m2. 00:53:01.750 --> 00:53:14.800 So m1 is at the coordinates x1, i, 0, and z1 k. 00:53:14.800 --> 00:53:24.640 And m2 is at x2, i, 0, and z2 k. 00:53:29.740 --> 00:53:32.280 Just points in a plane. 00:53:32.280 --> 00:53:39.040 And I want now to compute the-- I 00:53:39.040 --> 00:53:46.980 want to find the angular momentum of this object 00:53:46.980 --> 00:53:49.960 with respect to point a. 00:53:49.960 --> 00:53:52.620 Remember we compute angular momentum in respect to points. 00:53:52.620 --> 00:53:55.120 So I'm going to do it with respect to point A. 00:53:55.120 --> 00:53:57.640 And that's going to be the sum of the angular 00:53:57.640 --> 00:54:00.990 momentum of mass 1 with respect to A, 00:54:00.990 --> 00:54:31.850 plus the angular momentum of mass 2 with respect to A. 00:54:31.850 --> 00:54:37.410 So the angular momentum of any particle i with respect to A 00:54:37.410 --> 00:54:39.340 is r cross p. 00:54:39.340 --> 00:54:41.570 r cross the linear momentum. 00:54:41.570 --> 00:54:50.080 So it's r i with respect to A cross p i. 00:54:50.080 --> 00:54:53.520 Now the p, this is the momentum of the particle. 00:54:53.520 --> 00:54:57.170 That's always with respect to what kind of frame. 00:54:57.170 --> 00:54:59.820 When you compute angular momentum. 00:54:59.820 --> 00:55:01.370 Must be the inertial frame, right? 00:55:01.370 --> 00:55:04.460 So technically to start with, just remind you of that, 00:55:04.460 --> 00:55:05.950 we'd say oh. 00:55:05.950 --> 00:55:08.760 But we've already said our A is a fixed 00:55:08.760 --> 00:55:10.430 point in an inertial frame. 00:55:10.430 --> 00:55:16.840 So it's OK to write r i with respect to A cross, 00:55:16.840 --> 00:55:20.610 in this case, p i with respect to A. They're the same thing. 00:55:20.610 --> 00:55:22.450 These two things are exactly the same thing. 00:55:22.450 --> 00:55:24.230 The momentum measured at any two fixed 00:55:24.230 --> 00:55:26.422 points in an inertial frame is the same. 00:55:26.422 --> 00:55:28.338 Doesn't matter where you're measuring it from. 00:55:32.541 --> 00:55:33.040 OK. 00:55:40.390 --> 00:55:45.580 And we know that p i with respect to A 00:55:45.580 --> 00:55:51.940 now, we'll call it, is the mass i times the velocity of i 00:55:51.940 --> 00:55:56.650 with respect to A. That's just ordinary linear momentum. 00:56:01.890 --> 00:56:06.700 So I need an expression for the velocity of i with respect 00:56:06.700 --> 00:56:08.860 to A. Any point. 00:56:12.930 --> 00:56:14.130 So these are fixed now. 00:56:14.130 --> 00:56:16.400 These are fixed length things. 00:56:16.400 --> 00:56:21.500 The velocity of a moving point is just the derivative 00:56:21.500 --> 00:56:23.940 of the position vector. 00:56:23.940 --> 00:56:28.170 But you have this equation some people call a transport 00:56:28.170 --> 00:56:29.172 equation. 00:56:29.172 --> 00:56:31.130 So the length of this thing's not changing any, 00:56:31.130 --> 00:56:32.750 so it's just going to have one term in it. 00:56:32.750 --> 00:56:33.625 So what's a velocity? 00:56:36.980 --> 00:56:43.328 In vector notation, omega cross. 00:56:43.328 --> 00:56:43.828 All right. 00:56:51.650 --> 00:56:54.400 Right. 00:56:54.400 --> 00:56:55.970 And this could also. 00:56:55.970 --> 00:56:57.790 All right, these are vectors. 00:57:03.290 --> 00:57:10.350 And because I can say that, then I can say hi with respect to A 00:57:10.350 --> 00:57:19.835 is mi riA cross omega with respect 00:57:19.835 --> 00:57:28.090 to O cross ri with respect to A. OK. 00:57:28.090 --> 00:57:28.680 All vectors. 00:57:33.030 --> 00:57:36.472 So any rigid body. 00:57:36.472 --> 00:57:37.430 So here's the link now. 00:57:37.430 --> 00:57:44.010 Here's the jump from points particles to rigid bodies. 00:57:44.010 --> 00:57:47.480 Any rigid body is made up of the whole mess of particles, 00:57:47.480 --> 00:57:49.410 connected rigidly together. 00:57:49.410 --> 00:57:50.560 No relative motion. 00:57:50.560 --> 00:57:52.100 But a whole mass of particles. 00:57:52.100 --> 00:58:00.630 So I can compute the total momentum of a rigid body 00:58:00.630 --> 00:58:06.321 as the summation over all the little particles in it. 00:58:06.321 --> 00:58:06.820 mi. 00:58:10.270 --> 00:58:17.680 riA cross omega with respect to O cross riA. 00:58:17.680 --> 00:58:19.030 Just sum them all up. 00:58:21.740 --> 00:58:23.790 And when you have continuous bodies, 00:58:23.790 --> 00:58:28.110 these summations turn into integrals. 00:58:28.110 --> 00:58:30.400 So you'll find definitions for like there's 00:58:30.400 --> 00:58:34.060 a mass moment of inertia about this axis of this wheel. 00:58:34.060 --> 00:58:36.750 It's mr squared over 2. 00:58:36.750 --> 00:58:42.960 And it comes from the-- and that's 00:58:42.960 --> 00:58:45.190 the number that you have to multiply by omega 00:58:45.190 --> 00:58:47.140 to get the angular momentum. 00:58:47.140 --> 00:58:49.880 So it comes from summing up all these little particles 00:58:49.880 --> 00:58:53.700 in this thing is the total momentum, angular momentum, 00:58:53.700 --> 00:58:55.530 of the object. 00:58:55.530 --> 00:58:57.310 All right, let's do that. 00:58:57.310 --> 00:59:00.940 We're going to do that for our too little masses 00:59:00.940 --> 00:59:07.265 here and see what kind of things result. Oops. 00:59:17.580 --> 00:59:21.180 I want to get my h with respect to A 00:59:21.180 --> 00:59:26.630 is the sum of h1 with respect to A plus h2 with respect to A. 00:59:26.630 --> 00:59:28.340 And I'm just going to use that formula. 00:59:34.420 --> 00:59:36.730 So it's m1. 00:59:36.730 --> 00:59:39.890 So if I were just work out that little vector products there. 00:59:42.460 --> 00:59:43.660 m1. 00:59:43.660 --> 00:59:45.270 here's riA. 00:59:45.270 --> 01:00:02.375 It's x1 i times z1k cross omega zk cross x1i plus z1k. 01:00:06.120 --> 01:00:10.380 And then I have a second term, the m2 term. 01:00:10.380 --> 01:00:27.650 x2i plus z2k omega zk x2i plus z2k. 01:00:27.650 --> 01:00:30.740 So just a lot of little vector terms. 01:00:30.740 --> 01:00:35.290 That is that expression for our two little particles. 01:00:35.290 --> 01:00:40.020 With their specific positions at x1 and z1 and x2 and z2. 01:00:43.130 --> 01:00:46.630 So if I multiply all that out, then I'll 01:00:46.630 --> 01:00:52.445 get the following result. An h with respect to A here. 01:00:56.210 --> 01:01:13.890 It's m1 x1 squared omega zk minus m1 x1 z1 omega 01:01:13.890 --> 01:01:25.100 z in the i hat direction plus an m2 x2 squared omega 01:01:25.100 --> 01:01:38.040 zk minus m2 x2 z2 omega z in the i direction. 01:01:38.040 --> 01:01:41.110 So this is the angular momentum of particle one. 01:01:41.110 --> 01:01:42.940 This is angular momentum of a particle two. 01:01:47.610 --> 01:01:49.340 And I'm going to do a special case. 01:01:54.088 --> 01:01:59.950 And the special case I'm going to let 01:01:59.950 --> 01:02:07.280 m1 equal m2 equal m over 2. 01:02:07.280 --> 01:02:09.790 So they'll do sum to m. 01:02:09.790 --> 01:02:18.380 And x1 equals minus x2 and z1 equals minus z2. 01:02:18.380 --> 01:02:24.190 So they're nice and symmetrically opposite 01:02:24.190 --> 01:02:25.960 like drawn in the picture. 01:02:25.960 --> 01:02:28.600 That I'm making an equal masses in equal distances 01:02:28.600 --> 01:02:30.435 on either side of the origin. 01:02:30.435 --> 01:02:32.810 And that's going to make this thing simplify quite a bit. 01:03:37.330 --> 01:03:39.870 This is of the form. 01:03:39.870 --> 01:03:42.480 This angular momentum vector is of the form 01:03:42.480 --> 01:03:44.740 has three vector components. 01:03:44.740 --> 01:03:48.410 In this particular case, this one's 0. 01:03:48.410 --> 01:03:53.900 And we call the first component, this one here will be hx. 01:03:53.900 --> 01:03:58.274 And this one here is clearly hz, the component 01:03:58.274 --> 01:03:59.065 in the z direction. 01:04:02.690 --> 01:04:06.420 And if we draw, here's our system. 01:04:11.000 --> 01:04:12.500 Here's our coordinate system. 01:04:17.860 --> 01:04:20.070 The coordinate system attached to the body. 01:04:22.640 --> 01:04:28.165 It has a z component of angular momentum positive upwards. 01:04:30.750 --> 01:04:36.650 And it has an x component of angular momentum in the minus 01:04:36.650 --> 01:04:40.140 direction like that. 01:04:40.140 --> 01:04:43.650 When you add them together, you get that. 01:04:43.650 --> 01:04:50.053 So this is h with respect to A. This is hz, this is hx. 01:04:53.720 --> 01:04:55.702 Now, we found this before. 01:04:55.702 --> 01:04:58.160 We didn't talk anything about moments of inertia, anything. 01:04:58.160 --> 01:05:01.350 We just deal in particles earlier as we did problems. 01:05:01.350 --> 01:05:04.830 We found out that when you have this kind of unbalance, 01:05:04.830 --> 01:05:07.790 the direction of the angular momentum vector 01:05:07.790 --> 01:05:11.140 is not in the same direction as the rotation vector. 01:05:11.140 --> 01:05:14.060 In this case, the rotation doesn't make a zk. 01:05:14.060 --> 01:05:15.200 It's like that. 01:05:15.200 --> 01:05:17.470 The vector is going around it. 01:05:17.470 --> 01:05:18.230 Angular momentum. 01:05:46.300 --> 01:05:49.576 Now, in general you would write hx. 01:05:52.830 --> 01:05:54.017 General case. 01:05:54.017 --> 01:05:56.350 And this is what you can pull off, this little two sheet 01:05:56.350 --> 01:05:59.120 handout that you can download and you 01:05:59.120 --> 01:06:00.710 don't have to copy everything. 01:06:00.710 --> 01:06:04.500 This is going to look like an ixx omega 01:06:04.500 --> 01:06:13.505 x plus ixy omega y plus ixz omega z. 01:06:18.280 --> 01:06:21.210 So if we look at that and we look 01:06:21.210 --> 01:06:30.780 at this, this particular case the hx term is this, right? 01:06:30.780 --> 01:06:46.110 So this is the general expression for hx. 01:06:46.110 --> 01:06:48.920 And in this particular case, that 01:06:48.920 --> 01:07:00.140 will look like minus m x1 z1 omega z. 01:07:00.140 --> 01:07:02.870 And this is the piece that's in the i direction. 01:07:02.870 --> 01:07:04.610 That's why we call it hx. 01:07:04.610 --> 01:07:13.340 And this is then ixz omega z. 01:07:13.340 --> 01:07:19.440 So this piece here is what we call ixz. 01:07:19.440 --> 01:07:20.910 It's where it comes from. 01:07:25.320 --> 01:07:26.380 And we can write it. 01:07:26.380 --> 01:07:29.346 So this is our particular case. 01:07:39.560 --> 01:07:45.784 Get this result. And we find there's h in the h. 01:07:45.784 --> 01:07:48.350 y is 0. 01:07:48.350 --> 01:08:02.290 And hz is mx1 squared omega z omega z. 01:08:02.290 --> 01:08:07.050 And that's got to be of the form izz omega z. 01:08:10.710 --> 01:08:15.250 Now, how do you remember what the subscripts mean? 01:08:15.250 --> 01:08:23.050 ixz means this is the h component 01:08:23.050 --> 01:08:29.000 and this is the omega component it's multiplied by. 01:08:29.000 --> 01:08:35.090 So ixz is the product of inertia for hx. 01:08:35.090 --> 01:08:39.970 It's related to rotation in the z component rotation. 01:08:39.970 --> 01:08:41.590 That's what the subscripts mean. 01:08:52.394 --> 01:08:53.888 Maybe I'll do this. 01:09:35.750 --> 01:09:40.770 So in general, if you know what these constants are 01:09:40.770 --> 01:09:46.460 for your rigid body and you know your rotation rate, 01:09:46.460 --> 01:09:49.680 you instantly know your angular momentum. 01:09:49.680 --> 01:09:53.220 These things, the products in moments of inertia, 01:09:53.220 --> 01:09:57.080 are basically cataloged-- you'll find them 01:09:57.080 --> 01:09:59.330 in the back of your textbook-- for all sorts 01:09:59.330 --> 01:10:01.090 of different objects. 01:10:01.090 --> 01:10:11.620 So I know that if you have z in this direction and this thing's 01:10:11.620 --> 01:10:16.160 rotating around the z, hz is the total mass 01:10:16.160 --> 01:10:18.940 of the system times the radius squared 01:10:18.940 --> 01:10:24.890 divided by 2. mr squared over 2 would be izz for this object. 01:10:24.890 --> 01:10:26.840 And for all sorts of objects. 01:10:26.840 --> 01:10:30.150 These are just cataloged values. 01:10:30.150 --> 01:10:32.360 And then there's ways of moving the axes, called 01:10:32.360 --> 01:10:34.990 parallel axis theorems that you've probably run into, 01:10:34.990 --> 01:10:42.370 that allows you then to construct these values from one 01:10:42.370 --> 01:10:45.190 known point to moving the point to someplace else 01:10:45.190 --> 01:10:46.930 and having it move around that. 01:10:46.930 --> 01:10:51.650 So these values are tabulated, calculated, 01:10:51.650 --> 01:10:54.600 with respect to the centers of mass. 01:10:54.600 --> 01:10:57.130 And if you want to have the mass moments of inertia 01:10:57.130 --> 01:10:59.760 with respect to any other point, then you 01:10:59.760 --> 01:11:02.750 will use something which we call a parallel axis theorem, which 01:11:02.750 --> 01:11:04.520 we'll get to in due course. 01:11:08.130 --> 01:11:11.630 Pretty good on timing here. 01:11:11.630 --> 01:11:14.700 A note about textbooks. 01:11:20.840 --> 01:11:21.895 Textbook conventions. 01:11:28.980 --> 01:11:30.820 This I matrix. 01:11:30.820 --> 01:11:41.501 In some they write it ixx ixy ixz and so forth. 01:11:41.501 --> 01:11:42.000 ix. 01:11:45.470 --> 01:11:46.180 No, iyx. 01:11:51.560 --> 01:11:53.670 iyy. 01:11:53.670 --> 01:11:54.170 iyz. 01:11:57.001 --> 01:11:57.500 cx. 01:12:06.130 --> 01:12:07.620 Some write it like that. 01:12:07.620 --> 01:12:10.897 And others write it with all of these with minus signs 01:12:10.897 --> 01:12:11.980 on the off diagonal terms. 01:12:17.110 --> 01:12:26.994 So Hibbler uses the minus signs. 01:12:29.920 --> 01:12:34.540 Williams does not. 01:12:37.110 --> 01:12:40.032 So the diagonal terms are always positive. 01:12:40.032 --> 01:12:41.024 Yeah? 01:12:41.024 --> 01:12:43.780 AUDIENCE: [INAUDIBLE]. 01:12:43.780 --> 01:12:45.840 PROFESSOR: All the off diagonals are negative. 01:12:45.840 --> 01:12:47.020 So this is positive. 01:12:47.020 --> 01:12:49.290 Positive, positive, positive and then 01:12:49.290 --> 01:12:52.752 negative, negative, negative, negative, negative. 01:12:52.752 --> 01:12:54.460 Now there are actually negative-- they'll 01:12:54.460 --> 01:12:58.040 be negative-- the numbers will pop up negative and so forth. 01:12:58.040 --> 01:13:01.250 It's just that in the notation, some authors have adopted 01:13:01.250 --> 01:13:03.660 putting the minus signs here. 01:13:03.660 --> 01:13:07.690 Others have embedded them in the value itself. 01:13:07.690 --> 01:13:15.220 So Williams' notation, he would say that ixz 01:13:15.220 --> 01:13:19.390 is minus m x1 z1 for this body. 01:13:19.390 --> 01:13:23.530 Hibbler would say it's plus and he'd put the minus sign 01:13:23.530 --> 01:13:26.250 in the notation. 01:13:26.250 --> 01:13:27.400 So just beware of that. 01:13:27.400 --> 01:13:28.430 Because all your life you're going 01:13:28.430 --> 01:13:30.430 to run into people saying the product of inertia 01:13:30.430 --> 01:13:33.945 of this thing is and you got to know which way they define it. 01:13:33.945 --> 01:13:34.445 All right. 01:13:43.630 --> 01:13:44.850 Compute torques. 01:13:44.850 --> 01:13:48.290 You just take time to [INAUDIBLE] angular momentum. 01:13:48.290 --> 01:13:50.590 And we'll do that as a last little step next time. 01:13:50.590 --> 01:13:55.130 But you've got the essence of the movement 01:13:55.130 --> 01:13:57.700 from talking about particles to how we're going 01:13:57.700 --> 01:13:59.620 to talk about rigid bodies. 01:13:59.620 --> 01:14:00.830 So you have muddy cards. 01:14:00.830 --> 01:14:03.600 You have two or three minutes. 01:14:03.600 --> 01:14:05.970 Write down what was tough for you here. 01:14:05.970 --> 01:14:08.010 Write down what wasn't. 01:14:08.010 --> 01:14:11.165 And see you next Tuesday. 01:14:11.165 --> 01:14:16.620 Oh, I must say, so this stuff about-- the mass moment 01:14:16.620 --> 01:14:17.570 of inertia matrix. 01:14:17.570 --> 01:14:20.600 That stuff is not on the exam. 01:14:20.600 --> 01:14:26.960 But knowing about particles and particle moments of inertia is.