WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. 00:00:03.800 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.680 To make a donation or to view additional materials 00:00:12.680 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.251 at ocw.mit.edu. 00:00:21.210 --> 00:00:25.760 PROFESSOR: OK, let's get on with today's lecture. 00:00:25.760 --> 00:00:31.850 And I want to look at a variety of different problems, 00:00:31.850 --> 00:00:36.156 different classes of problems. 00:00:36.156 --> 00:00:38.530 We're going to look at four different classes of problems 00:00:38.530 --> 00:00:43.490 and talk about the way you'd go about approaching them. 00:00:43.490 --> 00:00:52.320 We need very few fundamental laws. 00:00:52.320 --> 00:00:55.370 We use the same fundamental laws again and again. 00:01:00.432 --> 00:01:02.890 And the issue is really, how do you go about applying them? 00:01:02.890 --> 00:01:05.660 One is the sum-- this fundamental law's 00:01:05.660 --> 00:01:08.145 for rigid bodies. 00:01:13.990 --> 00:01:17.470 So the fundamental laws, they're always true. 00:01:17.470 --> 00:01:20.190 The sum of the external forces, vectors. 00:01:26.290 --> 00:01:28.260 Time-rated change of the momentum 00:01:28.260 --> 00:01:31.500 of the system with respect to an inertial frame. 00:01:31.500 --> 00:01:36.090 And we recognize this mass times acceleration. 00:01:36.090 --> 00:01:37.740 Newton's second law. 00:01:37.740 --> 00:01:43.875 The other one is the summation of external torques. 00:01:47.340 --> 00:01:51.040 And that one is the time derivative 00:01:51.040 --> 00:01:55.790 of angular momentum with respect to a point which you choose-- 00:01:55.790 --> 00:02:06.960 this is an A-- plus the velocity of A with respect 00:02:06.960 --> 00:02:12.410 to the inertial frame crossed with the momentum with respect 00:02:12.410 --> 00:02:13.730 to the inertial frame. 00:02:13.730 --> 00:02:15.780 So that's the torque equation. 00:02:15.780 --> 00:02:19.650 Now we have two special cases of this where 00:02:19.650 --> 00:02:22.060 the second term goes away. 00:02:22.060 --> 00:02:26.860 One is when you are at the center of gravity, 00:02:26.860 --> 00:02:30.040 in which case by definition the velocity of that point 00:02:30.040 --> 00:02:32.020 and the momentum are in the same direction, 00:02:32.020 --> 00:02:33.910 the cross product disappears. 00:02:33.910 --> 00:02:37.120 So there's two special cases we use all the time 00:02:37.120 --> 00:02:39.980 when it simplifies to that. 00:02:39.980 --> 00:02:42.120 One is when you're at the center of gravity. 00:02:42.120 --> 00:02:44.130 The other is when, for whatever reason, 00:02:44.130 --> 00:02:46.190 this velocity is parallel to that, 00:02:46.190 --> 00:02:47.960 and the cross product goes to zero. 00:02:47.960 --> 00:02:50.510 And sometimes the velocity is just plain zero. 00:02:50.510 --> 00:02:52.410 So there's some certain cases that we 00:02:52.410 --> 00:02:54.359 use lots of times when that term goes away. 00:02:54.359 --> 00:02:55.400 But sometimes it doesn't. 00:02:55.400 --> 00:02:57.090 You just have to put up with it. 00:03:01.080 --> 00:03:02.690 So those are our two fundamental laws. 00:03:02.690 --> 00:03:05.840 And the question really is how to go about applying them. 00:03:08.890 --> 00:03:11.806 So I'm going to look at four classes of problems. 00:03:17.320 --> 00:03:19.090 Let's name them right now. 00:03:19.090 --> 00:03:28.780 Pure-- and we'll do simplest to hardest-- pure rotation 00:03:28.780 --> 00:03:42.150 about a fixed axis through G, through the center of mass. 00:03:42.150 --> 00:03:42.800 Pretty trivial. 00:03:42.800 --> 00:03:45.050 We can all do those kind of problems. 00:03:45.050 --> 00:04:05.630 A second class, pure rotation about a fixed axis at A, 00:04:05.630 --> 00:04:09.680 which is not equal to G not at the center of mass. 00:04:09.680 --> 00:04:11.675 Again, pretty simple problems. 00:04:15.880 --> 00:04:34.420 Third class, no external-- I'll call it 00:04:34.420 --> 00:04:35.516 no external constraints. 00:04:43.660 --> 00:04:46.410 We'll have to give an example to see what I really mean. 00:04:46.410 --> 00:04:51.790 Well, I'll give you an example right now, which we'll do. 00:04:51.790 --> 00:04:57.800 You have this hockey puck with a string attached to it force. 00:04:57.800 --> 00:05:01.780 And this whole thing is on a frictionless surface. 00:05:01.780 --> 00:05:05.760 So it's constrained so it can't go through the surface. 00:05:05.760 --> 00:05:08.670 So no external constraints in the direction 00:05:08.670 --> 00:05:09.840 that the motion can happen. 00:05:09.840 --> 00:05:12.340 So this is a 2D problem. 00:05:12.340 --> 00:05:15.050 Can move in the x, y, and rotation. 00:05:15.050 --> 00:05:16.050 But it's not touching. 00:05:16.050 --> 00:05:17.920 There's no things constraining it 00:05:17.920 --> 00:05:20.146 in the directions of movement, which 00:05:20.146 --> 00:05:21.270 are allowed in the problem. 00:05:21.270 --> 00:05:23.230 That's just too much words to write up here. 00:05:23.230 --> 00:05:24.690 So this kind of problem. 00:05:24.690 --> 00:05:51.935 And the fourth are problems with moving points of constraint. 00:06:07.690 --> 00:06:12.610 You won't see a textbook with sections broken out like this. 00:06:12.610 --> 00:06:14.070 I'm using my own terminology. 00:06:14.070 --> 00:06:15.460 I just made it up last night as I 00:06:15.460 --> 00:06:17.780 was finishing off the lecture. 00:06:17.780 --> 00:06:20.680 But I'm trying to give you some insight, 00:06:20.680 --> 00:06:24.040 and this is the way I think about these things. 00:06:24.040 --> 00:06:27.060 So let's quickly go through examples 00:06:27.060 --> 00:06:29.720 from the first couple of types because they're 00:06:29.720 --> 00:06:30.810 especially easy. 00:06:33.840 --> 00:06:35.110 So what are we saying? 00:06:35.110 --> 00:06:37.110 They're pure rotation about a fixed 00:06:37.110 --> 00:06:46.400 axis through G. Center of mass is somewhere along this axle. 00:06:46.400 --> 00:06:52.420 The axle is fixed so object can spin around it. 00:06:52.420 --> 00:06:57.210 And these kind of problems are particularly simple, 00:06:57.210 --> 00:06:58.850 so I'm not going to dwell on them. 00:06:58.850 --> 00:07:05.080 But here's class one, class one problems. 00:07:05.080 --> 00:07:11.500 And it's basically-- these are rotors, typically rotors, 00:07:11.500 --> 00:07:12.595 of almost all kinds. 00:07:17.120 --> 00:07:25.600 So for these problems, you use momentum, angular momentum, 00:07:25.600 --> 00:07:27.175 with respect to the center of mass. 00:07:29.910 --> 00:07:43.290 You can express it as IG times omega x, omega y, omega z. 00:07:43.290 --> 00:07:45.140 All these problems can be expressed 00:07:45.140 --> 00:07:49.270 with a moment of inertia matrix times the rotation vector. 00:07:49.270 --> 00:07:52.740 And it will give you HX, HY, HZ. 00:07:52.740 --> 00:07:56.080 And you can use that second formula up there. 00:07:56.080 --> 00:07:58.640 The torques are dH dt. 00:07:58.640 --> 00:08:01.890 So it's fixed through G. So the second term doesn't appear. 00:08:04.460 --> 00:08:07.540 So to do these problems, the sum of the-- see, 00:08:07.540 --> 00:08:09.520 I'll just give an example here. 00:08:09.520 --> 00:08:16.415 For 2D planar motion, they get especially simple. 00:08:21.880 --> 00:08:31.170 Then, for 2D planar motion, that means omega here is 0, 00:08:31.170 --> 00:08:37.590 0, and we'll let z be the rotation axis. 00:08:37.590 --> 00:08:49.050 And this H then with respect to G just becomes Izz omega z. 00:08:49.050 --> 00:08:54.090 All those 2D planar motion problems boil down to this. 00:08:54.090 --> 00:09:09.130 And dHG dt then is Izz with respect to G omega z do, 00:09:09.130 --> 00:09:12.705 or more familiar notation. 00:09:15.370 --> 00:09:19.240 So all those planar motion problems, axis 00:09:19.240 --> 00:09:20.820 passing through G like this. 00:09:23.590 --> 00:09:31.300 For those problems, does this matrix 00:09:31.300 --> 00:09:37.400 have to be with respect to principal axes? 00:09:37.400 --> 00:09:41.560 Do you have your XYZ coordinate system? 00:09:41.560 --> 00:09:44.940 To do this style of problem, does 00:09:44.940 --> 00:09:48.869 that actually have to be expressed in principle 00:09:48.869 --> 00:09:50.285 coordinates so that it's diagonal? 00:09:55.510 --> 00:09:57.510 It doesn't have to be. 00:09:57.510 --> 00:10:00.330 You're worried about rotation about z. 00:10:00.330 --> 00:10:02.950 You'll find that that statement's still always true. 00:10:02.950 --> 00:10:05.180 Now, there may be-- the thing could definitely 00:10:05.180 --> 00:10:08.560 have imbalances and have unusual other torques. 00:10:08.560 --> 00:10:11.180 That will fall out in the problem. 00:10:11.180 --> 00:10:16.710 But for the motion around the axis of spin, this [INAUDIBLE]. 00:10:20.650 --> 00:10:24.310 Where you only have a 1 omega z component, just 00:10:24.310 --> 00:10:28.210 the one component, you will get-- it'll work out just fine. 00:10:28.210 --> 00:10:28.892 Yeah. 00:10:28.892 --> 00:10:32.748 AUDIENCE: If you do have some kind of Ixz and Iyz terms, 00:10:32.748 --> 00:10:38.020 you would end up with more-- 00:10:38.020 --> 00:10:39.920 PROFESSOR: You will end up with-- first 00:10:39.920 --> 00:10:43.345 of all, if you have the off-diagonal z terms, xz, 00:10:43.345 --> 00:10:48.630 yz terms, when you multiply that out, you will find components 00:10:48.630 --> 00:10:54.160 of H that are in the z direction as well as perhaps in the x 00:10:54.160 --> 00:10:55.070 and y. 00:10:55.070 --> 00:10:58.790 And those other two components tell you 00:10:58.790 --> 00:11:02.610 that H is not pointing in the same direction as omega, right? 00:11:02.610 --> 00:11:06.900 And that it tells you instantly that the thing is dynamically 00:11:06.900 --> 00:11:11.530 unbalanced and will have other torques that 00:11:11.530 --> 00:11:15.060 are trying to bend that axle. 00:11:15.060 --> 00:11:16.750 So they'll always appear. 00:11:20.872 --> 00:11:21.790 All right. 00:11:21.790 --> 00:11:22.450 Let's move on. 00:11:22.450 --> 00:11:25.650 I want to make sure we get through this today. 00:11:25.650 --> 00:11:26.595 Class two problems. 00:11:32.060 --> 00:11:34.370 These are basically-- these are the pure rotation 00:11:34.370 --> 00:11:37.130 around some point that's not through G. 00:11:37.130 --> 00:11:49.030 And again now, this is a fixed-- key here is fixed axis at A. 00:11:49.030 --> 00:11:50.290 It's not moving. 00:11:50.290 --> 00:11:52.510 This is what makes these problems simpler. 00:11:52.510 --> 00:11:55.200 For these kinds of problems, you do 00:11:55.200 --> 00:11:57.350 the sum of the torques with respect 00:11:57.350 --> 00:11:58.935 to A, the external torques. 00:12:13.060 --> 00:12:16.510 And because that point's fixed, the second terms don't appear, 00:12:16.510 --> 00:12:19.740 the v cross p terms. 00:12:19.740 --> 00:12:27.320 And you can write these as I, a moment of inertia matrix, 00:12:27.320 --> 00:12:37.020 times whatever the rotations are. 00:12:40.410 --> 00:12:43.750 In order to define the mass moment of inertia matrix, 00:12:43.750 --> 00:12:48.930 you must have chosen a set of coordinates 00:12:48.930 --> 00:12:52.350 attached to the body. 00:12:52.350 --> 00:12:54.370 And then with those coordinates, you've 00:12:54.370 --> 00:12:57.490 computed the moments of inertia for the body. 00:12:57.490 --> 00:13:01.170 And if you chose wisely, you get principal coordinates 00:13:01.170 --> 00:13:03.720 and you only get diagonal entries on the matrix. 00:13:03.720 --> 00:13:07.960 If you chose unwisely, you will get other stuff. 00:13:07.960 --> 00:13:12.350 But it's still a valid mass moment of inertia matrix. 00:13:12.350 --> 00:13:14.120 It just gives rise-- you have to deal 00:13:14.120 --> 00:13:15.950 with a bunch of other terms. 00:13:15.950 --> 00:13:22.490 So this will still yield the same answer. 00:13:22.490 --> 00:13:25.170 How do you get I with respect to A? 00:13:29.410 --> 00:13:33.700 These are opportunities when you can use parallel axis. 00:13:33.700 --> 00:13:34.765 Yeah. 00:13:34.765 --> 00:13:40.545 AUDIENCE: Isn't I omega just H dot dH dt? 00:13:40.545 --> 00:13:41.420 PROFESSOR: Excuse me. 00:13:45.280 --> 00:13:46.730 You need to finish that out. 00:13:46.730 --> 00:13:50.250 This is H. So the time derivative of H 00:13:50.250 --> 00:13:52.220 is the time derivative of this expression. 00:13:52.220 --> 00:13:54.450 You have to figure out the moments of inertia 00:13:54.450 --> 00:13:55.970 and the rotation rates. 00:13:55.970 --> 00:14:01.230 And you may get multiple terms, only one of which-- 00:14:01.230 --> 00:14:08.920 let's say that this will work out. 00:14:08.920 --> 00:14:10.230 You will get multiple terms. 00:14:10.230 --> 00:14:15.130 You will get torques that are not in the direction of spin. 00:14:15.130 --> 00:14:18.280 Again, these might be unbalanced. 00:14:18.280 --> 00:14:23.130 On the other hand, it may be, for 2D problems, which 00:14:23.130 --> 00:14:31.960 is common-- so for the 2D planar motion, which 00:14:31.960 --> 00:14:37.430 most of the problems we do are, then what you would do 00:14:37.430 --> 00:14:43.070 is you're saying omega is, say, 0, 0, omega z, 00:14:43.070 --> 00:14:44.640 which simplifies that. 00:14:44.640 --> 00:14:47.800 We're multiplying this thing out quite a lot. 00:14:47.800 --> 00:15:03.990 And if I with respect to G is diagonal, 00:15:03.990 --> 00:15:07.535 then that means you your G you chose principal axes. 00:15:12.740 --> 00:15:16.640 But then how do you get to I with respect to A? 00:15:16.640 --> 00:15:21.690 For 2D problems, really simple ones, 00:15:21.690 --> 00:15:23.791 how do you get to I with respect to A? 00:15:23.791 --> 00:15:24.790 AUDIENCE: Parallel axis. 00:15:24.790 --> 00:15:27.690 PROFESSOR: That's the classic case 00:15:27.690 --> 00:15:31.670 for using the parallel axis theorem. 00:15:31.670 --> 00:15:36.320 So for these 2D planar motion problems-- and planar motion 00:15:36.320 --> 00:15:47.120 problems, then you can use parallel axis. 00:15:47.120 --> 00:15:48.040 I'll do an example. 00:15:48.040 --> 00:15:51.070 So this is kind of the set up for this. 00:15:51.070 --> 00:15:53.330 An example of this on the homework. 00:15:53.330 --> 00:15:57.845 What problem on the homework is just perfect for this? 00:15:57.845 --> 00:16:02.054 It's 2D, planar motion, about a fixed point. 00:16:02.054 --> 00:16:03.720 AUDIENCE: Circle with the square cutout. 00:16:03.720 --> 00:16:07.380 PROFESSOR: Yeah, the cylinder, this [INAUDIBLE] disk 00:16:07.380 --> 00:16:11.520 with the square cutout with the pin at the top turning it 00:16:11.520 --> 00:16:13.440 into a pendulum. 00:16:13.440 --> 00:16:14.470 That's the fixed point. 00:16:17.440 --> 00:16:18.770 You can figure this out. 00:16:18.770 --> 00:16:20.870 You can use parallel axis. 00:16:20.870 --> 00:16:23.600 And we're going to do an example right now that's almost 00:16:23.600 --> 00:16:26.120 identical to that problem. 00:16:26.120 --> 00:16:27.560 And we started it last time. 00:16:27.560 --> 00:16:32.190 So the example I want to work is very similar. 00:16:32.190 --> 00:16:35.190 It's basically this problem is that pendulum. 00:16:38.130 --> 00:16:41.170 So let's just work it through quickly. 00:16:41.170 --> 00:16:44.830 We had done the setup last time. 00:16:58.080 --> 00:17:01.580 So last time I basically derived an example 00:17:01.580 --> 00:17:05.589 of the parallel axis theorem for my little stick here. 00:17:05.589 --> 00:17:12.575 And I'll give you some geometry, some values. 00:17:17.609 --> 00:17:26.690 So here's G. There's the point I wanted to rotate about, A. 00:17:26.690 --> 00:17:31.650 The distance between these two points is d. 00:17:31.650 --> 00:17:33.920 That's going to pop up in my parallel axis theorem. 00:17:36.924 --> 00:17:39.920 We've got a set of coordinates here. 00:17:39.920 --> 00:17:42.140 My G is, of course, in the center of this thing, 00:17:42.140 --> 00:17:43.240 the geometric center. 00:17:43.240 --> 00:17:47.870 So I have a body fixed set of axes, 00:17:47.870 --> 00:17:51.000 which I'm going to call z. 00:17:51.000 --> 00:17:55.520 And my x is in this direction. 00:17:55.520 --> 00:18:00.860 So that would make my y going into the board. 00:18:00.860 --> 00:18:02.720 So the y is kind of like that. 00:18:06.520 --> 00:18:09.400 And this has some properties. 00:18:09.400 --> 00:18:12.870 The dimension in this direction means 00:18:12.870 --> 00:18:17.530 A. The direction in this dimension is B. 00:18:17.530 --> 00:18:30.630 So it's a width of A, a thickness B, and a length L. 00:18:30.630 --> 00:18:34.580 So when you compute, these are-- ah, symmetry now. 00:18:37.830 --> 00:18:40.120 So the axes that I've chosen for this problem 00:18:40.120 --> 00:18:41.175 are they principal axes. 00:18:47.070 --> 00:18:50.160 There's three planes of symmetry in this problem, 00:18:50.160 --> 00:18:52.615 and I've got one principal axis perpendicular 00:18:52.615 --> 00:18:53.840 to every one of them. 00:18:53.840 --> 00:18:56.680 And all three pass through the center of mass, 00:18:56.680 --> 00:18:59.300 and they're orthogonal to one another. 00:18:59.300 --> 00:19:02.160 So those conditions are all satisfied 00:19:02.160 --> 00:19:06.890 for these to be principal axes for this rectangular body 00:19:06.890 --> 00:19:07.973 and its uniform density. 00:19:12.820 --> 00:19:16.580 So I'll give you the-- for bodies like this, 00:19:16.580 --> 00:19:19.060 in your book or any book on dynamics, 00:19:19.060 --> 00:19:26.322 you can look up then the properties, the Izz. 00:19:26.322 --> 00:19:27.780 And we're going to spin this thing. 00:19:27.780 --> 00:19:32.610 This thing, we're going to have it rotating about-- which 00:19:32.610 --> 00:19:34.240 one am I going to use? 00:19:34.240 --> 00:19:36.180 Yeah, around the z-axis. 00:19:36.180 --> 00:19:37.320 That's how I'll set it up. 00:19:37.320 --> 00:19:39.190 That's the way I drilled my hole, 00:19:39.190 --> 00:19:41.760 so it's going back and forth. 00:19:41.760 --> 00:19:44.100 So that the wide part of it's this way. 00:19:47.500 --> 00:19:59.070 So Izz with respect to G M L squared plus a squared over 12. 00:20:03.740 --> 00:20:04.240 Iyy. 00:20:29.090 --> 00:20:35.190 OK, now just to make a point, the dimensions 00:20:35.190 --> 00:20:50.560 L 32.1 centimeters, a 4.71, b, 1.25. 00:20:50.560 --> 00:20:54.990 And eventually my d, this offset would 00:20:54.990 --> 00:21:02.470 be, for example, where I've drilled that hole, is at 10.2. 00:21:02.470 --> 00:21:04.140 The point I want to make here, lots 00:21:04.140 --> 00:21:06.660 of times it would be nice if I could just 00:21:06.660 --> 00:21:10.440 make the simplification to call this a slender rod 00:21:10.440 --> 00:21:14.410 and be able to ignore these a and b dimensions in this, 00:21:14.410 --> 00:21:15.960 just to get quick answers. 00:21:15.960 --> 00:21:19.600 Do you think that's slender enough? 00:21:19.600 --> 00:21:22.780 It's not even 10 times-- the length of the width 00:21:22.780 --> 00:21:23.630 here isn't even 10. 00:21:23.630 --> 00:21:27.330 It's probably six or seven. 00:21:27.330 --> 00:21:29.810 So the key issue then, if you look at this, 00:21:29.810 --> 00:21:33.870 is really what's the ratio of a squared to L squared? 00:21:33.870 --> 00:21:36.650 That would tell you something about the relative importance 00:21:36.650 --> 00:21:39.540 of the a squared term to the L squared. 00:21:39.540 --> 00:21:41.170 So let me tell you about that. 00:21:41.170 --> 00:21:51.350 So a squared over L squared is 0.022. 00:21:51.350 --> 00:21:59.500 b squared over L squared is 0.002. 00:21:59.500 --> 00:22:02.600 So even with this kind of fat stick, 00:22:02.600 --> 00:22:05.460 the approximation of ML squared over 12 is pretty good. 00:22:05.460 --> 00:22:07.500 It's only 2% off. 00:22:07.500 --> 00:22:09.920 And this approximation, L squared over 12, 00:22:09.920 --> 00:22:14.440 is less than 2/10 of a percent off. 00:22:14.440 --> 00:22:17.600 So for roughly slender things, we oftentimes 00:22:17.600 --> 00:22:20.880 just say ML squared over 12 for spin about their center. 00:22:26.410 --> 00:22:29.370 We now need to apply parallel axis. 00:22:29.370 --> 00:22:32.410 I want to spin this around, let this rotate around, 00:22:32.410 --> 00:22:36.000 not around G, but now around a point off to the side. 00:22:36.000 --> 00:22:43.790 So we worked out last time that Izz with respect to A 00:22:43.790 --> 00:22:50.055 is IzzG plus Md squared. 00:23:10.520 --> 00:23:12.730 So in this particular problem, this Izz about G 00:23:12.730 --> 00:23:16.100 is approximately ML squared over 12. 00:23:16.100 --> 00:23:20.500 So just by way of example, to see what we might find out 00:23:20.500 --> 00:23:25.780 here, is let's let d equal L/2. 00:23:25.780 --> 00:23:28.560 That would be if I move this-- if I 00:23:28.560 --> 00:23:30.300 put my hole right at the very top, 00:23:30.300 --> 00:23:31.466 how would this thing behave? 00:23:31.466 --> 00:23:34.810 I don't have a hole right at the top, but I have one close. 00:23:34.810 --> 00:23:38.310 So this is just because the numbers are easy to work. 00:23:38.310 --> 00:23:41.850 What happens if you put in L/2 into this formula? 00:23:41.850 --> 00:23:53.650 Well then IzzA is ML squared over 12 plus M. L over 2 00:23:53.650 --> 00:23:55.150 squared is L squared over 4. 00:23:59.600 --> 00:24:04.510 And that's ML squared over 3, which 00:24:04.510 --> 00:24:06.960 is a number you'll run into again and again and again 00:24:06.960 --> 00:24:09.680 in mechanical engineering because examples like this 00:24:09.680 --> 00:24:10.350 are used a lot. 00:24:10.350 --> 00:24:13.630 The mass moment of inertia about a slender rod 00:24:13.630 --> 00:24:15.980 pinned at its end, ML squared over 3. 00:24:34.170 --> 00:24:38.380 So let's take this problem a little more towards completion. 00:24:38.380 --> 00:24:47.275 The sum of the external moments with respect to A dHA dt. 00:24:52.580 --> 00:24:56.580 And that's going to be d by dt. 00:24:56.580 --> 00:25:02.420 In this problem, the only rotation is omega z. 00:25:02.420 --> 00:25:10.380 So this is going to be Izz with respect to A omega z. 00:25:13.090 --> 00:25:22.380 And that just gives us IzzA omega z dot, or more familiar, 00:25:22.380 --> 00:25:28.460 IzzA theta double bond, if you want. 00:25:28.460 --> 00:25:29.730 Here's our problem. 00:25:29.730 --> 00:25:33.770 It simplifies to this slender rod. 00:25:33.770 --> 00:25:38.600 And let me do the more general case. 00:25:38.600 --> 00:25:39.970 Here's my rod. 00:25:39.970 --> 00:25:44.300 The pivot point that it's going around is here. 00:25:44.300 --> 00:25:50.310 This is d still, and this is G. And it's swinging with respect 00:25:50.310 --> 00:25:51.100 to this point. 00:25:51.100 --> 00:25:56.230 So here's the angle theta. 00:25:56.230 --> 00:25:59.160 So it swings about this point that you've fixed, 00:25:59.160 --> 00:26:06.490 and that point is d above the center of gravity, 00:26:06.490 --> 00:26:07.370 center of mass. 00:26:12.710 --> 00:26:18.670 So what are the external moments about this point? 00:26:18.670 --> 00:26:20.960 There's no torque right at the point, 00:26:20.960 --> 00:26:26.180 but our free body diagram of drawing this as a-- 00:26:26.180 --> 00:26:28.840 here's our point of rotation. 00:26:28.840 --> 00:26:33.590 Here it is displaced through an angle theta. 00:26:33.590 --> 00:26:36.835 The weight of the object can all be concentrated, 00:26:36.835 --> 00:26:39.210 thought of, for the purposes of the free body 00:26:39.210 --> 00:26:42.950 diagram as acting through G. 00:26:42.950 --> 00:26:49.970 So here's the mass at G, gravity acting down on it. 00:26:49.970 --> 00:26:56.980 And the length of this arm here about which it's swinging is D. 00:26:56.980 --> 00:27:04.160 So the torque about this is-- and it's pulling it back-- 00:27:04.160 --> 00:27:12.297 minus Mgd sine theta. 00:27:12.297 --> 00:27:14.130 Probably you've seen that many times before, 00:27:14.130 --> 00:27:17.960 including the recent homework. 00:27:17.960 --> 00:27:25.650 And that must be equal to Izz about A theta double dot. 00:27:25.650 --> 00:27:29.190 So we have an equation of motion. 00:27:29.190 --> 00:27:30.695 Just collecting the terms together. 00:27:40.900 --> 00:27:43.900 So this is a oscillator that, for this problem, 00:27:43.900 --> 00:27:47.260 has no external excitation. 00:27:47.260 --> 00:27:50.570 This is its equation of motion. 00:27:50.570 --> 00:27:51.850 And I need theta double dot. 00:27:54.670 --> 00:27:55.610 But is it linear? 00:27:59.714 --> 00:28:00.500 Is it linear? 00:28:00.500 --> 00:28:03.230 No, it's not a linear equation of motion. 00:28:03.230 --> 00:28:05.260 But for sure, it is an oscillator. 00:28:08.180 --> 00:28:12.190 And for anything that vibrates, you 00:28:12.190 --> 00:28:14.630 can have lots of nonlinear problems 00:28:14.630 --> 00:28:16.940 that exhibit vibration. 00:28:16.940 --> 00:28:20.280 You can think of them-- you can pose problems 00:28:20.280 --> 00:28:25.280 where you say, OK, what's their static equilibrium position? 00:28:25.280 --> 00:28:28.440 And think of a very small motion about that static equilibrium 00:28:28.440 --> 00:28:29.270 position. 00:28:29.270 --> 00:28:33.570 You can always linearize about the static equilibrium position 00:28:33.570 --> 00:28:36.270 and be able to come up with a linearized equation of motion 00:28:36.270 --> 00:28:40.090 that at least from that you can calculate the natural frequency 00:28:40.090 --> 00:28:45.710 of the system for small motions around its static equilibrium 00:28:45.710 --> 00:28:46.210 position. 00:28:46.210 --> 00:28:48.420 So in this case, it's pretty easy to do. 00:28:48.420 --> 00:28:55.215 And you've seen it before for small theta. 00:28:58.450 --> 00:29:01.070 Sine theta is approximately equal to theta. 00:29:01.070 --> 00:29:04.630 So we are going to linearize the equation. 00:29:04.630 --> 00:29:09.140 This theta equals 0 is a static equilibrium position. 00:29:09.140 --> 00:29:11.710 So we're linearizing around 0. 00:29:11.710 --> 00:29:15.280 And around 0, that's the approximation. 00:29:15.280 --> 00:29:21.230 So you just substitute that in, IzzA theta double 00:29:21.230 --> 00:29:26.960 dot plus Mgd theta equals 0. 00:29:26.960 --> 00:29:29.160 There's your linearized equation of motion. 00:29:29.160 --> 00:29:31.980 I want an estimate of the natural frequency. 00:29:31.980 --> 00:29:36.300 So find omega n. 00:29:41.010 --> 00:29:42.870 So this is basically entering into solving 00:29:42.870 --> 00:29:44.220 differential equations. 00:29:44.220 --> 00:29:47.800 But I let mother nature tell me what the answer is. 00:29:47.800 --> 00:29:50.230 I do the example, and I say it oscillates. 00:29:50.230 --> 00:29:54.297 Looks a lot like sine omega t to me. 00:29:54.297 --> 00:29:55.630 Plug in [INAUDIBLE] to make a t. 00:29:55.630 --> 00:29:56.390 Let's find out. 00:30:00.450 --> 00:30:06.080 Some theta amplitude sine omega t. 00:30:06.080 --> 00:30:08.110 Plug it in. 00:30:08.110 --> 00:30:13.720 So you plug that in [INAUDIBLE] and you get minus omega squared 00:30:13.720 --> 00:30:21.860 IzzA plus Mgd. 00:30:21.860 --> 00:30:24.700 And all of this, you can factor out the theta 00:30:24.700 --> 00:30:30.400 0 sine omega t equals 0. 00:30:30.400 --> 00:30:32.150 That's what you get. 00:30:32.150 --> 00:30:34.240 And in general, theta, that's not 0, or else 00:30:34.240 --> 00:30:37.590 you'd have a trivial problem, not moving at all. 00:30:37.590 --> 00:30:41.840 But in order for this equation-- and this can be anything. 00:30:41.840 --> 00:30:45.360 We're doing the 0 minus 1 and plus 1 depending on the time. 00:30:45.360 --> 00:30:47.650 So in order to satisfy this equation, 00:30:47.650 --> 00:30:54.150 this part inside of the parentheses has to be 0. 00:30:54.150 --> 00:30:58.120 And when you just solve that, you 00:30:58.120 --> 00:31:04.365 find that omega squared equals Mgd/IzzA. 00:31:12.600 --> 00:31:17.490 And the reason I've gone to the bother of working this out 00:31:17.490 --> 00:31:22.900 in detail right to the end is that every one degree 00:31:22.900 --> 00:31:28.400 of freedom rotational oscillator that you will ever encounter-- 00:31:28.400 --> 00:31:40.080 sticks, wheels with static imbalances. 00:31:40.080 --> 00:31:41.148 Let me show you this one. 00:31:45.730 --> 00:31:47.650 It's an oscillator too. 00:31:47.650 --> 00:31:51.300 Any one degree of freedom oscillator, 00:31:51.300 --> 00:31:53.840 rotational oscillator, pendulum-- basically 00:31:53.840 --> 00:31:56.412 all pendula-- this is the formula 00:31:56.412 --> 00:31:57.495 for the natural frequency. 00:32:02.610 --> 00:32:06.781 So it's going to be of that form for any pendulum. 00:32:06.781 --> 00:32:11.465 So any 1dof pendulum. 00:32:14.740 --> 00:32:17.000 This is the generic answer. 00:32:19.520 --> 00:32:23.920 So that cutout problem for today has to come down to this. 00:32:23.920 --> 00:32:27.370 Or this is the distance between the mass center 00:32:27.370 --> 00:32:30.320 and the point of rotation. 00:32:30.320 --> 00:32:31.960 And that's your mass moment of inertia 00:32:31.960 --> 00:32:34.096 about the point of rotation. 00:32:34.096 --> 00:32:38.110 So that's worth knowing that one. 00:32:38.110 --> 00:32:39.770 OK, keep moving. 00:32:49.913 --> 00:32:53.120 Now things will begin to get interesting. 00:32:53.120 --> 00:32:59.340 These latter two classes are harder conceptually, 00:32:59.340 --> 00:33:01.590 but once you have a solution method for them, 00:33:01.590 --> 00:33:03.690 they're not all that hard. 00:33:03.690 --> 00:33:08.080 This one is the problem I described at the beginning. 00:33:08.080 --> 00:33:10.030 We've got this hockey puck like thing. 00:33:13.270 --> 00:33:16.640 And the string wrapped around it pulling on it 00:33:16.640 --> 00:33:17.840 with a known force. 00:33:17.840 --> 00:33:21.680 In this problem, they call it 150 newtons. 00:33:21.680 --> 00:33:27.086 The mass of this thing is 75 kilograms. 00:33:30.870 --> 00:33:33.135 It's on a frictionless surface. 00:33:36.550 --> 00:33:46.350 And we want to find its acceleration 00:33:46.350 --> 00:33:51.800 of the center of mass and the rotation 00:33:51.800 --> 00:33:54.380 around the center of mass. 00:33:54.380 --> 00:33:58.280 So find theta double dot and find the linear acceleration. 00:33:58.280 --> 00:34:00.960 That's basically the name of the problem. 00:34:00.960 --> 00:34:06.630 And they give you that it's 75 kilograms, 150 newtons, 00:34:06.630 --> 00:34:15.449 and kappa, the radius of gyration, is 0.15 meters. 00:34:15.449 --> 00:34:21.404 This is defined as the radius of gyration. 00:34:24.870 --> 00:34:28.320 So what's that? 00:34:28.320 --> 00:34:29.420 It's a radius of gyration. 00:34:29.420 --> 00:34:34.320 It's really appropriate, really only useful, 00:34:34.320 --> 00:34:39.949 for mostly 2D rotational problems 00:34:39.949 --> 00:34:41.960 around an axis of rotation. 00:34:41.960 --> 00:34:51.480 And what it means is this is the distance away from the center 00:34:51.480 --> 00:34:56.139 of rotation at which you could concentrate all of the mass 00:34:56.139 --> 00:35:01.580 and have the same mass moment of inertia. 00:35:01.580 --> 00:35:06.260 So this has-- here's G here at the center, uniform disk. 00:35:09.040 --> 00:35:17.330 We know that there is an Izz about G for this problem. 00:35:17.330 --> 00:35:19.260 And I need to pick a coordinate system so we 00:35:19.260 --> 00:35:20.870 can talk about things here. 00:35:20.870 --> 00:35:23.570 Let me get M out of the center. 00:35:23.570 --> 00:35:27.510 So I'm going to let z be upwards. 00:35:27.510 --> 00:35:30.690 And because I'm looking ahead and want 00:35:30.690 --> 00:35:36.140 to keep the equation simple, I'm going to make my x-axis 00:35:36.140 --> 00:35:41.590 here parallel to f so I only have to deal with one vector 00:35:41.590 --> 00:35:43.100 component equation. 00:35:43.100 --> 00:35:45.830 And then that makes the y-axis this way. 00:35:51.550 --> 00:35:56.280 So I'm interested in Izz because I'm spinning around the z-axis. 00:35:56.280 --> 00:35:59.100 I know that for a uniform disk, that's a principal axis, 00:35:59.100 --> 00:36:01.870 is the vertical one. 00:36:01.870 --> 00:36:05.830 And what I'm saying is that you can then set find. 00:36:05.830 --> 00:36:12.990 There's an IzzG that can be expressed as M kappa squared. 00:36:12.990 --> 00:36:21.700 So kappa, in effect, is just IzzG over M square root. 00:36:21.700 --> 00:36:23.450 Now, why do we use that kind of thing? 00:36:23.450 --> 00:36:28.009 Well, the way this problem was set up-- I 00:36:28.009 --> 00:36:29.300 actually took it out of a book. 00:36:29.300 --> 00:36:32.170 It wasn't a uniform disk. 00:36:32.170 --> 00:36:34.230 It's a pulley wheel or something. 00:36:34.230 --> 00:36:40.760 And it's got spokes in here and a rim. 00:36:40.760 --> 00:36:47.577 It's still axially-- it still has some axial symmetries. 00:36:47.577 --> 00:36:48.910 But it's getting a little messy. 00:36:48.910 --> 00:36:50.326 It's hard for you-- you can't just 00:36:50.326 --> 00:36:52.620 say that's MR squared over 2. 00:36:52.620 --> 00:36:56.250 It's got some other mass moment of inertia about the center. 00:36:56.250 --> 00:36:59.670 But it's got holes and stuff in it because of the spokes. 00:36:59.670 --> 00:37:06.010 So oftentimes, you'll be given the radius of gyration 00:37:06.010 --> 00:37:08.650 because it's a little difficult to give you 00:37:08.650 --> 00:37:13.316 a mathematical description of what the actual Izz is. 00:37:13.316 --> 00:37:14.440 That's often why you do it. 00:37:14.440 --> 00:37:17.346 And the thing is, you can measure. 00:37:17.346 --> 00:37:18.720 Rather than try to calculate, you 00:37:18.720 --> 00:37:21.320 can actually just measure the mass moment 00:37:21.320 --> 00:37:22.960 of inertia of something. 00:37:22.960 --> 00:37:26.060 So how would I measure-- let's say I didn't know any formulas, 00:37:26.060 --> 00:37:28.510 but how would I measure the mass moment of inertia 00:37:28.510 --> 00:37:30.052 of this in the z direction? 00:37:36.450 --> 00:37:38.055 What experiment would you do? 00:37:38.055 --> 00:37:41.310 AUDIENCE: [INAUDIBLE] angular acceleration. 00:37:41.310 --> 00:37:42.740 PROFESSOR: She says apply torque. 00:37:42.740 --> 00:37:44.830 Measure the angular acceleration. 00:37:44.830 --> 00:37:46.620 Hang a weight off of it, known weight. 00:37:46.620 --> 00:37:48.590 Wrap a string around it. 00:37:48.590 --> 00:37:49.400 Known mass. 00:37:49.400 --> 00:37:52.120 Known G. Known torque around the center. 00:37:52.120 --> 00:37:54.620 Measure the angular acceleration. 00:37:54.620 --> 00:37:56.955 I theta double dot equals the torque. 00:37:56.955 --> 00:37:59.505 You know the torque, you know the measure 00:37:59.505 --> 00:38:00.810 of the theta double dot. 00:38:00.810 --> 00:38:03.910 Calculate I. I equals M kappa squared, and you could just 00:38:03.910 --> 00:38:06.910 say, well, the kappa for this system is. 00:38:06.910 --> 00:38:08.900 That's how you use it. 00:38:08.900 --> 00:38:11.380 I'll give you a very common example, really hard 00:38:11.380 --> 00:38:13.180 to calculate mass moment of inertia. 00:38:13.180 --> 00:38:15.679 A marine propeller. 00:38:15.679 --> 00:38:17.970 You actually do want to know the mass moment of inertia 00:38:17.970 --> 00:38:21.400 about its center for purposes of torsional oscillations 00:38:21.400 --> 00:38:23.290 on the shaft, et cetera. 00:38:23.290 --> 00:38:24.620 Hard to calculate. 00:38:24.620 --> 00:38:25.590 Pretty easy to measure. 00:38:28.140 --> 00:38:32.700 So how many degrees of freedom does this problem has? 00:38:32.700 --> 00:38:36.980 Well, when we say it's 2D, it's a rigid body. 00:38:36.980 --> 00:38:40.870 But it's 2D, which means it's lying on a plane. 00:38:40.870 --> 00:38:42.290 It's a planar motion problem. 00:38:42.290 --> 00:38:44.390 Only allowed to rotate in z. 00:38:44.390 --> 00:38:48.280 Not allowed to rotate in around the x-axis or the y-axis. 00:38:48.280 --> 00:38:52.570 So for the rigid body, six degrees of freedom possible. 00:38:52.570 --> 00:38:55.340 There's two immediately that you said it can't rotate. 00:38:55.340 --> 00:38:57.770 So we're down to four. 00:38:57.770 --> 00:39:06.070 It cannot translate in the z direction. 00:39:06.070 --> 00:39:09.750 We're down to three. 00:39:09.750 --> 00:39:11.300 So that leaves us what? 00:39:14.120 --> 00:39:18.340 So the degrees of freedom for this problem is 00:39:18.340 --> 00:39:20.822 6 minus 3 constraints is 3. 00:39:20.822 --> 00:39:23.030 That means we have to have three equations of motion. 00:39:23.030 --> 00:39:25.520 And they would account for what are the possible-- 00:39:25.520 --> 00:39:26.895 in other words, saying this, what 00:39:26.895 --> 00:39:29.170 are the possible motions now of this problem? 00:39:29.170 --> 00:39:30.520 AUDIENCE: [INAUDIBLE]. 00:39:30.520 --> 00:39:31.935 PROFESSOR: x, y, and z. 00:39:31.935 --> 00:39:33.030 Or rotation in z. 00:39:52.260 --> 00:39:54.140 So notice we've set the problem up. 00:39:54.140 --> 00:40:01.160 Now to go about solving it, we need a free body diagram. 00:40:01.160 --> 00:40:03.530 So here's my disk. 00:40:03.530 --> 00:40:04.650 Here's the force. 00:40:08.660 --> 00:40:13.420 Any other-- and there certainly has weight in the z direction, 00:40:13.420 --> 00:40:17.320 but there's no z acceleration. 00:40:17.320 --> 00:40:20.240 So in the plane of the board, that's 00:40:20.240 --> 00:40:21.970 the only external forces acting on this. 00:40:25.540 --> 00:40:28.690 And there's our G. 00:40:28.690 --> 00:40:31.510 Now this problem-- and I'll say generalize 00:40:31.510 --> 00:40:32.580 on this in a few minutes. 00:40:32.580 --> 00:40:40.030 This problem can always be restated as-- recast, 00:40:40.030 --> 00:40:42.710 let me put it that way-- as, here's your point 00:40:42.710 --> 00:40:49.060 G with a force acting on this center of mass. 00:40:49.060 --> 00:40:52.209 See, this force doesn't go through the center of mass. 00:40:52.209 --> 00:40:54.000 This force goes through the center of mass. 00:40:54.000 --> 00:40:58.070 I'm going to replace that problem with this problem. 00:40:58.070 --> 00:41:02.760 A pure moment acting about the center of mass. 00:41:02.760 --> 00:41:04.600 You can always make this transition. 00:41:04.600 --> 00:41:07.490 And I'll do the general case for you in just a minute. 00:41:07.490 --> 00:41:08.690 But you can always do this. 00:41:11.230 --> 00:41:14.330 So that's kind of my second point here. 00:41:14.330 --> 00:41:19.810 Third point, we need then-- this is our free body diagram. 00:41:19.810 --> 00:41:25.211 We need apply our laws of motion. 00:41:25.211 --> 00:41:31.420 So sum of the forces in the y are-- now remember, 00:41:31.420 --> 00:41:33.880 this is z coming out of the board. 00:41:33.880 --> 00:41:36.930 y this way, x this way. 00:41:36.930 --> 00:41:40.210 Sum of the forces in the y? 00:41:40.210 --> 00:41:45.570 Zero, M acceleration of G in the y direction, zero. 00:41:45.570 --> 00:41:48.020 So you know there's no acceleration in the y. 00:41:48.020 --> 00:41:49.910 So it has three degrees of freedom. 00:41:49.910 --> 00:41:52.340 That's the first equation of motion. 00:41:52.340 --> 00:41:57.020 It gives you a trivial result. So y double dot is zero. 00:41:57.020 --> 00:41:59.647 That's your first equation of motion. 00:41:59.647 --> 00:42:01.480 Then you have the second equation of motion, 00:42:01.480 --> 00:42:04.620 sum of the forces in the x direction 00:42:04.620 --> 00:42:09.090 equals just our F i hat. 00:42:09.090 --> 00:42:11.480 It's positive x direction because we 00:42:11.480 --> 00:42:15.310 were clever in how we set up the coordinate system. 00:42:15.310 --> 00:42:21.450 And that's got to be Mx double dot, i hat direction. 00:42:21.450 --> 00:42:26.020 So we know right away that x double dot is the force 00:42:26.020 --> 00:42:33.410 F divided by M. And that's 150 newtons over 75 kilograms, 00:42:33.410 --> 00:42:37.740 or 2 meters per second squared. 00:42:37.740 --> 00:42:38.920 All right. 00:42:38.920 --> 00:42:48.600 The third one, then, is the moment 00:42:48.600 --> 00:42:52.580 of inertia with respect to G in this problem-- 00:42:52.580 --> 00:43:02.254 excuse me, the angular momentum in some IzzG times omega z. 00:43:02.254 --> 00:43:03.670 And that's what we're looking for. 00:43:03.670 --> 00:43:10.920 So this is another way of-- IzzG theta dot k hat direction. 00:43:16.800 --> 00:43:20.680 And we're going to apply that the external torques, some 00:43:20.680 --> 00:43:28.000 of the torques, is dH respect to G dt. 00:43:28.000 --> 00:43:33.620 And that's going to give us IzzG theta double dot. 00:43:38.720 --> 00:43:40.690 So this third class of problems you 00:43:40.690 --> 00:43:45.110 are best just working with respect to the center of mass. 00:43:45.110 --> 00:43:46.360 That's kind of the point here. 00:43:46.360 --> 00:43:51.550 There's no points of contact. 00:43:51.550 --> 00:43:54.110 There's just known external forces. 00:43:54.110 --> 00:43:55.650 You have to deal with them. 00:43:55.650 --> 00:43:59.360 Do your work with respect to the center of mass. 00:43:59.360 --> 00:44:02.280 So we have force equations. 00:44:02.280 --> 00:44:04.720 We have moment equations. 00:44:04.720 --> 00:44:10.350 And basically you know Izz for this problem 00:44:10.350 --> 00:44:12.460 is kappa squared M. And you're given M, 00:44:12.460 --> 00:44:14.200 and you're given kappa. 00:44:14.200 --> 00:44:18.790 So you can now-- and what we have to-- actually, 00:44:18.790 --> 00:44:21.090 the last thing left here is to figure out the torque. 00:44:21.090 --> 00:44:24.400 What's the torque? 00:44:24.400 --> 00:44:29.410 Well, it's R in this direction crossed 00:44:29.410 --> 00:44:30.990 with F in that direction. 00:44:30.990 --> 00:44:41.670 So it's Rj cross with Fi, j cross i, minus RF, 00:44:41.670 --> 00:44:48.320 minus RF, k hat. 00:44:48.320 --> 00:44:57.160 So you can now solve for theta double dot as RF over Izz, 00:44:57.160 --> 00:44:59.480 or RF over M kappa squared. 00:45:22.620 --> 00:45:26.890 And that says-- minus says it's rotating this way, which 00:45:26.890 --> 00:45:28.430 is what you'd expect. 00:45:28.430 --> 00:45:30.570 Right? 00:45:30.570 --> 00:45:32.990 Now I meant to ask you a question before we started. 00:45:32.990 --> 00:45:34.222 But think about this. 00:45:34.222 --> 00:45:36.180 If I had, right at the beginning, had said, OK, 00:45:36.180 --> 00:45:37.020 this is a problem. 00:45:39.850 --> 00:45:43.220 If I grab this string and pull in this direction, 00:45:43.220 --> 00:45:45.717 will there be any motion in the y direction? 00:45:45.717 --> 00:45:46.800 I meant to start that way. 00:45:46.800 --> 00:45:48.930 I'm really kicking myself for not doing that. 00:45:48.930 --> 00:45:50.680 Because a lot of people think that there's possible, 00:45:50.680 --> 00:45:52.420 that it could move off in the y direction 00:45:52.420 --> 00:45:56.480 because it's not being loaded symmetrically. 00:45:56.480 --> 00:45:58.100 You're pulling on a side. 00:45:58.100 --> 00:46:01.127 Some people think it'll kind of try to move away like that. 00:46:01.127 --> 00:46:01.960 It doesn't, does it? 00:46:21.200 --> 00:46:24.026 So an important generalization. 00:46:24.026 --> 00:46:25.315 We've got a rigid body. 00:46:28.780 --> 00:46:30.310 You have a force acting on it. 00:46:33.390 --> 00:46:34.910 Has a mass center here. 00:46:37.460 --> 00:46:43.150 So perpendicular to that force is some distance. 00:46:43.150 --> 00:46:43.980 We'll call it d. 00:46:46.880 --> 00:46:53.450 You can always equate this problem to-- and set it up 00:46:53.450 --> 00:46:59.306 as-- a force. 00:46:59.306 --> 00:47:00.680 Conceptually, you can think of it 00:47:00.680 --> 00:47:02.544 as equal and opposite forces. 00:47:02.544 --> 00:47:03.710 But it's cancel one another. 00:47:03.710 --> 00:47:06.580 It's just like I've done nothing to this problem. 00:47:06.580 --> 00:47:12.880 And then a force acting at this distance F. 00:47:12.880 --> 00:47:14.060 This is our distance d. 00:47:14.060 --> 00:47:16.520 So this problem is identical to that problem. 00:47:16.520 --> 00:47:19.095 I've just added and taken away two more forces. 00:47:23.510 --> 00:47:25.450 So the total forces on the system 00:47:25.450 --> 00:47:31.030 are still F. And there's still an F operating at a lever arm 00:47:31.030 --> 00:47:32.310 d. 00:47:32.310 --> 00:47:36.980 But now, if I had put these two together, 00:47:36.980 --> 00:47:38.600 they are equal and opposite. 00:47:38.600 --> 00:47:43.630 And they form a couple, a moment, acting like that. 00:47:43.630 --> 00:47:53.170 So this is equivalent to-- there's G with an F on it 00:47:53.170 --> 00:47:59.480 and a moment M0. 00:47:59.480 --> 00:48:04.750 And this M0 is my D cross F. 00:48:04.750 --> 00:48:08.240 So that's the generalization for what I did up here. 00:48:08.240 --> 00:48:10.730 I went from that to that. 00:48:10.730 --> 00:48:12.260 And this is why you can do that. 00:48:15.400 --> 00:48:22.180 And so now if you have an object and lots of different forces 00:48:22.180 --> 00:48:26.570 acting on it, and this is Fi down here and here's G, 00:48:26.570 --> 00:48:32.770 you can draw a radius from G to this point. 00:48:32.770 --> 00:48:40.480 So I'll call that Ri with respect to G. Then 00:48:40.480 --> 00:48:48.840 the way to generalize this is that this is equal to some F 00:48:48.840 --> 00:49:00.050 total and a moment acting at G. 00:49:00.050 --> 00:49:05.170 And all that you have to do there is F total 00:49:05.170 --> 00:49:09.430 is the summation of the Fi's, vector sum. 00:49:09.430 --> 00:49:13.400 And the MG, the M with respect to G here, 00:49:13.400 --> 00:49:21.010 is the summation of the Ri with respect to G cross Fi. 00:49:21.010 --> 00:49:26.070 So that's the generalization for multiple forces on a body. 00:49:26.070 --> 00:49:30.940 So you are making an equivalent force acting at G and a moment 00:49:30.940 --> 00:49:33.300 acting at G. I shouldn't call this little g. 00:49:33.300 --> 00:49:34.300 That's really confusing. 00:49:36.980 --> 00:49:38.660 So that's the generalization when you 00:49:38.660 --> 00:49:43.080 need to do problems like this. 00:50:03.060 --> 00:50:08.950 Catch your breath while I scrub a board here. 00:50:08.950 --> 00:50:12.365 Now we've got to move on to these class four problems. 00:50:16.500 --> 00:50:17.155 Moving points. 00:50:30.070 --> 00:50:32.624 An example is the truck problem that you had. 00:50:38.700 --> 00:50:39.535 Known acceleration. 00:50:45.990 --> 00:50:50.670 So there's two common ways of doing this problem. 00:50:50.670 --> 00:50:53.470 You can do this problem by summing forces 00:50:53.470 --> 00:50:57.800 at the center of mass of that pipe 00:50:57.800 --> 00:51:01.650 and summing moments around it. 00:51:01.650 --> 00:51:04.900 But the moment around this comes from a friction force 00:51:04.900 --> 00:51:07.490 here, which you don't know. 00:51:07.490 --> 00:51:09.490 So that introduces an unknown that you 00:51:09.490 --> 00:51:10.830 have to then solve for. 00:51:10.830 --> 00:51:15.830 So if you work around G for this pipe, you can do it. 00:51:15.830 --> 00:51:18.760 You can work around G. You could say sum of the moments 00:51:18.760 --> 00:51:20.700 around G, sum of the force with respect to G. 00:51:20.700 --> 00:51:23.245 But you have to deal with unknown forces. 00:51:29.060 --> 00:51:41.260 So you're working with respect to G implies unknown forces, e, 00:51:41.260 --> 00:51:42.460 for example, friction. 00:51:45.760 --> 00:51:49.760 So you'd really maybe rather around the point of contact, 00:51:49.760 --> 00:51:54.370 A. Because if you sum your moments about A, 00:51:54.370 --> 00:51:55.980 the friction force has no moment arm, 00:51:55.980 --> 00:51:57.570 and it doesn't appear in the answer. 00:52:02.250 --> 00:52:04.940 But this gets trickier. 00:52:04.940 --> 00:52:10.670 This is a little more sophisticated, I'll just 00:52:10.670 --> 00:52:12.740 call it, step. 00:52:12.740 --> 00:52:18.680 And you need-- to do this, you need a little theorem. 00:52:18.680 --> 00:52:27.260 So to work with respect to A, you 00:52:27.260 --> 00:52:31.770 need to be able to say that the angular momentum with respect 00:52:31.770 --> 00:52:37.110 to A-- you now are working around a moving point, maybe 00:52:37.110 --> 00:52:38.880 accelerating. 00:52:38.880 --> 00:52:40.750 It's very handy to be able to say 00:52:40.750 --> 00:52:44.210 it's the angular momentum around G, which is easier 00:52:44.210 --> 00:52:52.980 to calculate, plus RGA, the distance 00:52:52.980 --> 00:52:56.490 from the center of mass to the point you're working on, 00:52:56.490 --> 00:53:00.280 cross the linear momentum of the system with respect 00:53:00.280 --> 00:53:02.720 to the inertial frame. 00:53:02.720 --> 00:53:03.630 We need this. 00:53:03.630 --> 00:53:05.670 This is a formula we need. 00:53:05.670 --> 00:53:17.690 And see why this is true? 00:53:17.690 --> 00:53:20.910 This is sort of thing-- this is a formula that's 00:53:20.910 --> 00:53:22.750 come out of the blue here. 00:53:22.750 --> 00:53:24.430 And why is it true? 00:53:24.430 --> 00:53:27.760 So I don't usually like to do proofs, but the proofs of this 00:53:27.760 --> 00:53:29.090 [INAUDIBLE] on the board. 00:53:29.090 --> 00:53:31.555 But the proof of this is really quite simple. 00:53:37.320 --> 00:53:40.430 Here's a little mass point Mi. 00:53:40.430 --> 00:53:45.940 And this radius is R of i with respect to A. 00:53:45.940 --> 00:53:53.160 This is R of G with respect to A. 00:53:53.160 --> 00:54:00.290 And therefore, this is R of i with respect to G is this one. 00:54:00.290 --> 00:54:04.960 And we know that this plus this gives us that. 00:54:04.960 --> 00:54:16.120 We can say Ri particle i with respect to A is RG with respect 00:54:16.120 --> 00:54:20.775 to A plus Ri with respect to G, all vectors. 00:54:45.170 --> 00:54:52.730 So the angular momentum of that body from the basic definition 00:54:52.730 --> 00:54:54.530 of angular momentum is the summation 00:54:54.530 --> 00:54:58.137 of all the little mass bits of the Ri with respect to A 00:54:58.137 --> 00:55:01.920 cross the linear momentum of each little mass bit. 00:55:04.560 --> 00:55:07.880 But we can expand that with that sum. 00:55:07.880 --> 00:55:13.370 So this is the summation of my RG with respect 00:55:13.370 --> 00:55:22.440 to A plus Ri with respect to G cross Pio, summation 00:55:22.440 --> 00:55:25.190 over all the mass bits. 00:55:25.190 --> 00:55:28.570 I'm going to expand this. 00:55:28.570 --> 00:55:33.010 And I can expand this into this times that, summations 00:55:33.010 --> 00:55:35.230 of this times that, and this times that. 00:55:35.230 --> 00:55:50.610 So that becomes a RGA cross summation of-- what's Pio? 00:55:50.610 --> 00:55:55.780 This is a little Mi's, Vi with respect to o. 00:55:55.780 --> 00:55:56.830 Each one has velocity. 00:55:56.830 --> 00:55:58.570 Each one has a mass. 00:55:58.570 --> 00:56:08.340 So this is Mi Vi with respect to o plus the summation 00:56:08.340 --> 00:56:13.982 RiG's cross Mi Vio's. 00:56:16.890 --> 00:56:19.580 Now, notice I pulled this one outside the summation. 00:56:19.580 --> 00:56:21.052 That's because this is a single-- 00:56:21.052 --> 00:56:22.010 this is a fixed number. 00:56:22.010 --> 00:56:23.468 It doesn't change in the summation. 00:56:23.468 --> 00:56:27.500 It's just the distance from my starting point to G. 00:56:27.500 --> 00:56:29.280 So I can pull it out and do the summation 00:56:29.280 --> 00:56:30.530 and then do the cross product. 00:56:33.680 --> 00:56:39.155 What is the summation of all the MVi's for the body? 00:56:44.590 --> 00:56:46.490 That's the momentum of each little particle. 00:56:46.490 --> 00:56:49.400 Add them all up, what do you get? 00:56:49.400 --> 00:56:54.520 This is RGA cross P with respect to o. 00:56:54.520 --> 00:56:56.620 That's this term. 00:56:56.620 --> 00:57:00.730 And this is all the little distances 00:57:00.730 --> 00:57:08.120 from the center of mass to the cross with the momentum 00:57:08.120 --> 00:57:09.100 of each little one. 00:57:13.460 --> 00:57:15.370 What's that? 00:57:15.370 --> 00:57:20.190 Well, this looks like a definition of angular momentum. 00:57:20.190 --> 00:57:24.580 This is the angular momentum of every little mass particle 00:57:24.580 --> 00:57:28.530 with respect to G added up. 00:57:28.530 --> 00:57:33.900 This is H with respect to G, which 00:57:33.900 --> 00:57:36.780 is what we set out to prove. 00:57:36.780 --> 00:57:43.920 So the H with respect to A is H with respect to G plus RGA 00:57:43.920 --> 00:57:46.570 cross the linear momentum of the body. 00:57:46.570 --> 00:57:47.928 Yeah. 00:57:47.928 --> 00:57:52.374 AUDIENCE: Can you also do this by writing Pi with respect 00:57:52.374 --> 00:57:54.350 to o, I mean the velocity part of that, 00:57:54.350 --> 00:57:57.808 as the sum of the velocities? 00:57:57.808 --> 00:58:00.344 PROFESSOR: You got to keep the M's in there. 00:58:00.344 --> 00:58:01.760 AUDIENCE: Well, yeah. [INAUDIBLE]. 00:58:01.760 --> 00:58:03.736 But instead of writing out R as a sum, 00:58:03.736 --> 00:58:08.676 you can write out the velocity as the sum of the velocities 00:58:08.676 --> 00:58:11.991 with respect to the origin. 00:58:11.991 --> 00:58:13.990 PROFESSOR: Are you talking about this term here? 00:58:13.990 --> 00:58:19.310 AUDIENCE: No, the definition of angular momentum. 00:58:19.310 --> 00:58:19.810 Yeah. 00:58:24.175 --> 00:58:25.130 The velocity-- 00:58:25.130 --> 00:58:26.020 PROFESSOR: If you could figure out-- 00:58:26.020 --> 00:58:27.880 if you had the angular momentum [INAUDIBLE] 00:58:27.880 --> 00:58:30.520 and multiplied by Ri, that is the angular momentum 00:58:30.520 --> 00:58:32.940 with respect to A. But I broke it 00:58:32.940 --> 00:58:37.110 apart so that I could show you that this formula I want to use 00:58:37.110 --> 00:58:37.875 has two pieces. 00:58:40.490 --> 00:58:42.430 I want to use that. 00:58:42.430 --> 00:58:49.020 So that if I can use that-- it's easy to get H with respect to G 00:58:49.020 --> 00:58:49.650 sometimes. 00:58:49.650 --> 00:58:52.160 It's really hard to know what to do with things that 00:58:52.160 --> 00:58:53.980 happen around this point A. 00:58:53.980 --> 00:58:55.832 So now let's go back. 00:58:55.832 --> 00:58:57.790 I think, to understand this, we need to go back 00:58:57.790 --> 00:59:00.190 to our truck problem. 00:59:00.190 --> 00:59:03.200 We now have a formula that you know where it comes from. 00:59:03.200 --> 00:59:08.890 We have our truck that is accelerating at x1 double dot. 00:59:08.890 --> 00:59:16.820 And we want to find out what's theta double dot for that pipe. 00:59:16.820 --> 00:59:18.272 What's it doing? 00:59:18.272 --> 00:59:19.730 So first we needed some kinematics. 00:59:25.780 --> 00:59:34.710 And in particular, what is the x2-- did I label this 00:59:34.710 --> 00:59:35.210 very well? 00:59:35.210 --> 00:59:36.390 I didn't. 00:59:36.390 --> 00:59:38.770 So here's my pipe. 00:59:38.770 --> 00:59:42.310 Here's my truck bed that it's in contact with. 00:59:42.310 --> 00:59:46.450 Truck's moving at x1 double dot, we know. 00:59:46.450 --> 00:59:49.310 In an inertial frame, the movement 00:59:49.310 --> 00:59:54.210 of the center of mass of my pipe is x2. 00:59:54.210 --> 01:00:02.630 And it has some angular rotation I'll call theta. 01:00:02.630 --> 01:00:06.750 So the movement of this guy I need 01:00:06.750 --> 01:00:10.260 to be able to express in terms of x1 and theta. 01:00:10.260 --> 01:00:14.590 Well, if this is fixed to the truck and the truck move 01:00:14.590 --> 01:00:18.540 forward, then x2 would be equal to x1. 01:00:18.540 --> 01:00:22.360 But in fact, it rolls back a little bit. 01:00:22.360 --> 01:00:23.820 And the distance this point moves 01:00:23.820 --> 01:00:25.790 if it rolls through an angle theta 01:00:25.790 --> 01:00:28.310 is it rolls backwards an amount R theta. 01:00:32.140 --> 01:00:35.180 And we're going to need to take two derivatives of that. 01:00:35.180 --> 01:00:42.200 x2 double dot equals x1 double dot minus r theta double dot. 01:00:42.200 --> 01:00:43.950 So that's a little kinematic relationship 01:00:43.950 --> 01:00:45.075 we need to do this problem. 01:00:55.440 --> 01:00:58.830 Next we need to apply a physical law, which 01:00:58.830 --> 01:01:03.720 is the one I've derived. 01:01:03.720 --> 01:01:06.760 So this is our physics here now, our physical law. 01:01:11.020 --> 01:01:16.680 And that's the external torques, dH with respect 01:01:16.680 --> 01:01:24.310 to A dt, plus VAo cross Po. 01:01:27.360 --> 01:01:34.656 And in this case, is VAo 0? 01:01:34.656 --> 01:01:38.140 No, in fact, it's x1 dot, right? 01:01:38.140 --> 01:01:40.480 It's in the i direction. 01:01:40.480 --> 01:01:44.164 How about what direction is P, the momentum of the pipe? 01:01:49.250 --> 01:01:52.355 Does it move in the y direction? 01:01:52.355 --> 01:01:52.855 Up, down? 01:01:52.855 --> 01:01:53.355 No. 01:01:53.355 --> 01:01:54.490 It only moves in the x. 01:01:54.490 --> 01:01:56.920 So this velocity is only in the x. 01:01:56.920 --> 01:01:59.740 This is in the x, or the i hat direction. 01:01:59.740 --> 01:02:01.910 This cross product is zero. 01:02:01.910 --> 01:02:03.730 So it just happens that they're parallel. 01:02:03.730 --> 01:02:04.890 So this thing goes to zero. 01:02:04.890 --> 01:02:07.025 We don't have to deal with it. 01:02:07.025 --> 01:02:08.650 That's because these guys are parallel. 01:02:13.350 --> 01:02:15.380 Parallel motion. 01:02:15.380 --> 01:02:17.080 But you did have to consider it. 01:02:17.080 --> 01:02:18.371 You did have to think about it. 01:02:18.371 --> 01:02:20.120 It's not just trivially 0. 01:02:20.120 --> 01:02:24.790 OK, so that means that the torques about A is just dHA dt. 01:02:24.790 --> 01:02:33.070 And HA for this problem is i-- well, 01:02:33.070 --> 01:02:41.030 it's HG, which is IzzG theta double dot. 01:02:41.030 --> 01:02:44.480 But I have to put in this-- theta dot, excuse me. 01:02:44.480 --> 01:02:46.550 This is just the H. I have to put 01:02:46.550 --> 01:02:51.690 in the second term, RG with respect 01:02:51.690 --> 01:02:55.747 to A cross P with respect to o. 01:02:59.330 --> 01:03:01.280 So here's my HA. 01:03:04.234 --> 01:03:05.400 I'll write it again up here. 01:03:05.400 --> 01:03:13.150 This is IzzG theta dot in the k. 01:03:13.150 --> 01:03:15.020 And now this second term. 01:03:15.020 --> 01:03:19.990 R is Rj. 01:03:19.990 --> 01:03:21.920 My y-axis is up. 01:03:21.920 --> 01:03:26.990 The radius of this thing is R. Here's the radius. 01:03:26.990 --> 01:03:37.050 So the moment arm is Rj crossed with the linear momentum 01:03:37.050 --> 01:03:42.070 of that piece of pipe, which is the mass of the pipe times 01:03:42.070 --> 01:03:48.140 x2 dot in the i direction. 01:03:48.140 --> 01:03:50.350 j cross i is minus k. 01:03:50.350 --> 01:03:56.330 So Izz with respect to G theta double dot-- theta dot. 01:03:56.330 --> 01:03:59.580 I keep taking the derivative a little too soon. 01:03:59.580 --> 01:04:06.050 k minus RM x2 dot k. 01:04:06.050 --> 01:04:12.300 And now some of the torques, d by dt of H with respect to A. 01:04:12.300 --> 01:04:14.380 Take the derivatives. 01:04:14.380 --> 01:04:27.456 IzzG theta double dot k minus RM x2 double dot k. 01:04:27.456 --> 01:04:29.830 And fortunately, none of these unit vectors are rotating. 01:04:29.830 --> 01:04:32.070 So we don't have to deal with any of that. 01:04:32.070 --> 01:04:34.710 And I'm almost to the end, but now I 01:04:34.710 --> 01:04:39.410 have to go back to that original kinematic relationship, which 01:04:39.410 --> 01:04:46.380 allows me to express x2 in terms of x1 and R theta. 01:04:46.380 --> 01:04:51.100 And if I substitute that in here and solve for it, 01:04:51.100 --> 01:05:10.620 I get theta double dot equals MR over IzzG plus MR squared. 01:05:13.870 --> 01:05:20.310 x double dot whole thing, x1 double dot. 01:05:20.310 --> 01:05:22.860 So you've accelerated x1 double dot. 01:05:22.860 --> 01:05:25.740 The thing starts rolling. 01:05:25.740 --> 01:05:28.440 It's actually rolling backwards, which is a plus theta 01:05:28.440 --> 01:05:31.820 direction, at this rate. 01:05:31.820 --> 01:05:37.010 We did this using this formula for HA. 01:05:37.010 --> 01:05:39.700 Now, the beauty of this formula is 01:05:39.700 --> 01:05:43.170 that it works for any points that 01:05:43.170 --> 01:05:46.870 are moving, even can be accelerating, 01:05:46.870 --> 01:05:48.925 all sorts of nasty conditions, it's true. 01:05:48.925 --> 01:05:54.380 It's based on the fundamental definition of angular momentum. 01:05:54.380 --> 01:05:55.965 So the book-- yeah. 01:05:55.965 --> 01:05:58.000 AUDIENCE: [INAUDIBLE] final point here. 01:05:58.000 --> 01:06:00.160 I just want to make sure i understood. 01:06:00.160 --> 01:06:04.620 In this first line here, [INAUDIBLE]. 01:06:04.620 --> 01:06:07.390 That stroke doesn't mean that V sub A was 0. 01:06:07.390 --> 01:06:10.940 It meant that that term is 0. 01:06:10.940 --> 01:06:13.370 PROFESSOR: This term's zero because these 01:06:13.370 --> 01:06:16.772 happen to be parallel, but the product is zero in this case. 01:06:16.772 --> 01:06:20.390 If it had not turned out to be-- they were different directions. 01:06:20.390 --> 01:06:21.650 It had to be a non-zero term. 01:06:21.650 --> 01:06:25.715 And you would have to bring it along and take its derivative 01:06:25.715 --> 01:06:26.840 along with the other stuff. 01:06:29.870 --> 01:06:32.430 But this is a really powerful method. 01:06:32.430 --> 01:06:36.750 Now, the book is reasonably good in lots of points. 01:06:36.750 --> 01:06:39.750 But in chapter 17, when it does problems 01:06:39.750 --> 01:06:43.910 that are kind of like this, it introduces something 01:06:43.910 --> 01:06:46.910 that I find it hard to digest what they're doing. 01:06:46.910 --> 01:06:48.525 There's particularly-- where's Matt? 01:06:48.525 --> 01:06:53.830 There's equation 1715. 01:06:53.830 --> 01:06:56.340 When you get to that bit of the book, just ignore it. 01:06:56.340 --> 01:06:59.510 Use this method. 01:06:59.510 --> 01:07:02.360 The author tries to give you a little trick that you can use. 01:07:02.360 --> 01:07:05.420 But the problem with tricks is you have to memorize them. 01:07:05.420 --> 01:07:09.800 So what I've shown you today is based on basic definition 01:07:09.800 --> 01:07:10.740 of angular momentum. 01:07:10.740 --> 01:07:15.950 That expression at the top is always usable, not just 01:07:15.950 --> 01:07:20.250 special little conditions, which is what the formula in the book 01:07:20.250 --> 01:07:21.500 is generated for. 01:07:25.710 --> 01:07:27.010 We've got a couple of minutes. 01:07:29.540 --> 01:07:31.780 Let you ask questions, and then I'll 01:07:31.780 --> 01:07:34.100 just pose a conceptual problem or two 01:07:34.100 --> 01:07:35.650 and ask you what method you'd use. 01:07:35.650 --> 01:07:37.420 Yeah. 01:07:37.420 --> 01:07:40.892 AUDIENCE: So [INAUDIBLE] when you [INAUDIBLE] dHA dt, 01:07:40.892 --> 01:07:44.690 is that equal to 0 because there is no torque in the system? 01:07:44.690 --> 01:07:49.605 PROFESSOR: Oh, you know, boy, I'm glad you caught that. 01:07:49.605 --> 01:07:52.080 Yeah, in order to be able to do this, 01:07:52.080 --> 01:07:56.550 we've got to know something here, right? 01:07:56.550 --> 01:07:58.020 Important catch. 01:07:58.020 --> 01:07:59.280 Thank you. 01:07:59.280 --> 01:08:02.580 Why is this true? 01:08:02.580 --> 01:08:05.490 AUDIENCE: [INAUDIBLE]. 01:08:05.490 --> 01:08:08.890 PROFESSOR: So we chose-- that's why 01:08:08.890 --> 01:08:16.590 we chose to work around point A. With respect to that point, 01:08:16.590 --> 01:08:19.350 there are no external forces that 01:08:19.350 --> 01:08:23.460 create moments on that pipe with respect to that point. 01:08:23.460 --> 01:08:26.250 And that's why you can say that the time 01:08:26.250 --> 01:08:28.310 derivative of this angular momentum 01:08:28.310 --> 01:08:32.060 is equal to zero because there are no external torques. 01:08:32.060 --> 01:08:35.660 If you had picked G to do this problem, 01:08:35.660 --> 01:08:39.437 would the sum of the torques about G be zero? 01:08:39.437 --> 01:08:41.520 No, you'd have to put that friction force in there 01:08:41.520 --> 01:08:44.600 and have RF and figure out F is. 01:08:44.600 --> 01:08:47.640 We completely avoided having to calculate the friction force. 01:08:47.640 --> 01:08:50.880 That's the point of being able to use techniques like this 01:08:50.880 --> 01:08:54.990 and make your computations around points of contact. 01:09:05.890 --> 01:09:10.390 So textbooks have lots of problems like this. 01:09:10.390 --> 01:09:14.774 You've got a box on a cart. 01:09:14.774 --> 01:09:17.149 And your kid's pushing it, and he gets a little exuberant 01:09:17.149 --> 01:09:19.689 and pushes a little too hard, accelerates the cart 01:09:19.689 --> 01:09:22.710 a little too fast, and the box falls over and breaks 01:09:22.710 --> 01:09:25.300 the lamp or whatever's in it. 01:09:25.300 --> 01:09:28.580 And if it's this way, and I accelerate it, 01:09:28.580 --> 01:09:30.620 it's much more tolerant. 01:09:30.620 --> 01:09:32.859 Falls over easier this way. 01:09:32.859 --> 01:09:35.640 But if I asked you, gave you a problem 01:09:35.640 --> 01:09:40.899 and said, calculate the maximum acceleration 01:09:40.899 --> 01:09:44.950 that I can put on this object such that it 01:09:44.950 --> 01:09:49.149 just barely-- just right at the edge of tipping over, 01:09:49.149 --> 01:09:50.149 but doesn't tip it over. 01:09:50.149 --> 01:09:52.648 What's that maximum acceleration that you don't tip it over? 01:09:52.648 --> 01:09:57.050 What method would you use? 01:09:57.050 --> 01:10:01.100 I gave you four classes of approaches to problems. 01:10:07.412 --> 01:10:09.740 AUDIENCE: [INAUDIBLE]. 01:10:09.740 --> 01:10:11.230 PROFESSOR: I hear a four. 01:10:11.230 --> 01:10:12.890 Anybody else want to bid here? 01:10:16.090 --> 01:10:17.050 More fours. 01:10:17.050 --> 01:10:20.168 Would three work? 01:10:20.168 --> 01:10:22.580 AUDIENCE: [INAUDIBLE]. 01:10:22.580 --> 01:10:23.270 PROFESSOR: Why? 01:10:23.270 --> 01:10:26.060 He says three would be complicated. 01:10:26.060 --> 01:10:28.400 Three means taking moments and forces with respect 01:10:28.400 --> 01:10:31.400 to the center of mass. 01:10:31.400 --> 01:10:34.378 If you do that, what do you have to deal with in this problem? 01:10:34.378 --> 01:10:35.294 AUDIENCE: [INAUDIBLE]. 01:10:36.857 --> 01:10:39.190 PROFESSOR: You have to, then-- around the center of mass 01:10:39.190 --> 01:10:40.910 there's a friction force. 01:10:40.910 --> 01:10:44.210 There's a normal force pushing up here. 01:10:44.210 --> 01:10:47.240 The way you do this problem is where it's just barely 01:10:47.240 --> 01:10:51.940 about to go, all of the force is pushing on this corner. 01:10:51.940 --> 01:10:52.440 Think of it. 01:10:52.440 --> 01:10:55.540 It's just lifting up a fraction. 01:10:55.540 --> 01:10:57.710 All of that contact point moves to right here. 01:10:57.710 --> 01:11:00.230 You have an upward force and a friction force, 01:11:00.230 --> 01:11:02.500 and they create a moment about G. 01:11:02.500 --> 01:11:04.120 But if you do the forces around G, 01:11:04.120 --> 01:11:06.970 you have to solve for those two. 01:11:06.970 --> 01:11:11.680 You do the forces around A. The trick here is figuring out 01:11:11.680 --> 01:11:15.540 where's A. But if you realize A is right here, 01:11:15.540 --> 01:11:19.350 and you do what we just did, this problem's 01:11:19.350 --> 01:11:21.500 just a piece of cake.