WEBVTT 00:00:00.070 --> 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 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.260 at ocw.mit.edu. 00:00:22.861 --> 00:00:25.560 PROFESSOR: OK, what I thought I would do 00:00:25.560 --> 00:00:32.549 is I'm going to go quickly through the problems that 00:00:32.549 --> 00:00:33.745 were assigned for practice. 00:00:33.745 --> 00:00:35.513 There's seven or eight of them. 00:00:35.513 --> 00:00:36.846 I'm not going to do all of them. 00:00:36.846 --> 00:00:38.304 I'm going to go through and kind of 00:00:38.304 --> 00:00:43.570 talk about what some of the key issues are with each problem 00:00:43.570 --> 00:00:51.000 and maybe make some points about identifying 00:00:51.000 --> 00:00:55.410 how you do problems, not necessarily specifically focus 00:00:55.410 --> 00:00:57.620 on the exact question that was asked here. 00:00:57.620 --> 00:01:01.670 So here was this first-- this was a quiz from last year. 00:01:01.670 --> 00:01:07.290 And you were asked to find an equation of motion 00:01:07.290 --> 00:01:09.530 for this thing. 00:01:09.530 --> 00:01:12.190 And it can be a bit of a puzzling problem, 00:01:12.190 --> 00:01:15.570 so how many degrees of freedom, what are the constraints, what 00:01:15.570 --> 00:01:16.920 equation should you use. 00:01:19.510 --> 00:01:21.660 So first of all, I look at a problem like this. 00:01:21.660 --> 00:01:24.070 Is it planar motion or not? 00:01:24.070 --> 00:01:26.970 OK, so maximum three degrees of freedom. 00:01:26.970 --> 00:01:29.463 What are the constraints, though? 00:01:34.411 --> 00:01:37.700 STUDENT: So G is [INAUDIBLE]. 00:01:37.700 --> 00:01:39.970 PROFESSOR: So G is R/2 from the center. 00:01:39.970 --> 00:01:43.380 So that's true. 00:01:43.380 --> 00:01:49.520 And so what you're really saying is it's fixed in radial motion, 00:01:49.520 --> 00:01:50.290 right? 00:01:50.290 --> 00:01:51.730 And you're right. 00:01:51.730 --> 00:01:52.820 So that's one constraint. 00:01:52.820 --> 00:01:54.780 What's another constraint in this problem? 00:01:57.760 --> 00:02:01.050 How many degrees of freedom do you expect to end up with? 00:02:01.050 --> 00:02:02.920 How many equations do you need? 00:02:02.920 --> 00:02:04.370 How many coordinates? 00:02:04.370 --> 00:02:05.368 One, right? 00:02:05.368 --> 00:02:07.243 So how do we articulate the other constraint? 00:02:10.650 --> 00:02:12.280 So it can't move radially. 00:02:12.280 --> 00:02:13.310 We figured that out. 00:02:26.780 --> 00:02:29.370 So this is actually kind of an important point 00:02:29.370 --> 00:02:30.220 in this problem. 00:02:30.220 --> 00:02:32.480 Because if you can get this down, 00:02:32.480 --> 00:02:34.515 then the problem becomes very simple. 00:02:37.930 --> 00:02:39.090 So talk to a neighbor. 00:02:39.090 --> 00:02:43.060 How would you describe the other constraint here? 00:02:43.060 --> 00:02:45.130 I'm going to hold it a second. 00:02:45.130 --> 00:02:47.070 STUDENT: [INAUDIBLE] 00:02:47.070 --> 00:02:47.820 PROFESSOR: Pardon? 00:02:47.820 --> 00:02:48.560 STUDENT: Gravity is just [INAUDIBLE]. 00:02:48.560 --> 00:02:49.670 PROFESSOR: Well, there is gravity, yes. 00:02:49.670 --> 00:02:51.260 But gravity is not a constraint. 00:02:51.260 --> 00:02:53.555 So this is a pendulum, actually, of sorts. 00:02:56.830 --> 00:02:58.360 So talk to a neighbor. 00:02:58.360 --> 00:03:00.275 How would you describe the other constraint? 00:03:41.870 --> 00:03:46.310 OK, how would you describe it? 00:03:46.310 --> 00:03:49.250 What's the other constraint here? 00:03:57.770 --> 00:04:00.840 So sometimes if something is pinned, 00:04:00.840 --> 00:04:02.895 we say, well, then it's fixed in x and y. 00:04:02.895 --> 00:04:04.680 And that's two constraints. 00:04:04.680 --> 00:04:06.290 It's only left to rotate. 00:04:06.290 --> 00:04:08.290 But this one doesn't seem to-- it's kind of hard 00:04:08.290 --> 00:04:12.140 to understand where this one is pinned. 00:04:12.140 --> 00:04:14.057 You know this thing is constrained in x and y. 00:04:14.057 --> 00:04:16.390 Because you want to use theta for the coordinate, right? 00:04:16.390 --> 00:04:18.070 But how do you say that it's constrained 00:04:18.070 --> 00:04:18.903 in that other thing? 00:04:18.903 --> 00:04:20.264 Yeah. 00:04:20.264 --> 00:04:22.650 STUDENT: [INAUDIBLE] 00:04:29.860 --> 00:04:31.985 PROFESSOR: OK, so he's saying that you can define x 00:04:31.985 --> 00:04:34.185 and y in terms of r and theta. 00:04:34.185 --> 00:04:36.176 And that's true. 00:04:36.176 --> 00:04:38.050 And I'm going to say it slightly differently. 00:04:38.050 --> 00:04:39.591 So the way I think about this problem 00:04:39.591 --> 00:04:42.040 is we've said that it's constrained 00:04:42.040 --> 00:04:43.440 in the radial direction. 00:04:43.440 --> 00:04:44.880 So that's one. 00:04:44.880 --> 00:04:48.040 It can move in the tangential direction, right? 00:04:48.040 --> 00:04:50.060 This problem is really exactly the same 00:04:50.060 --> 00:04:54.090 as the skateboard problem in a bowl. 00:04:54.090 --> 00:04:55.759 You shorten that thing a little bit, 00:04:55.759 --> 00:04:57.050 and it looks like a skateboard. 00:04:57.050 --> 00:04:59.190 And we're ignoring the inertia of the wheels. 00:04:59.190 --> 00:05:01.500 So it's just a stick sliding up and down. 00:05:01.500 --> 00:05:06.370 So what's the relationship between the tangential motion 00:05:06.370 --> 00:05:07.860 and theta? 00:05:07.860 --> 00:05:10.690 Is it a fixed relationship? 00:05:10.690 --> 00:05:15.370 So how far does it move in the theta? 00:05:15.370 --> 00:05:17.350 If you have a little motion, delta theta, 00:05:17.350 --> 00:05:20.290 how far does it move tangentially? 00:05:20.290 --> 00:05:27.400 R delta theta is the delta r tangential, 00:05:27.400 --> 00:05:29.210 the distance it moves. 00:05:29.210 --> 00:05:32.980 So this is the way to say the other constraint, 00:05:32.980 --> 00:05:36.610 is if you know theta, you know how far it's 00:05:36.610 --> 00:05:39.520 moved in the radial, in the tangential direction. 00:05:39.520 --> 00:05:42.180 So that's a way of saying that other constraint. 00:05:42.180 --> 00:05:45.080 OK, then we can say we completely describe 00:05:45.080 --> 00:05:49.060 this problem by one coordinate. 00:05:49.060 --> 00:05:51.660 And that's theta. 00:05:51.660 --> 00:05:54.440 And the second you can do that, this problem then 00:05:54.440 --> 00:05:59.270 has a center of rotation right in the center of the circle. 00:05:59.270 --> 00:06:03.440 And anytime you have a body which 00:06:03.440 --> 00:06:07.350 rotates around a central point, then 00:06:07.350 --> 00:06:09.960 you know that you can describe the angular 00:06:09.960 --> 00:06:13.300 momentum of the body as some I with respect 00:06:13.300 --> 00:06:16.250 to A times-- and it's planar motion. 00:06:16.250 --> 00:06:22.520 So then it's just omega z, as long as this is a principal, 00:06:22.520 --> 00:06:26.760 as long as you have the mass moments of inertia in terms 00:06:26.760 --> 00:06:30.410 of principal axes here. 00:06:30.410 --> 00:06:33.460 And it's nice when you can write it that simply. 00:06:33.460 --> 00:06:37.070 The second you can identify a fixed point about which 00:06:37.070 --> 00:06:39.510 something rotates, then the angular momentum 00:06:39.510 --> 00:06:41.532 simplifies to that. 00:06:41.532 --> 00:06:48.370 OK, and then this problem, of course 00:06:48.370 --> 00:06:51.580 the sum of the torques with respect to the center 00:06:51.580 --> 00:06:59.990 here has got to be d H with respect to A dt plus vAO 00:06:59.990 --> 00:07:02.490 cross PGO. 00:07:02.490 --> 00:07:06.960 But this is-- what's the velocity at point A? 00:07:09.820 --> 00:07:12.380 0, so you don't have to worry about this term. 00:07:12.380 --> 00:07:14.580 And it's just that. 00:07:14.580 --> 00:07:17.580 And you know that the sum of the external torques 00:07:17.580 --> 00:07:22.930 in this problem comes from gravity. 00:07:22.930 --> 00:07:26.202 So you put in the gravity term and compute the torques 00:07:26.202 --> 00:07:26.910 about the center. 00:07:29.640 --> 00:07:32.330 So our object really looks like that. 00:07:32.330 --> 00:07:38.885 And here's G. And in the problem, the way it was posed, 00:07:38.885 --> 00:07:42.340 the theta is drawn from this line. 00:07:42.340 --> 00:07:47.530 So this is theta. 00:07:47.530 --> 00:07:49.702 So what's the moment arm? 00:07:49.702 --> 00:07:55.510 It looks like whatever this distance is, which is R/2. 00:07:55.510 --> 00:07:58.490 It's given. 00:07:58.490 --> 00:08:04.910 R/2 cosine theta is the length of this side. 00:08:04.910 --> 00:08:11.391 So the moment that gravity makes is some Mg. 00:08:16.860 --> 00:08:20.980 And that's going to be equal to some Izz 00:08:20.980 --> 00:08:23.990 about A theta double dot. 00:08:23.990 --> 00:08:25.910 OK, good. 00:08:25.910 --> 00:08:33.370 Let's take a look at the next problem. 00:08:33.370 --> 00:08:34.969 And now so my intention here, I'm 00:08:34.969 --> 00:08:36.760 going to go kind of quickly one by another. 00:08:36.760 --> 00:08:38.843 And I'm just trying to hit the important concepts. 00:08:38.843 --> 00:08:41.789 So if you have a question about the concepts, ask. 00:08:41.789 --> 00:08:44.610 If I've left you wondering, I really 00:08:44.610 --> 00:08:47.060 want this to be kind of a conversation here. 00:08:47.060 --> 00:08:48.180 So that's the whole point. 00:08:48.180 --> 00:08:53.610 I've given you the essence of what makes these problems work. 00:08:53.610 --> 00:08:57.930 OK, this one, we just find a location of the center of mass. 00:08:57.930 --> 00:09:01.790 I think you're pretty good at that sort of thing. 00:09:01.790 --> 00:09:05.710 Well, remember just a couple points about center of mass. 00:09:05.710 --> 00:09:07.820 This one it says, find it. 00:09:07.820 --> 00:09:10.455 And you're given the two objects. 00:09:10.455 --> 00:09:12.210 The important point is that you can 00:09:12.210 --> 00:09:15.520 pick any coordinate at all in order 00:09:15.520 --> 00:09:18.430 to use it to compute the center of mass. 00:09:18.430 --> 00:09:22.050 So I pick S here, the M1. 00:09:22.050 --> 00:09:24.930 And some distance down here is where the center of mass 00:09:24.930 --> 00:09:26.060 is that we're looking for. 00:09:26.060 --> 00:09:30.200 And so this is the G we're looking for. 00:09:30.200 --> 00:09:38.426 So M1 plus M2 is the total mass times the position of G. 00:09:38.426 --> 00:09:43.050 I guess that ought to be a capital-- SG here 00:09:43.050 --> 00:09:48.610 must be equal to the sum of the parts times their positions, M1 00:09:48.610 --> 00:09:54.680 SG1 plus M2 SG2. 00:09:54.680 --> 00:09:56.430 And you can solve for SG. 00:09:56.430 --> 00:09:58.090 And that tells you where you're at. 00:09:58.090 --> 00:10:01.320 So you can pick any coordinate at all to calculate it. 00:10:01.320 --> 00:10:04.600 And once you find it, you know where it is. 00:10:04.600 --> 00:10:08.360 So you just solve here for SG. 00:10:08.360 --> 00:10:12.600 It's that divided by the total mass, obviously. 00:10:12.600 --> 00:10:21.370 OK, the next question was, draw principal axes. 00:10:21.370 --> 00:10:27.400 So does this object have some planes of symmetry? 00:10:27.400 --> 00:10:28.510 Tell me one. 00:10:28.510 --> 00:10:29.714 STUDENT: This one. 00:10:29.714 --> 00:10:31.630 PROFESSOR: Slice down through it this way, OK. 00:10:31.630 --> 00:10:34.210 And that means you have a principal axis where? 00:10:34.210 --> 00:10:36.000 Perpendicular to every plane of symmetry. 00:10:36.000 --> 00:10:39.360 So you have a principal axis coming out of the board. 00:10:39.360 --> 00:10:40.670 That's one. 00:10:40.670 --> 00:10:43.127 And give me another plane of symmetry. 00:10:43.127 --> 00:10:43.960 STUDENT: [INAUDIBLE] 00:10:43.960 --> 00:10:45.043 PROFESSOR: Yeah, this one. 00:10:45.043 --> 00:10:46.850 So that means you've got a plane of axis, 00:10:46.850 --> 00:10:48.910 principal axis going that way. 00:10:48.910 --> 00:10:52.680 And so then the third one has to be perpendicular to that. 00:10:52.680 --> 00:10:57.640 So you have one this way, this way, and this way. 00:11:09.369 --> 00:11:17.220 So I'd asked for-- so if this is G, then principal axes 00:11:17.220 --> 00:11:20.850 with respect to G, one there, one coming out of the board, 00:11:20.850 --> 00:11:23.250 and one going up like that. 00:11:28.300 --> 00:11:31.050 So what is it? 00:11:31.050 --> 00:11:35.840 Let's remind ourselves what it means to be a principal axis. 00:11:35.840 --> 00:11:41.370 So if you know the principal axes of an object, 00:11:41.370 --> 00:11:46.620 if you rotate, and you've chosen them to go through G, 00:11:46.620 --> 00:11:48.140 do they have to go through G? 00:11:48.140 --> 00:11:51.150 Do all principal axes have to go through G? 00:11:51.150 --> 00:11:52.800 No, not at all. 00:11:52.800 --> 00:11:56.650 The principal axes just have to give you, when you work it out, 00:11:56.650 --> 00:11:58.635 a diagonal inertia matrix. 00:12:04.730 --> 00:12:07.630 But what does it mean to have body 00:12:07.630 --> 00:12:09.920 coordinates that we know are principal axes? 00:12:09.920 --> 00:12:17.730 Well, one of the things we know, that if you rotate about G, 00:12:17.730 --> 00:12:21.340 rotate about a principal axis passing through G, 00:12:21.340 --> 00:12:23.440 if you rotate about any one of them, 00:12:23.440 --> 00:12:24.775 is it dynamically balanced? 00:12:27.880 --> 00:12:30.010 Guaranteed. 00:12:30.010 --> 00:12:33.760 You must rotate about the principal axis, though. 00:12:33.760 --> 00:12:35.550 Now, I'll ask you a second question. 00:12:35.550 --> 00:12:38.840 If you rotate about any axis passing through G, 00:12:38.840 --> 00:12:40.730 is it statically balanced? 00:12:40.730 --> 00:12:44.530 Right, if you rotate about any axis passing through G, 00:12:44.530 --> 00:12:46.960 is it dynamically balanced? 00:12:46.960 --> 00:12:48.140 Maybe, right? 00:12:48.140 --> 00:12:49.880 Not necessarily. 00:12:49.880 --> 00:12:53.600 It might be dynamically balanced if you pass through G, 00:12:53.600 --> 00:12:59.100 and it is a principal axis, right? 00:12:59.100 --> 00:13:07.350 OK, can you have a principal axis not going through G 00:13:07.350 --> 00:13:11.560 and rotate about that axis and have 00:13:11.560 --> 00:13:13.388 it be dynamically balanced? 00:13:13.388 --> 00:13:18.990 Yeah, OK, so we'll talk about an example in a second. 00:13:18.990 --> 00:13:25.860 OK, so I think the second question here said, 00:13:25.860 --> 00:13:30.530 OK, we want to rotate this about this point G2. 00:13:30.530 --> 00:13:35.317 Now, G2 is right in the middle at this body. 00:13:35.317 --> 00:13:36.775 That's where its center of mass is. 00:13:36.775 --> 00:13:39.150 And there's a G1 up here that's the center of mass 00:13:39.150 --> 00:13:40.430 of the upper body. 00:13:40.430 --> 00:13:42.350 And there's a composite center of mass right 00:13:42.350 --> 00:13:45.710 here, which we just computed what it was. 00:13:45.710 --> 00:13:51.740 And we want-- there's going to be an axis of rotation passing 00:13:51.740 --> 00:13:53.125 through G2. 00:13:53.125 --> 00:13:56.210 And we're rotating this body around it. 00:13:56.210 --> 00:13:59.475 So first question is, this axis coming through G2, 00:13:59.475 --> 00:14:01.333 is it a principal axis? 00:14:01.333 --> 00:14:02.280 Yes. 00:14:02.280 --> 00:14:05.255 If we rotate about that axis only, 00:14:05.255 --> 00:14:08.520 do you expect it to be dynamically balanced? 00:14:08.520 --> 00:14:09.450 Right. 00:14:09.450 --> 00:14:11.520 Do you expect it to be statically balanced? 00:14:11.520 --> 00:14:12.020 No. 00:14:25.820 --> 00:14:46.900 So we're spinning it around this G2, going like that around G2 00:14:46.900 --> 00:14:50.270 at some omega z. 00:14:50.270 --> 00:14:52.390 We know that this is G2. 00:14:52.390 --> 00:14:56.650 We know that G for the object is here. 00:14:56.650 --> 00:14:58.720 That's the center of mass of the whole thing. 00:14:58.720 --> 00:15:03.185 So there's some distance between those two, which I don't know. 00:15:03.185 --> 00:15:04.310 But we could figure it out. 00:15:04.310 --> 00:15:05.760 We'll call it e. 00:15:05.760 --> 00:15:10.730 So what's the force required at this pin 00:15:10.730 --> 00:15:12.370 that its axle is going around? 00:15:12.370 --> 00:15:15.890 What's the force required to allow this thing 00:15:15.890 --> 00:15:20.070 to spin around, to hold it, essentially, the force required 00:15:20.070 --> 00:15:22.620 to hold it in place? 00:15:22.620 --> 00:15:26.570 And you ought to intuitively have an idea of what it is. 00:15:26.570 --> 00:15:28.250 Why is there a force? 00:15:28.250 --> 00:15:30.260 That thing will spin around this central axle. 00:15:30.260 --> 00:15:32.210 And there's a force required to keep 00:15:32.210 --> 00:15:33.720 that thing from moving away. 00:15:33.720 --> 00:15:36.142 What is it? 00:15:36.142 --> 00:15:39.130 STUDENT: Centrifugal force. 00:15:39.130 --> 00:15:41.650 PROFESSOR: I hear centrifugal. 00:15:41.650 --> 00:15:45.130 So you're accelerating the center of mass of that thing. 00:15:45.130 --> 00:15:46.960 You're making it go in a circle. 00:15:46.960 --> 00:15:48.800 Mass times acceleration is a force. 00:15:48.800 --> 00:15:51.269 And that's the tension in the string when you're 00:15:51.269 --> 00:15:52.310 swinging the ball around. 00:15:52.310 --> 00:15:55.550 This is no different from swinging the ball on a string. 00:15:55.550 --> 00:15:57.100 And you're essentially calculating 00:15:57.100 --> 00:16:00.970 the tension required to keep that thing from flying off. 00:16:00.970 --> 00:16:02.580 And we could probably just guess. 00:16:02.580 --> 00:16:10.990 We can almost guess what it is, minus the total mass times 00:16:10.990 --> 00:16:14.174 the acceleration, which is inward. 00:16:14.174 --> 00:16:15.590 That's where the minus comes from. 00:16:15.590 --> 00:16:16.320 And what is it? 00:16:19.122 --> 00:16:20.530 STUDENT: [INAUDIBLE] 00:16:20.530 --> 00:16:21.648 PROFESSOR: Louder. 00:16:21.648 --> 00:16:23.239 STUDENT: [INAUDIBLE] 00:16:23.239 --> 00:16:24.280 PROFESSOR: Omega squared. 00:16:24.280 --> 00:16:25.680 I hear an omega z squared. 00:16:25.680 --> 00:16:27.540 We need something else. 00:16:27.540 --> 00:16:28.440 Pardon? 00:16:28.440 --> 00:16:29.872 STUDENT: [INAUDIBLE] 00:16:29.872 --> 00:16:30.830 PROFESSOR: Yeah, the r. 00:16:30.830 --> 00:16:32.577 And the r is this little e here. 00:16:32.577 --> 00:16:33.993 It's called the eccentricity, when 00:16:33.993 --> 00:16:36.230 you have something unbalanced. e omega squared, 00:16:36.230 --> 00:16:39.720 and it's in the r hat direction. 00:16:39.720 --> 00:16:40.910 We'll call it inwards. 00:16:40.910 --> 00:16:42.850 And the inwards is the minus. 00:16:42.850 --> 00:16:44.080 So that's the acceleration. 00:16:44.080 --> 00:16:46.430 Mass times acceleration is the force. 00:16:46.430 --> 00:16:53.950 Also, you could compute this by computing d by dt of P of G 00:16:53.950 --> 00:16:59.350 with respect to O. And that's then 00:16:59.350 --> 00:17:07.270 M1 plus M2 times v of G. You have to take the d by dt of it. 00:17:07.270 --> 00:17:10.380 And that gives you that, which is the acceleration. 00:17:10.380 --> 00:17:16.750 So the forces you get directly from computing dP/dt, 00:17:16.750 --> 00:17:18.730 if you need to remember a way to do that. 00:17:18.730 --> 00:17:26.950 All right, so conceptually, if I don't know what e is, 00:17:26.950 --> 00:17:29.080 then I don't want to have to figure it out. 00:17:33.380 --> 00:17:36.090 So I'm going to describe to you a way in which you could 00:17:36.090 --> 00:17:37.870 get directly at this answer, which 00:17:37.870 --> 00:17:41.320 reminds us something about angular 00:17:41.320 --> 00:17:43.530 momentum and these bodies. 00:17:43.530 --> 00:17:48.830 This object is made up of two objects. 00:17:51.480 --> 00:17:55.970 And we're spinning it about the center of one of them. 00:17:55.970 --> 00:17:58.890 So let's take them one at a time. 00:17:58.890 --> 00:18:09.350 How much force is required to keep this spinning 00:18:09.350 --> 00:18:14.170 about that axle that goes right through its center of mass? 00:18:14.170 --> 00:18:15.670 None, right? 00:18:15.670 --> 00:18:18.270 Because there's no r omega squared. 00:18:18.270 --> 00:18:19.520 There's no acceleration. 00:18:19.520 --> 00:18:21.980 You're not making the center of mass of this object 00:18:21.980 --> 00:18:23.010 go in a circle. 00:18:23.010 --> 00:18:28.340 So this piece doesn't enter into the solution. 00:18:28.340 --> 00:18:32.320 The actual force, the mass, this mass here, 00:18:32.320 --> 00:18:35.080 is M2 in the problem. 00:18:35.080 --> 00:18:39.350 M2 doesn't actually have anything to do with the answer. 00:18:39.350 --> 00:18:40.820 Because it's perfectly statically 00:18:40.820 --> 00:18:42.630 balanced around that point. 00:18:42.630 --> 00:18:45.500 All of this force making the thing go in a circle 00:18:45.500 --> 00:18:49.730 is making this piece go in a circle. 00:18:49.730 --> 00:18:54.380 So in fact, this force up here is minus M1. 00:18:56.910 --> 00:19:02.790 And now we need the distance from the axis of rotation 00:19:02.790 --> 00:19:06.240 to its center of mass. 00:19:06.240 --> 00:19:07.840 Well, that's some r. 00:19:07.840 --> 00:19:09.390 We need to know what that r is. 00:19:09.390 --> 00:19:13.970 And I think that's basically-- this was L/2. 00:19:13.970 --> 00:19:17.250 And this was b, so plus b/2. 00:19:17.250 --> 00:19:19.860 So the distance from this rotation point 00:19:19.860 --> 00:19:34.280 to that center of mass is L/2 plus b/2, omega z squared. 00:19:34.280 --> 00:19:37.430 Mass-- this is acceleration, and it's inwards. 00:19:37.430 --> 00:19:38.740 And that's the total force. 00:19:41.270 --> 00:19:43.020 All right, let's look at the next problem. 00:19:47.580 --> 00:19:49.320 Vicente, can you pull up the next one? 00:20:06.380 --> 00:20:08.530 OK, this problem-- this problem and the next 00:20:08.530 --> 00:20:09.470 are rather similar. 00:20:09.470 --> 00:20:11.859 This is the elevator problem with the pendulum 00:20:11.859 --> 00:20:12.525 in the elevator. 00:20:16.604 --> 00:20:18.580 There's a couple subtleties here. 00:20:18.580 --> 00:20:21.580 The elevator, you're told that it's moving upwards 00:20:21.580 --> 00:20:23.740 at some particular rate of acceleration, 00:20:23.740 --> 00:20:29.220 some y double dot. 00:20:29.220 --> 00:20:30.140 That's given. 00:20:30.140 --> 00:20:32.963 So how many degrees of freedom does this problem have? 00:20:36.810 --> 00:20:38.230 OK, two or one? 00:20:38.230 --> 00:20:39.530 I hear one. 00:20:39.530 --> 00:20:42.230 I see a two. 00:20:42.230 --> 00:20:43.960 How many coordinates is it going to take 00:20:43.960 --> 00:20:46.485 to completely describe the motion, 00:20:46.485 --> 00:20:48.390 or how many equations of motion do 00:20:48.390 --> 00:20:49.681 you think you're going to need? 00:20:52.280 --> 00:20:53.400 I see ones and twos. 00:20:53.400 --> 00:20:55.950 So there's a little not-- two, OK. 00:20:59.340 --> 00:21:04.290 If we know the acceleration of the thing-- we start 00:21:04.290 --> 00:21:06.110 with some initial conditions, at time 0, 00:21:06.110 --> 00:21:08.140 it's sitting on the ground, and y equals 0, 00:21:08.140 --> 00:21:11.770 and now it takes off from the first floor-- we 00:21:11.770 --> 00:21:15.300 know for all time its position. 00:21:15.300 --> 00:21:17.570 So that's actually a given. 00:21:17.570 --> 00:21:19.950 It's a specified thing. 00:21:19.950 --> 00:21:22.610 And we don't have to write a separate equation of motion 00:21:22.610 --> 00:21:26.540 for that that we then will have to solve. 00:21:26.540 --> 00:21:27.940 So it's actually given. 00:21:27.940 --> 00:21:30.580 So the actually only dynamic equation 00:21:30.580 --> 00:21:32.730 of motion we have to write is about the pendulum. 00:21:32.730 --> 00:21:34.540 Now, will it involve y double dot? 00:21:34.540 --> 00:21:35.610 Absolutely. 00:21:35.610 --> 00:21:36.657 But it's a given number. 00:21:36.657 --> 00:21:37.990 It's just a number you're given. 00:21:37.990 --> 00:21:40.239 You don't have to write a separate equation of motion. 00:21:40.239 --> 00:21:42.571 So actually you've got just one degree of freedom. 00:21:42.571 --> 00:21:43.071 Yeah. 00:21:43.071 --> 00:21:45.230 STUDENT: [INAUDIBLE] 00:21:45.230 --> 00:21:51.680 PROFESSOR: OK, so put up the next problem-- that one. 00:21:51.680 --> 00:21:55.090 So this problem has how many degrees of freedom? 00:21:55.090 --> 00:21:56.540 This is definitely two. 00:21:56.540 --> 00:21:57.850 Because you don't know. 00:21:57.850 --> 00:22:01.570 The motion in the up and down of the mass 00:22:01.570 --> 00:22:04.104 and the slider with a spring on it is unknown. 00:22:04.104 --> 00:22:06.520 You're going to have to write an equation of motion, which 00:22:06.520 --> 00:22:11.290 would have to be solved to get it, so a distinction, 00:22:11.290 --> 00:22:12.790 a subtlety that can trip you up. 00:22:12.790 --> 00:22:15.390 When the motion is actually specified, 00:22:15.390 --> 00:22:17.280 and you know what it is for all time, 00:22:17.280 --> 00:22:20.890 you don't need a separate equation for that. 00:22:20.890 --> 00:22:23.060 So this problem requires two. 00:22:23.060 --> 00:22:25.150 The behavior of the pendulum part 00:22:25.150 --> 00:22:29.390 is essentially the same in both. 00:22:29.390 --> 00:22:31.840 When you write the equation-- well, let's talk about it. 00:22:31.840 --> 00:22:34.650 How will we go about solving this problem? 00:22:34.650 --> 00:22:36.630 We need two equations, two coordinates. 00:22:36.630 --> 00:22:39.293 What would you pick for your coordinates? 00:22:39.293 --> 00:22:40.126 STUDENT: [INAUDIBLE] 00:22:43.920 --> 00:22:45.630 PROFESSOR: Theta, and then a coordinate. 00:22:45.630 --> 00:22:47.950 This one I guess is called y, so y 00:22:47.950 --> 00:22:50.520 for the square block sliding up and down, 00:22:50.520 --> 00:22:53.510 and a theta for the pendulum, two equations of motion. 00:22:53.510 --> 00:22:56.670 And to get the y equation of motion, 00:22:56.670 --> 00:22:58.600 you would use-- we're not using Lagrange here. 00:22:58.600 --> 00:23:00.725 We're going to do a little review of direct method. 00:23:00.725 --> 00:23:05.670 How would you write the equation of motion using-- up there 00:23:05.670 --> 00:23:08.380 I kind of wrote the key to everything 00:23:08.380 --> 00:23:11.740 that we do with the direct method, some of the forces, 00:23:11.740 --> 00:23:14.280 some of the torques. 00:23:14.280 --> 00:23:19.730 Every rigid body has at max how many degrees of freedom? 00:23:19.730 --> 00:23:21.530 Six for rigid body. 00:23:21.530 --> 00:23:23.910 In planar motion, this is reduced to three. 00:23:23.910 --> 00:23:27.640 But every rigid body in 3D space has 00:23:27.640 --> 00:23:32.180 six degrees of freedom-- three positions, three rotations. 00:23:32.180 --> 00:23:35.070 So you write some of the forces on the body. 00:23:35.070 --> 00:23:37.590 That is equal to the mass times acceleration. 00:23:37.590 --> 00:23:40.810 That gives you three equations, one in each vector, component 00:23:40.810 --> 00:23:41.710 direction. 00:23:41.710 --> 00:23:43.990 You write the sum of the torques. 00:23:43.990 --> 00:23:45.490 That's also a vector equation. 00:23:45.490 --> 00:23:47.780 You can get as many as three equations out of it 00:23:47.780 --> 00:23:49.060 if you need it. 00:23:49.060 --> 00:23:50.960 Then we keep reducing down how many 00:23:50.960 --> 00:23:54.390 equations we need to get by figuring out the constraints 00:23:54.390 --> 00:23:58.170 until we get down to the number of actual degrees of freedom. 00:23:58.170 --> 00:24:01.440 And that's the remaining equations, the number 00:24:01.440 --> 00:24:03.150 of equations you actually need. 00:24:03.150 --> 00:24:06.060 So this one requires two. 00:24:06.060 --> 00:24:09.000 And if you're going to write an equation of motion 00:24:09.000 --> 00:24:14.910 about the main mass that slides up and down, what law would 00:24:14.910 --> 00:24:15.480 you use? 00:24:15.480 --> 00:24:16.440 STUDENT: [INAUDIBLE] 00:24:16.440 --> 00:24:18.010 PROFESSOR: Yeah, sum of the forces 00:24:18.010 --> 00:24:20.219 equal to the mass times acceleration. 00:24:20.219 --> 00:24:21.760 Then we need to get another question. 00:24:21.760 --> 00:24:23.936 What law would you use for the second one? 00:24:23.936 --> 00:24:24.810 STUDENT: Torque. 00:24:24.810 --> 00:24:26.660 PROFESSOR: Torque about where? 00:24:26.660 --> 00:24:27.500 STUDENT: A. 00:24:27.500 --> 00:24:30.310 PROFESSOR: Torque about A. So let's look at that. 00:24:30.310 --> 00:24:32.790 So that's the second equation up there. 00:24:32.790 --> 00:24:36.530 Torque about A is the time derivative 00:24:36.530 --> 00:24:38.710 of the angular momentum with respect to A plus 00:24:38.710 --> 00:24:43.750 that vA cross momentum term, which is 00:24:43.750 --> 00:24:46.780 a little annoying to work out. 00:24:46.780 --> 00:24:50.360 So the simpler version is the third equation. 00:24:50.360 --> 00:24:53.120 It's a pretty quick derivation to go from that second equation 00:24:53.120 --> 00:24:54.290 to the third. 00:24:54.290 --> 00:24:57.970 And it's much less work, generally, 00:24:57.970 --> 00:25:01.270 to get that mass times acceleration 00:25:01.270 --> 00:25:04.160 term, a lot less work actually. 00:25:04.160 --> 00:25:07.340 And if I've got time, I'll go back 00:25:07.340 --> 00:25:09.440 and we'll work out a problem like this. 00:25:09.440 --> 00:25:11.160 And we'll do that part. 00:25:11.160 --> 00:25:14.110 But let's go on to the other questions. 00:25:14.110 --> 00:25:19.220 OK, this problem-- two rigid bodies. 00:25:19.220 --> 00:25:21.810 Is it planar motion? 00:25:21.810 --> 00:25:22.492 STUDENT: Yes. 00:25:22.492 --> 00:25:23.200 PROFESSOR: Right? 00:25:23.200 --> 00:25:27.180 How many possible degrees of freedom for each rigid body? 00:25:27.180 --> 00:25:31.970 Times 2, 6, now how many constraints? 00:25:31.970 --> 00:25:33.880 So let's take them one rigid body at a time. 00:25:33.880 --> 00:25:36.551 It's a big roller. 00:25:36.551 --> 00:25:37.050 Pardon? 00:25:37.050 --> 00:25:37.730 STUDENT: No y. 00:25:37.730 --> 00:25:38.850 PROFESSOR: No y, she says. 00:25:38.850 --> 00:25:41.120 OK, that's one constraint in it-- two left, 00:25:41.120 --> 00:25:42.940 two possible things left. 00:25:42.940 --> 00:25:43.720 STUDENT: No slip. 00:25:43.720 --> 00:25:46.780 PROFESSOR: Ahh, so I think this one's no slip. 00:25:46.780 --> 00:25:49.970 So how does a no slip then give you a constraint? 00:25:49.970 --> 00:25:54.570 STUDENT: So x dot equals r omega of the wheel turning. 00:25:54.570 --> 00:25:56.920 PROFESSOR: So r, there's an r. 00:25:56.920 --> 00:25:59.530 X equals r theta, or minus r theta 00:25:59.530 --> 00:26:02.570 depending on how you define the theta, right? 00:26:02.570 --> 00:26:08.239 So x is not independent of theta. 00:26:08.239 --> 00:26:09.530 And that's a second constraint. 00:26:09.530 --> 00:26:14.340 So you only need one equation, one generalized coordinate, 00:26:14.340 --> 00:26:15.872 to describe the motion of the wheel. 00:26:15.872 --> 00:26:16.830 And what would you use? 00:26:19.620 --> 00:26:20.495 STUDENT: Theta? 00:26:20.495 --> 00:26:21.370 PROFESSOR: OK, theta. 00:26:21.370 --> 00:26:22.203 You could use theta. 00:26:22.203 --> 00:26:23.850 Or you could use x. 00:26:23.850 --> 00:26:27.290 All right, and now the other object, how many constraints 00:26:27.290 --> 00:26:30.075 does it have? 00:26:30.075 --> 00:26:31.142 And what are they? 00:26:31.142 --> 00:26:32.725 What are the constraints on the T bar? 00:26:36.032 --> 00:26:39.860 STUDENT: Can't translate the y. 00:26:39.860 --> 00:26:41.490 PROFESSOR: OK, that's true. 00:26:41.490 --> 00:26:44.940 Let me give you a little hint about how I process 00:26:44.940 --> 00:26:46.500 this when I'm looking at it. 00:26:46.500 --> 00:26:53.230 Remember when you're picking generalized coordinates, 00:26:53.230 --> 00:26:55.530 they need to be independent. 00:26:55.530 --> 00:26:57.880 And independent means if you freeze all but one, 00:26:57.880 --> 00:26:59.810 that last one can still move. 00:26:59.810 --> 00:27:01.360 OK, so we've picked one. 00:27:01.360 --> 00:27:03.410 You said theta or x. 00:27:03.410 --> 00:27:05.710 Let's freeze it. 00:27:05.710 --> 00:27:09.154 What motion is left of that T bar? 00:27:09.154 --> 00:27:10.819 STUDENT: [INAUDIBLE] 00:27:10.819 --> 00:27:12.360 PROFESSOR: It can only rotate, right? 00:27:12.360 --> 00:27:15.800 And that means it has a pin, a point 00:27:15.800 --> 00:27:20.170 about at which it rotates, which you have just fixed. 00:27:20.170 --> 00:27:21.210 So it's constrained. 00:27:21.210 --> 00:27:24.005 That pin constrains it in how many directions? 00:27:24.005 --> 00:27:25.250 STUDENT: Two. 00:27:25.250 --> 00:27:26.940 PROFESSOR: Two, right off the bat. 00:27:26.940 --> 00:27:29.050 You have the two constraints right at the pin. 00:27:29.050 --> 00:27:31.840 But it's not so obvious until you say, let's freeze 00:27:31.840 --> 00:27:33.594 at other coordinates. 00:27:33.594 --> 00:27:35.010 And then we see that, ahh, there's 00:27:35.010 --> 00:27:37.470 only one possible motion left, the rotation. 00:27:37.470 --> 00:27:40.970 OK, so we have a rotation and a translation, or two rotations. 00:27:40.970 --> 00:27:44.720 That's the way we could write equations of motion. 00:27:44.720 --> 00:27:47.910 If we did translation of the main disk, 00:27:47.910 --> 00:27:51.710 what equation would you use to write the equation of motion 00:27:51.710 --> 00:27:55.140 for the main disk? 00:27:55.140 --> 00:27:56.600 STUDENT: Force. 00:27:56.600 --> 00:27:57.950 PROFESSOR: Force, yeah. 00:27:57.950 --> 00:28:01.190 Newton's second law, sum of the forces in what direction? 00:28:03.985 --> 00:28:05.607 It can only move in-- 00:28:05.607 --> 00:28:06.440 STUDENT: [INAUDIBLE] 00:28:06.440 --> 00:28:07.630 PROFESSOR: Horizontal, yeah. 00:28:07.630 --> 00:28:09.740 So sum of the forces in the x direction, 00:28:09.740 --> 00:28:12.120 sum of external forces in the x direction, 00:28:12.120 --> 00:28:14.160 equals mass times acceleration. 00:28:14.160 --> 00:28:16.870 Now this, though, has some problematic external forces, 00:28:16.870 --> 00:28:18.300 right? 00:28:18.300 --> 00:28:21.499 What are they, the difficult ones? 00:28:21.499 --> 00:28:22.397 STUDENT: [INAUDIBLE] 00:28:22.397 --> 00:28:23.480 PROFESSOR: Well, friction. 00:28:23.480 --> 00:28:24.521 There's a friction force. 00:28:24.521 --> 00:28:31.020 And there's also the internal forces at the pin. 00:28:31.020 --> 00:28:33.310 So you're going to have to sort out how 00:28:33.310 --> 00:28:34.920 to work your way through that. 00:28:34.920 --> 00:28:37.710 If you don't want to mess with external forces at the pin, 00:28:37.710 --> 00:28:42.133 you have to sum rotations at the-- 00:28:42.133 --> 00:28:42.966 STUDENT: [INAUDIBLE] 00:28:46.626 --> 00:28:48.000 PROFESSOR: Yeah, so this problem, 00:28:48.000 --> 00:28:50.220 there's no escaping something messy. 00:28:53.290 --> 00:28:59.850 So let's say one equation will be 00:28:59.850 --> 00:29:03.010 the sum of the forces on that main roller 00:29:03.010 --> 00:29:04.610 in the horizontal direction. 00:29:04.610 --> 00:29:06.880 And that's going to force us to deal with a friction 00:29:06.880 --> 00:29:11.720 force and an internal, two reaction forces at the pin. 00:29:11.720 --> 00:29:13.360 Well, you just write them down. 00:29:13.360 --> 00:29:18.200 Then go and write the same expression, sum of the forces 00:29:18.200 --> 00:29:19.580 on the other object. 00:29:19.580 --> 00:29:22.809 And it'll also show up with those two internal forces. 00:29:22.809 --> 00:29:24.350 You add those two equations together. 00:29:24.350 --> 00:29:25.183 Those forces cancel. 00:29:27.700 --> 00:29:30.190 And that gives you the first equation. 00:29:30.190 --> 00:29:33.880 And it'll involve acceleration of both masses. 00:29:33.880 --> 00:29:36.220 And you'll have to figure out the acceleration 00:29:36.220 --> 00:29:39.164 of that second mass around its own. 00:29:39.164 --> 00:29:40.330 But you know how to do that. 00:29:40.330 --> 00:29:41.288 That's just kinematics. 00:29:44.340 --> 00:29:46.780 And then you've got to do the torque equation. 00:29:46.780 --> 00:29:50.300 And you'd probably do it about A. 00:29:50.300 --> 00:29:56.752 And does this involve-- what about the velocity of A 00:29:56.752 --> 00:29:57.680 in this problem? 00:29:57.680 --> 00:29:59.610 Is it 0? 00:29:59.610 --> 00:30:02.400 No, so you may have to deal with that second term. 00:30:02.400 --> 00:30:04.740 And again, I would go use the third expression. 00:30:04.740 --> 00:30:08.460 Because you need to find the accelerations anyway. 00:30:08.460 --> 00:30:10.480 And I would use that. 00:30:10.480 --> 00:30:11.302 Yeah. 00:30:11.302 --> 00:30:14.719 STUDENT: Could you also take the sum of the torques about d? 00:30:14.719 --> 00:30:16.260 PROFESSOR: Some of the torques about? 00:30:16.260 --> 00:30:16.560 STUDENT: d? 00:30:16.560 --> 00:30:18.310 PROFESSOR: Then d is the point of contact. 00:30:22.480 --> 00:30:27.291 Yeah, and that would give you a way of getting at, 00:30:27.291 --> 00:30:29.040 I think, an equation of motion that mostly 00:30:29.040 --> 00:30:30.450 describes the big roller. 00:30:36.840 --> 00:30:38.910 I don't know. 00:30:38.910 --> 00:30:41.880 In terms of the other one, I'm not-- 00:30:41.880 --> 00:30:45.270 summing it, then talking about the second rigid body? 00:30:45.270 --> 00:30:47.460 I don't think I would do it. 00:30:47.460 --> 00:30:49.146 It gets a little complicated. 00:30:49.146 --> 00:30:50.020 But it's interesting. 00:30:50.020 --> 00:30:53.057 It might be worth a try. 00:30:53.057 --> 00:30:55.390 I'd probably do it about A, is the point I would choose, 00:30:55.390 --> 00:30:57.980 and deal with the fact that you need 00:30:57.980 --> 00:31:01.070 to know the acceleration of A. But in this problem, 00:31:01.070 --> 00:31:02.320 that's pretty straightforward. 00:31:02.320 --> 00:31:04.100 What's the acceleration of point A? 00:31:04.100 --> 00:31:08.380 And let's say our horizontal coordinate is x. 00:31:08.380 --> 00:31:10.520 What's the acceleration of point A? 00:31:10.520 --> 00:31:13.310 x double dot in the I direction-- pretty easy 00:31:13.310 --> 00:31:17.210 to stick that in this third expression up here. 00:31:17.210 --> 00:31:20.710 And you know Rg with respect to A 00:31:20.710 --> 00:31:25.290 is just the length of the distance down 00:31:25.290 --> 00:31:28.730 to the center of mass of that T. So that term's 00:31:28.730 --> 00:31:29.930 pretty easy to figure out. 00:31:29.930 --> 00:31:32.810 And then the lead term is just everything with respect to g. 00:31:32.810 --> 00:31:35.360 And that's pretty straightforward. 00:31:35.360 --> 00:31:38.240 OK, good, how are we doing on time? 00:31:38.240 --> 00:31:39.180 Not a lot left. 00:31:39.180 --> 00:31:41.880 And is that the last problem? 00:31:41.880 --> 00:31:42.519 Yeah. 00:31:42.519 --> 00:31:43.518 STUDENT: Quick question. 00:31:43.518 --> 00:31:46.018 The previous problem, we don't need to worry about friction, 00:31:46.018 --> 00:31:48.427 do we? 00:31:48.427 --> 00:31:50.010 STUDENT: If it was slipping, we would. 00:31:50.010 --> 00:31:53.410 PROFESSOR: Yeah, will friction ultimately 00:31:53.410 --> 00:31:57.565 end up in the answer to this one? 00:31:57.565 --> 00:31:58.065 No. 00:32:01.073 --> 00:32:01.906 STUDENT: [INAUDIBLE] 00:32:06.790 --> 00:32:08.880 PROFESSOR: Well, speaking in Lagrange terms, 00:32:08.880 --> 00:32:10.256 does it do any work? 00:32:10.256 --> 00:32:11.994 STUDENT: [INAUDIBLE] 00:32:11.994 --> 00:32:14.160 PROFESSOR: I know, but you know about it now, right? 00:32:14.160 --> 00:32:16.350 Constraints that do no work usually 00:32:16.350 --> 00:32:19.045 don't end up in the final answer. 00:32:27.220 --> 00:32:31.050 I don't have a simple way of explaining why it doesn't. 00:32:31.050 --> 00:32:35.115 But if in the first case, instead of doing Newton's law 00:32:35.115 --> 00:32:38.730 on the first one, we had done torques about d, 00:32:38.730 --> 00:32:40.850 it would have given us an equation of motion 00:32:40.850 --> 00:32:43.040 for the roller. 00:32:43.040 --> 00:32:44.540 And then you could have done torques 00:32:44.540 --> 00:32:46.629 about A, a different equation. 00:32:46.629 --> 00:32:48.920 And we'd have got an equation of motion largely dealing 00:32:48.920 --> 00:32:51.910 with T. You still have some internal forces 00:32:51.910 --> 00:32:54.470 and torques that you're going to have to eliminate. 00:32:54.470 --> 00:32:59.280 So this is not a simple problem, but straightforward if you 00:32:59.280 --> 00:33:04.080 know what laws to apply to some of the forces, some 00:33:04.080 --> 00:33:04.820 of the torques. 00:33:04.820 --> 00:33:08.360 And be careful with the extra term. 00:33:08.360 --> 00:33:11.680 If this object were just given an initial deflection, 00:33:11.680 --> 00:33:14.275 like the T bar is picked up and let go, and it's just there, 00:33:14.275 --> 00:33:16.800 and it's just doing its thing, what 00:33:16.800 --> 00:33:20.225 can you say about the center of mass of the system? 00:33:23.342 --> 00:33:26.150 STUDENT: [INAUDIBLE] 00:33:26.150 --> 00:33:30.215 PROFESSOR: Well, it's not that it doesn't necessarily move. 00:33:35.500 --> 00:33:36.780 Well, let's put it this way. 00:33:36.780 --> 00:33:39.830 What if I gave this thing a push to start with, 00:33:39.830 --> 00:33:41.751 and now it's going to roll along? 00:33:41.751 --> 00:33:42.250 Huh? 00:33:42.250 --> 00:33:43.600 STUDENT: [INAUDIBLE] 00:33:43.600 --> 00:33:46.160 PROFESSOR: OK, in that case, you're 00:33:46.160 --> 00:33:50.550 saying that the center of mass moves with constant velocity. 00:33:50.550 --> 00:33:52.550 And Newton would agree with you. 00:33:52.550 --> 00:33:53.159 Why? 00:33:53.159 --> 00:33:53.992 STUDENT: [INAUDIBLE] 00:33:57.349 --> 00:33:59.390 PROFESSOR: Because once it gets rolling, is there 00:33:59.390 --> 00:34:00.300 any friction force? 00:34:00.300 --> 00:34:01.012 STUDENT: No. 00:34:01.012 --> 00:34:03.262 PROFESSOR: So you've got to draw the free body diagram 00:34:03.262 --> 00:34:06.100 and decide whether or not there are any forces on the system. 00:34:06.100 --> 00:34:08.500 If there are no external forces on the system, 00:34:08.500 --> 00:34:13.710 Newton says no acceleration, no change in momentum. 00:34:13.710 --> 00:34:15.659 So once you get it rolling, there's 00:34:15.659 --> 00:34:17.070 actually no friction force. 00:34:17.070 --> 00:34:20.790 Because it isn't trying to either speed up or slow down 00:34:20.790 --> 00:34:21.870 the rolling. 00:34:21.870 --> 00:34:23.449 So it goes actually to 0. 00:34:23.449 --> 00:34:25.086 Is there slip? 00:34:25.086 --> 00:34:26.610 No, it still isn't slipping. 00:34:26.610 --> 00:34:28.250 But it's just happily rolling along. 00:34:28.250 --> 00:34:30.020 There are no external forces acting on it. 00:34:30.020 --> 00:34:33.010 It means the mass times the acceleration of the center 00:34:33.010 --> 00:34:34.469 of mass of the system is zero. 00:34:34.469 --> 00:34:36.625 So the center of mass does not accelerate. 00:34:36.625 --> 00:34:38.790 That means it has constant velocity. 00:34:38.790 --> 00:34:41.639 Now, if I contrive to have the initial velocity of the center 00:34:41.639 --> 00:34:46.760 of mass be 0, just cause that thing to swing back and forth, 00:34:46.760 --> 00:34:51.290 what would you see the center of mass do? 00:34:51.290 --> 00:34:54.159 It has no linear momentum. 00:34:54.159 --> 00:34:56.120 The linear momentum of the system is 0. 00:34:56.120 --> 00:35:01.470 So the center of mass sits still. 00:35:01.470 --> 00:35:02.850 Its velocity now is 0. 00:35:02.850 --> 00:35:04.170 There's still no acceleration. 00:35:04.170 --> 00:35:07.410 It's just this T bar is rocking back and forth. 00:35:07.410 --> 00:35:09.696 So what must the roller be doing? 00:35:09.696 --> 00:35:11.010 STUDENT: [INAUDIBLE] 00:35:11.010 --> 00:35:13.810 PROFESSOR: So the roller has to move to the left 00:35:13.810 --> 00:35:15.740 when the T bar is going to the right. 00:35:15.740 --> 00:35:17.940 So the two things are going like this. 00:35:17.940 --> 00:35:21.970 In the center of mass, wherever it is, it's just sitting still. 00:35:21.970 --> 00:35:24.240 All right, is there a last problem? 00:35:26.820 --> 00:35:31.150 This forces you to go back to the fundamental definition 00:35:31.150 --> 00:35:33.060 of angular momentum. 00:35:33.060 --> 00:35:39.440 It's R cross, this little summation of each RI cross PI. 00:35:39.440 --> 00:35:42.030 And add them up, and you get the angular momentum with respect 00:35:42.030 --> 00:35:47.790 to the rotational point, 1 degree of freedom, 00:35:47.790 --> 00:35:48.691 has a center of mass. 00:35:48.691 --> 00:35:50.440 Where is the center of mass of the system? 00:35:50.440 --> 00:35:51.273 Can you describe it? 00:35:54.437 --> 00:35:55.820 STUDENT: Right in between-- 00:35:55.820 --> 00:35:57.440 PROFESSOR: Somewhere in between the two masses. 00:35:57.440 --> 00:35:59.065 There are two point masses, center mass 00:35:59.065 --> 00:36:00.610 has to be someplace in between. 00:36:00.610 --> 00:36:02.390 So once you figure out where that is, then 00:36:02.390 --> 00:36:06.440 you can concentrate all the mass there and go from there. 00:36:06.440 --> 00:36:08.400 Does this have mass moment of inertia? 00:36:13.810 --> 00:36:15.970 Can you write the equation of motion? 00:36:15.970 --> 00:36:16.840 This is a pendulum. 00:36:16.840 --> 00:36:25.580 Can you write I about A theta double dot equals minus Mg 00:36:25.580 --> 00:36:29.570 something sine theta? 00:36:29.570 --> 00:36:30.880 This is a pendulum. 00:36:30.880 --> 00:36:34.670 The restoring torque on it comes from gravity. 00:36:34.670 --> 00:36:37.150 And you'll end up with an expression 00:36:37.150 --> 00:36:39.000 that looks like some I with respect 00:36:39.000 --> 00:36:41.050 to the point of rotation. 00:36:41.050 --> 00:36:42.760 So basically, you just need to figure out 00:36:42.760 --> 00:36:48.290 what is I with respect to A. And they're particles. 00:36:48.290 --> 00:36:52.500 So each particle has an I with respect to its center of mass. 00:36:52.500 --> 00:36:59.020 It's equal to, for a particle, concentrated point mass? 00:36:59.020 --> 00:36:59.994 STUDENT: Mr squared? 00:36:59.994 --> 00:37:01.410 PROFESSOR: Yeah, but there's no r. 00:37:01.410 --> 00:37:06.890 So I for a particle about its center mass is 0. 00:37:06.890 --> 00:37:10.700 So this only has parallel axis theorem components. 00:37:10.700 --> 00:37:12.960 It has M1 times this distance squared 00:37:12.960 --> 00:37:15.480 from the center plus M2 times this distance squared 00:37:15.480 --> 00:37:16.210 from the center. 00:37:16.210 --> 00:37:19.180 And that's the total Izz. 00:37:19.180 --> 00:37:20.960 So then you can write it out. 00:37:20.960 --> 00:37:26.460 OK, so I think we've run out of time.