WEBVTT 00:00:00.070 --> 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 --> 00:00:12.690 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 --> 00:00:17.251 at ocw.mit.edu. 00:00:21.040 --> 00:00:23.540 PROFESSOR: Last, kind of under the announcements category, 00:00:23.540 --> 00:00:25.990 is I want to talk about the muddy cards. 00:00:25.990 --> 00:00:28.510 So I've used those many times in the past. 00:00:28.510 --> 00:00:30.870 Last time was the first time I handed them out. 00:00:30.870 --> 00:00:32.790 And your comments were great. 00:00:32.790 --> 00:00:38.445 This was really a good lecture to have handed them out. 00:00:38.445 --> 00:00:42.340 We covered lots of interesting important concepts. 00:00:42.340 --> 00:00:45.170 And so I'm going to review a couple of things that 00:00:45.170 --> 00:00:47.800 came up in the muddy cards. 00:00:47.800 --> 00:00:50.730 A couple of the most positive comments 00:00:50.730 --> 00:00:52.840 is people really like the demos, and they really 00:00:52.840 --> 00:00:56.535 like the explanations, especially with examples. 00:00:59.550 --> 00:01:01.320 People particularly commented it was 00:01:01.320 --> 00:01:05.180 really helpful to compute angular momentum from two 00:01:05.180 --> 00:01:07.010 different points. 00:01:07.010 --> 00:01:08.780 And you get the revelation that you 00:01:08.780 --> 00:01:12.030 get two very different answers. 00:01:12.030 --> 00:01:16.440 So that's a really important point. 00:01:16.440 --> 00:01:19.090 And somebody then asked the question, 00:01:19.090 --> 00:01:22.060 said, well, I thought that vectors 00:01:22.060 --> 00:01:25.450 were independent of the coordinate system 00:01:25.450 --> 00:01:27.270 that you select. 00:01:27.270 --> 00:01:28.080 It's true. 00:01:28.080 --> 00:01:29.620 A velocity ought to be a velocity 00:01:29.620 --> 00:01:32.660 no matter whether it's r theta or x, y, z. 00:01:32.660 --> 00:01:33.800 But why? 00:01:33.800 --> 00:01:37.990 That seems to kind of violate that notion that vectors should 00:01:37.990 --> 00:01:40.770 be independent of coordinates. 00:01:40.770 --> 00:01:43.800 And yet we computed an angular momentum with respect 00:01:43.800 --> 00:01:45.930 to one place and respect to a different place, 00:01:45.930 --> 00:01:47.260 and we got different answers. 00:01:51.460 --> 00:01:52.850 How do you resolve that? 00:01:55.836 --> 00:01:56.336 Yeah. 00:01:56.336 --> 00:01:59.822 AUDIENCE: It's sort of like since it's moving, 00:01:59.822 --> 00:02:01.316 the coordinates shouldn't matter. 00:02:01.316 --> 00:02:05.300 Like if it was equilibrium, it wouldn't matter where 00:02:05.300 --> 00:02:06.794 you would put the [INAUDIBLE]. 00:02:14.264 --> 00:02:17.750 PROFESSOR: OK, well, you're getting close. 00:02:17.750 --> 00:02:21.380 Here was the problem I think we did. 00:02:21.380 --> 00:02:24.700 And we chose this point, which I'll call 2, 00:02:24.700 --> 00:02:27.410 and this point, which I'll call 1. 00:02:27.410 --> 00:02:34.960 And we computed h1. 00:02:34.960 --> 00:02:37.740 I'll just call h1 with respect to point 1. 00:02:37.740 --> 00:02:40.970 We'll call this A. We computed with respect to 1. 00:02:40.970 --> 00:02:43.510 And we computed the angular momentum of A 00:02:43.510 --> 00:02:45.250 with respect to 2. 00:02:45.250 --> 00:02:50.250 But in this case, angular momentum of a particle with 00:02:50.250 --> 00:02:57.710 respect to some location, origin of a coordinate system, 00:02:57.710 --> 00:03:04.440 is defined as r of the partial with respect to the coordinate 00:03:04.440 --> 00:03:10.920 system crossed with the linear momentum of the particle that 00:03:10.920 --> 00:03:14.095 you're-- we'll call it B here, just the name of the particle. 00:03:17.430 --> 00:03:20.520 The definition-- these are both vectors. 00:03:20.520 --> 00:03:22.740 You arch changing the vector. 00:03:22.740 --> 00:03:24.030 It's a different vector. 00:03:24.030 --> 00:03:28.960 Because this changes in the two parts. 00:03:28.960 --> 00:03:31.860 So it's not a constant vector at all. 00:03:31.860 --> 00:03:33.960 By its definition, it is just something 00:03:33.960 --> 00:03:37.180 different when you move to a different place and this piece 00:03:37.180 --> 00:03:37.680 changes. 00:03:37.680 --> 00:03:40.700 This piece is invariant, but this piece is not. 00:03:40.700 --> 00:03:43.390 And that's the answer to that one. 00:03:43.390 --> 00:03:46.730 Lots of people were still not clear about Coriolis. 00:03:46.730 --> 00:03:50.090 We'll work on that as time goes by. 00:03:50.090 --> 00:03:56.530 And people were interested in how 00:03:56.530 --> 00:03:59.900 to pick reference frames and so forth. 00:03:59.900 --> 00:04:02.487 Somebody made the suggestion, try using some colored chalk. 00:04:02.487 --> 00:04:03.070 It would help. 00:04:03.070 --> 00:04:04.319 I don't own any colored chalk. 00:04:04.319 --> 00:04:06.594 My assistant just walked in with some. 00:04:06.594 --> 00:04:08.010 She found some at the last minute. 00:04:08.010 --> 00:04:08.980 So I'll try doing that. 00:04:08.980 --> 00:04:10.032 That's a good idea. 00:04:10.032 --> 00:04:11.990 And someone else says, take a break in an hour. 00:04:11.990 --> 00:04:13.850 And that's a pretty good idea, too. 00:04:13.850 --> 00:04:15.842 I'll try to remember to do that. 00:04:15.842 --> 00:04:17.050 So the muddy cards are great. 00:04:17.050 --> 00:04:21.890 Please today we'll do the same thing. 00:04:21.890 --> 00:04:27.690 So let's start with this topic. 00:04:31.750 --> 00:04:34.570 It's a subject which we constantly 00:04:34.570 --> 00:04:39.160 use throughout the course doing dynamics. 00:04:39.160 --> 00:04:40.880 You have to be able to figure out 00:04:40.880 --> 00:04:43.710 coordinate systems, degrees of freedom, 00:04:43.710 --> 00:04:45.070 drawing free body diagrams. 00:04:45.070 --> 00:04:52.930 So I'm going to do a few quick examples-- coordinates, fbd's. 00:04:56.140 --> 00:04:58.960 I picked some examples here just to emphasize 00:04:58.960 --> 00:05:02.010 a few different points. 00:05:02.010 --> 00:05:04.010 Here we have a slope. 00:05:04.010 --> 00:05:07.130 I've got a wheel. 00:05:07.130 --> 00:05:07.970 It's a rigid body. 00:05:12.230 --> 00:05:14.170 I pick a preliminary coordinate system. 00:05:14.170 --> 00:05:16.420 Sometimes you do that just to help you think about it. 00:05:19.050 --> 00:05:20.900 And now let's talk about degrees of freedom. 00:05:20.900 --> 00:05:22.730 What do we mean by degrees of freedom? 00:05:27.340 --> 00:05:39.640 I'm going to define it as the number of independent 00:05:39.640 --> 00:06:00.585 coordinates necessary to describe the motion. 00:06:05.640 --> 00:06:08.380 So it's the number of independent coordinates 00:06:08.380 --> 00:06:10.690 that you need. 00:06:10.690 --> 00:06:19.740 Now, with few exceptions, I can compute that 00:06:19.740 --> 00:06:28.270 by multiplying 6 times the number of rigid bodies 00:06:28.270 --> 00:06:33.750 in the problem plus 3 times the number of particles 00:06:33.750 --> 00:06:37.140 minus the constraints, the number of constraints. 00:06:37.140 --> 00:06:40.445 So this is the number of rigid bodies. 00:06:44.840 --> 00:06:47.885 This is the number of particles. 00:06:51.280 --> 00:06:53.205 And C, this is the constraints. 00:07:01.650 --> 00:07:03.650 So take a look at this problem. 00:07:03.650 --> 00:07:06.750 A wheel's a rigid body. 00:07:06.750 --> 00:07:08.970 The difference between a rigid body and a particle 00:07:08.970 --> 00:07:10.440 is a rigid body is big enough. 00:07:10.440 --> 00:07:12.600 It has mass at some extent. 00:07:12.600 --> 00:07:14.910 Its rotational inertia will matter. 00:07:14.910 --> 00:07:16.790 So here we've got a wheel. 00:07:16.790 --> 00:07:19.300 So we've got certainly one rigid body but no particles. 00:07:19.300 --> 00:07:21.676 So in this case, n is 1, m is 0. 00:07:21.676 --> 00:07:23.300 And so the number of degrees of freedom 00:07:23.300 --> 00:07:25.216 that we should come up with in this problem is 00:07:25.216 --> 00:07:28.620 going to look like 6 minus C. 00:07:28.620 --> 00:07:30.060 And so then the problem becomes-- 00:07:30.060 --> 00:07:32.470 let's identify what the constraints are in the problem. 00:07:36.070 --> 00:07:38.010 So I drew my initial little coordinate system 00:07:38.010 --> 00:07:42.150 just so that I can use language like x, y, and z direction. 00:07:42.150 --> 00:07:43.910 So z's coming out of the board. 00:07:43.910 --> 00:07:46.000 So in this problem, let's figure out 00:07:46.000 --> 00:07:47.330 how many constraints there are. 00:08:00.400 --> 00:08:02.555 How about the y direction? 00:08:05.709 --> 00:08:07.500 The simplest kind of constraints are things 00:08:07.500 --> 00:08:08.800 that just allow no motion. 00:08:08.800 --> 00:08:12.230 Can it move in the y direction? 00:08:12.230 --> 00:08:19.019 No, OK, so y, y dot, y double dot, are 0. 00:08:19.019 --> 00:08:20.185 So that's a hard constraint. 00:08:20.185 --> 00:08:22.555 You can't move through the wall in the y direction. 00:08:25.920 --> 00:08:30.971 You have to make assumptions when you're doing problems. 00:08:30.971 --> 00:08:33.179 You have to try to simplify things as much as you can 00:08:33.179 --> 00:08:34.303 so that you make them easy. 00:08:34.303 --> 00:08:37.500 So really I'm going to assume in this problem-- I 00:08:37.500 --> 00:08:41.210 haven't shown any constraints or wheels or guides or rollers. 00:08:41.210 --> 00:08:46.312 But I'm going to assume that it won't move in the z direction. 00:08:46.312 --> 00:08:47.520 So that's another constraint. 00:08:47.520 --> 00:08:49.740 So this implies 1. 00:08:49.740 --> 00:08:51.300 This implies another constraint. 00:08:56.090 --> 00:08:59.110 If I don't do this, if I don't assume that, 00:08:59.110 --> 00:09:02.370 then I just end up with another equation of motion. 00:09:02.370 --> 00:09:04.830 For every degree of freedom, you end up with a problem. 00:09:04.830 --> 00:09:07.180 You're going to need an equation of motion. 00:09:07.180 --> 00:09:09.020 So if I did not make this assumption, 00:09:09.020 --> 00:09:13.300 I'd say that the summation of the forces in the z direction 00:09:13.300 --> 00:09:18.000 is equal to 0. 00:09:18.000 --> 00:09:21.980 And that's equal to the mass times the acceleration 00:09:21.980 --> 00:09:25.670 of the body in the z acceleration. 00:09:25.670 --> 00:09:28.327 It's a vector, vector component. 00:09:28.327 --> 00:09:29.910 And then you just say, oh well, that's 00:09:29.910 --> 00:09:31.470 a trivial equation of motion. 00:09:31.470 --> 00:09:33.095 And so you could deal with it that way. 00:09:33.095 --> 00:09:36.110 But we'll just assume that we have no motion in the z. 00:09:36.110 --> 00:09:39.710 And that gives us another constraint. 00:09:39.710 --> 00:09:42.840 Now we can assume again that there's 00:09:42.840 --> 00:09:45.580 constraints in the problem, or it's well behaved, 00:09:45.580 --> 00:09:48.000 in that the thing won't fall over. 00:09:48.000 --> 00:09:49.910 It won't roll over. 00:09:49.910 --> 00:09:53.400 And it won't change direction running down the hill. 00:09:53.400 --> 00:10:06.410 So we'll assume no rotation about the x or y axes. 00:10:09.450 --> 00:10:11.665 So that implies two more constraints. 00:10:14.500 --> 00:10:19.560 And finally, the constraints can come in many flavors. 00:10:19.560 --> 00:10:22.800 Finally we know that in this problem 00:10:22.800 --> 00:10:36.570 that I need to think in terms of a rotation. 00:10:36.570 --> 00:10:40.120 So there's a positive rotation in that coordinate system. 00:10:40.120 --> 00:10:42.270 So now I have a rotational coordinate 00:10:42.270 --> 00:10:46.910 that I can think about. 00:10:46.910 --> 00:10:52.710 But this now says that if there's no slip, 00:10:52.710 --> 00:10:57.030 I can say that the distance it rolls down the hill 00:10:57.030 --> 00:11:00.490 is minus r theta. 00:11:00.490 --> 00:11:03.540 That's a constraint. 00:11:03.540 --> 00:11:07.640 x and theta are not independent of one another. 00:11:07.640 --> 00:11:11.840 We're looking for the number of independent coordinates 00:11:11.840 --> 00:11:15.040 required to completely describe the motion. 00:11:15.040 --> 00:11:17.840 x and r are not independent because of this no slip 00:11:17.840 --> 00:11:19.090 condition. 00:11:19.090 --> 00:11:22.500 And so that implies yet another. 00:11:22.500 --> 00:11:30.620 OK, so we've got one, two, three, four, five constraints. 00:11:30.620 --> 00:11:32.890 And we said that the number of degrees 00:11:32.890 --> 00:11:38.670 of freedom in this problem is equal to 6 minus 5, which is 1. 00:11:38.670 --> 00:11:40.120 So you take the single coordinate. 00:11:40.120 --> 00:11:42.305 You could have told me that long ago that that's 00:11:42.305 --> 00:11:43.570 what it's going to take. 00:11:43.570 --> 00:11:45.180 But this is the sort of thinking you 00:11:45.180 --> 00:11:48.195 have to go through to come up with all these constraints. 00:11:48.195 --> 00:11:50.070 So this is going to take a single coordinate. 00:11:50.070 --> 00:11:51.520 It could be x. 00:11:51.520 --> 00:11:54.520 It could be theta. 00:11:54.520 --> 00:11:56.220 But you don't actually need them both. 00:11:56.220 --> 00:11:58.946 You use them both for a while, because it's convenient. 00:11:58.946 --> 00:12:00.320 But in the final analysis, you'll 00:12:00.320 --> 00:12:01.903 be able to write an equation of motion 00:12:01.903 --> 00:12:05.320 just in terms of x or just in terms of theta. 00:12:05.320 --> 00:12:20.150 OK, free body diagram of our wheel-- we said no slip. 00:12:20.150 --> 00:12:22.660 So here's your slope. 00:12:22.660 --> 00:12:24.140 You know you've got mg. 00:12:26.970 --> 00:12:30.140 You know there's going to be a normal force from the slope. 00:12:30.140 --> 00:12:32.890 There may also be some tangential force 00:12:32.890 --> 00:12:38.660 that makes it impossible for it not to slip, some f. 00:12:38.660 --> 00:12:42.150 So there's the free body diagram, in this case. 00:12:42.150 --> 00:12:51.960 What if-- I'll do e here, or do a case ii here, slip allowed. 00:12:57.130 --> 00:13:05.160 Then how many constraints do we have? 00:13:05.160 --> 00:13:05.920 Just four. 00:13:05.920 --> 00:13:08.392 Because now you can no longer say that this is true. 00:13:08.392 --> 00:13:09.850 They're independent of one another. 00:13:09.850 --> 00:13:13.800 They take on values not controlled by this formulation. 00:13:13.800 --> 00:13:16.255 So 6 minus 4 gives you 2. 00:13:16.255 --> 00:13:17.630 And you're going to end up having 00:13:17.630 --> 00:13:20.300 to have both x and theta probably as your chosen 00:13:20.300 --> 00:13:22.030 coordinates to do the problem. 00:13:22.030 --> 00:13:24.312 And you'll end up with two equations of motion. 00:13:41.700 --> 00:13:43.320 So I've got a hockey puck here. 00:13:46.140 --> 00:13:49.270 I've drawn kind of a 3D perspective of this. 00:13:49.270 --> 00:13:51.660 So it has a coordinate system out here. 00:13:51.660 --> 00:13:53.500 The z-axis is going like that. 00:13:53.500 --> 00:13:55.220 So here's my z. 00:13:55.220 --> 00:13:56.230 Here's my x. 00:13:56.230 --> 00:14:04.930 Here's my y in the plane of the ice 00:14:04.930 --> 00:14:07.896 that this thing is sliding on. 00:14:07.896 --> 00:14:11.490 So this is my z-coordinate. 00:14:11.490 --> 00:14:13.560 And I have string wrapped around it. 00:14:13.560 --> 00:14:17.224 And I've got a piece of string coming off like that. 00:14:17.224 --> 00:14:18.890 And I'm pulling on it with some tension. 00:14:21.950 --> 00:14:26.360 Because otherwise the only constraints 00:14:26.360 --> 00:14:29.540 are it's sitting on this icy surface, which 00:14:29.540 --> 00:14:31.490 I'm going to assume is-- well, I don't even 00:14:31.490 --> 00:14:33.130 have to assume it's frictionless. 00:14:33.130 --> 00:14:33.930 I could. 00:14:33.930 --> 00:14:36.170 So let's figure out how many equations of motion 00:14:36.170 --> 00:14:37.600 we're going to need here. 00:14:37.600 --> 00:14:42.230 So the number of degrees of freedom, 6 times-- this 00:14:42.230 --> 00:14:43.502 is a rigid body. 00:14:43.502 --> 00:14:44.335 It's not a particle. 00:14:44.335 --> 00:14:46.150 And there's only one of them. 00:14:46.150 --> 00:14:51.540 So it's 6 times 1 minus C and the number of constraints. 00:14:51.540 --> 00:14:55.400 So can it move into the table, into the surface? 00:14:55.400 --> 00:14:58.680 No, so that's a constraint in z. 00:14:58.680 --> 00:15:02.530 Are there any constraints in the x or y? 00:15:02.530 --> 00:15:05.360 It can rotate about z. 00:15:05.360 --> 00:15:08.680 But it can't rotate about the x or y-axes. 00:15:08.680 --> 00:15:11.340 All right, so constraint into z is one. 00:15:11.340 --> 00:15:14.385 Can't rotate about x or y-- two, three. 00:15:14.385 --> 00:15:15.580 Are there any others? 00:15:23.110 --> 00:15:25.800 Who thinks I may have missed one? 00:15:25.800 --> 00:15:29.920 OK, I'd say we've got 6 minus 3. 00:15:29.920 --> 00:15:32.665 We're going to need three equations of motion 00:15:32.665 --> 00:15:35.040 to be able to actually describe the motion of this thing. 00:15:37.560 --> 00:15:42.220 And we'd probably use-- this is a pretty good 00:15:42.220 --> 00:15:43.050 coordinate system. 00:15:43.050 --> 00:15:50.973 We'd probably use an x, a y, and some theta with respect 00:15:50.973 --> 00:15:51.473 to z-axis. 00:15:54.640 --> 00:16:10.400 All right, another quick example-- 00:16:10.400 --> 00:16:12.985 so we've got a rod leaning against a wall. 00:16:12.985 --> 00:16:20.192 It's length L. Actually, I don't want to do that. 00:16:23.030 --> 00:16:23.880 It's L long. 00:16:23.880 --> 00:16:28.690 It's got a center of mass here at the middle, uniform rod. 00:16:31.840 --> 00:16:36.230 Let this be x, y, z coming out of the board, 00:16:36.230 --> 00:16:40.940 maybe give it an angle here just to help us describe motions. 00:16:40.940 --> 00:16:43.410 You might start off with a preliminary little coordinate 00:16:43.410 --> 00:16:45.880 system just so you could think that, OK, no motion 00:16:45.880 --> 00:16:47.660 in the x, no rotation in this. 00:16:47.660 --> 00:16:49.160 But once you get the problem, you're 00:16:49.160 --> 00:16:50.900 ready to set up the equations of motion, 00:16:50.900 --> 00:16:53.930 you might decide, OK, I know I need two coordinates. 00:16:53.930 --> 00:16:56.860 And the ones I've preliminarily chosen aren't too good. 00:16:56.860 --> 00:16:59.060 Then you change, and you pick the really good ones. 00:16:59.060 --> 00:17:01.018 But I'm just making the preliminary assessments 00:17:01.018 --> 00:17:04.680 so I can assess the problem here. 00:17:04.680 --> 00:17:11.280 Again, the degrees of freedom-- 6 times 1 minus C. 00:17:11.280 --> 00:17:14.089 Because I only have one rigid body. 00:17:14.089 --> 00:17:19.975 Now, the constraints here are a little more subtle. 00:17:26.740 --> 00:17:30.530 Let me just discuss these two points. 00:17:30.530 --> 00:17:33.840 What can you say about the motion at point A? 00:17:33.840 --> 00:17:36.921 This is right here where it touches the wall. 00:17:36.921 --> 00:17:38.420 This thing is sliding down the wall. 00:17:38.420 --> 00:17:40.580 It might be frictionless, might not. 00:17:40.580 --> 00:17:42.820 Whether or not friction acts doesn't really 00:17:42.820 --> 00:17:45.000 change the number of degrees of freedom 00:17:45.000 --> 00:17:49.050 unless you invoke things like no slip. 00:17:49.050 --> 00:17:52.180 What can you say kinematically about motion, 00:17:52.180 --> 00:17:53.610 about the motion at A? 00:17:53.610 --> 00:17:55.660 Does the wall restrict the motion? 00:17:55.660 --> 00:17:56.890 AUDIENCE: Yes. 00:17:56.890 --> 00:17:58.987 PROFESSOR: In what direction? 00:17:58.987 --> 00:18:00.671 AUDIENCE: x. 00:18:00.671 --> 00:18:01.921 PROFESSOR: In which direction? 00:18:01.921 --> 00:18:02.410 AUDIENCE: x. 00:18:02.410 --> 00:18:03.785 PROFESSOR: In x direction, right. 00:18:03.785 --> 00:18:07.159 So is this body constrained in the x direction? 00:18:07.159 --> 00:18:09.380 AUDIENCE: [INAUDIBLE]. 00:18:09.380 --> 00:18:11.270 PROFESSOR: Pardon? 00:18:11.270 --> 00:18:12.260 I hear a no. 00:18:12.260 --> 00:18:14.890 I hear some yeses. 00:18:14.890 --> 00:18:18.466 What about at B? 00:18:18.466 --> 00:18:20.490 Is it constrained at B? 00:18:20.490 --> 00:18:21.281 In the-- 00:18:21.281 --> 00:18:21.780 AUDIENCE: y. 00:18:21.780 --> 00:18:24.720 PROFESSOR: y direction, OK. 00:18:24.720 --> 00:18:29.970 I didn't bring my big foam disk. 00:18:29.970 --> 00:18:33.960 But earlier in the term, I said that the definition 00:18:33.960 --> 00:18:40.820 of translation is that all points on a rigid body 00:18:40.820 --> 00:18:46.980 do what if you're rectilinear or curvilinear translation as 00:18:46.980 --> 00:18:48.713 opposed to rotation? 00:18:48.713 --> 00:18:50.040 AUDIENCE: Move in parallel. 00:18:50.040 --> 00:18:55.380 PROFESSOR: All points move in parallel, exactly right. 00:18:55.380 --> 00:18:59.560 That means that if we use real strict definitions 00:18:59.560 --> 00:19:03.520 of translation and rotation, that if I constrain 00:19:03.520 --> 00:19:07.680 the motion of any point on that object, 00:19:07.680 --> 00:19:12.260 that object is now not allowed to translate in that direction 00:19:12.260 --> 00:19:14.710 by the definition of translation. 00:19:14.710 --> 00:19:18.340 So this point constrains it in the x direction. 00:19:18.340 --> 00:19:22.170 This constrains it in the y direction. 00:19:22.170 --> 00:19:32.786 And we're going to-- so it's constrained in translation. 00:19:41.210 --> 00:19:44.410 And that implies 2. 00:19:44.410 --> 00:19:58.416 And b, we'll assume that z motion is 0. 00:19:58.416 --> 00:19:59.790 We'll just assume there's nothing 00:19:59.790 --> 00:20:02.290 going on out of the plane. 00:20:02.290 --> 00:20:04.040 So that gives me another one. 00:20:07.470 --> 00:20:13.440 And I'll assume no rotation. 00:20:13.440 --> 00:20:17.170 I'm not going to allow any rotation in this problem. 00:20:17.170 --> 00:20:34.350 I'm not interested in rotation about the x or about the y, 00:20:34.350 --> 00:20:35.750 about these axes. 00:20:35.750 --> 00:20:37.800 And that implies two more. 00:20:37.800 --> 00:20:41.705 So we have two, three, four, five. 00:20:43.842 --> 00:20:46.050 And I better not have any more than that or the thing 00:20:46.050 --> 00:20:47.970 can't move. 00:20:47.970 --> 00:20:51.930 So in this case, this is 6 minus 5 equals 1. 00:20:51.930 --> 00:20:54.250 I need a single coordinate to describe the motion. 00:20:54.250 --> 00:20:55.350 And if you look at it, you say, well, 00:20:55.350 --> 00:20:56.891 that's kind of intuitive and obvious. 00:20:56.891 --> 00:20:59.731 If I specify the x position here, 00:20:59.731 --> 00:21:00.980 I could figure everything out. 00:21:00.980 --> 00:21:03.605 If I know the length and the x, I could figure out where it is. 00:21:03.605 --> 00:21:05.615 If I know the y position and the length, 00:21:05.615 --> 00:21:07.120 I could figure out where it is. 00:21:07.120 --> 00:21:08.850 If I know theta, I could figure it out. 00:21:08.850 --> 00:21:10.467 I only need one. 00:21:10.467 --> 00:21:12.175 But that strict definition of translation 00:21:12.175 --> 00:21:13.280 is really helpful here. 00:21:13.280 --> 00:21:18.700 This thing, this object, is in pure rotation. 00:21:18.700 --> 00:21:23.260 And if it's in pure rotation, it must rotate about some point. 00:21:23.260 --> 00:21:24.172 Where? 00:21:24.172 --> 00:21:25.255 You know how to find that? 00:21:28.500 --> 00:21:31.440 What's the velocity here? 00:21:31.440 --> 00:21:34.460 It's got gravity acting on it, so it's probably down. 00:21:34.460 --> 00:21:35.860 But it's parallel to the wall. 00:21:35.860 --> 00:21:36.990 It has to be. 00:21:36.990 --> 00:21:38.840 What's the velocity here? 00:21:38.840 --> 00:21:42.460 It's got to be parallel to the wall. 00:21:42.460 --> 00:21:45.370 If I draw perpendicular to that-- if I'm saying, 00:21:45.370 --> 00:21:47.600 this is rotation. 00:21:47.600 --> 00:21:51.036 All points in the body rotate at the same rate. 00:21:51.036 --> 00:21:52.910 But their speed is determined by the distance 00:21:52.910 --> 00:21:56.110 away from the center of rotation. 00:21:56.110 --> 00:21:58.520 But if it's pure rotation, there must be 00:21:58.520 --> 00:22:00.270 a center of rotation somewhere. 00:22:00.270 --> 00:22:03.610 And it must be perpendicular to any velocity vector. 00:22:03.610 --> 00:22:04.950 So you draw the perpendicular. 00:22:04.950 --> 00:22:07.680 You draw the perpendicular. 00:22:07.680 --> 00:22:12.159 And here is the instantaneous center of rotation, the ICR. 00:22:12.159 --> 00:22:14.284 There's a little short section in the book on that. 00:22:17.250 --> 00:22:21.840 When this thing drops down to here, same kind of arguments 00:22:21.840 --> 00:22:24.320 hold. 00:22:24.320 --> 00:22:30.050 But the center of rotation has changed locations. 00:22:30.050 --> 00:22:31.021 Yeah. 00:22:31.021 --> 00:22:36.320 AUDIENCE: So is this not translating x and y? 00:22:36.320 --> 00:22:41.910 PROFESSOR: So now let's talk about the center of mass. 00:22:41.910 --> 00:22:44.900 So she asked-- excuse me, I should repeat the question. 00:22:44.900 --> 00:22:48.500 You guys aren't holding me to that very well. 00:22:48.500 --> 00:22:52.540 Raise your hands if I don't repeat an important question. 00:22:52.540 --> 00:22:53.920 She says, is it not translating? 00:22:56.720 --> 00:23:01.750 We've determined that-- let me ask you, 00:23:01.750 --> 00:23:04.170 does the center of mass move? 00:23:07.060 --> 00:23:10.984 Does Newton's second law apply to the motion 00:23:10.984 --> 00:23:11.900 of the center of mass? 00:23:15.246 --> 00:23:16.680 AUDIENCE: [INAUDIBLE]? 00:23:16.680 --> 00:23:18.114 PROFESSOR: Yeah, it's got to. 00:23:18.114 --> 00:23:20.050 So the center of mass translates-- 00:23:20.050 --> 00:23:21.240 no doubt about that. 00:23:21.240 --> 00:23:24.530 Newton's second law applies to it. 00:23:24.530 --> 00:23:27.430 So we're not saying that there isn't motion 00:23:27.430 --> 00:23:29.236 of the system in the x and y. 00:23:29.236 --> 00:23:31.610 We're just saying that that motion is caused by rotation. 00:23:34.800 --> 00:23:38.717 It's not caused by what is strictly 00:23:38.717 --> 00:23:39.675 defined as translation. 00:23:44.540 --> 00:23:55.872 Free body diagram, the reason-- let's 00:23:55.872 --> 00:23:57.580 think about a free body diagram for this. 00:23:57.580 --> 00:24:00.210 Here's our rod. 00:24:00.210 --> 00:24:03.930 There must be-- I'll let it be frictionless to keep 00:24:03.930 --> 00:24:05.310 the problem simple for a moment. 00:24:05.310 --> 00:24:07.610 That means there must be just a normal force 00:24:07.610 --> 00:24:09.020 in the x direction here. 00:24:09.020 --> 00:24:10.710 I'll call it Nx. 00:24:10.710 --> 00:24:15.360 And there must be a normal force in the y, call it Ny. 00:24:15.360 --> 00:24:18.035 The center of mass, there must be an Mg. 00:24:22.550 --> 00:24:25.300 So now that I set the problem up this way, 00:24:25.300 --> 00:24:27.530 how many unknowns are there in the problem? 00:24:27.530 --> 00:24:31.514 If I want to calculate literally the motion of this thing, 00:24:31.514 --> 00:24:33.180 find an equation of motion and solve it, 00:24:33.180 --> 00:24:34.500 how many unknowns do I have? 00:24:42.190 --> 00:24:43.443 How many do you think? 00:24:43.443 --> 00:24:44.044 AUDIENCE: One. 00:24:44.044 --> 00:24:44.960 PROFESSOR: I hear one. 00:24:44.960 --> 00:24:46.105 I see two on the board. 00:24:48.900 --> 00:24:51.409 But is two the right answer? 00:24:51.409 --> 00:24:51.950 I hear three. 00:24:51.950 --> 00:24:53.280 Who said three? 00:24:53.280 --> 00:24:55.010 All right, what's the third one? 00:24:55.010 --> 00:24:56.850 AUDIENCE: Acceleration? 00:24:56.850 --> 00:24:58.820 PROFESSOR: How do you describe acceleration 00:24:58.820 --> 00:25:04.518 with a coordinate of some kind? 00:25:04.518 --> 00:25:07.520 So there's yet another unknown. 00:25:07.520 --> 00:25:09.890 It's probably the thing that you're trying to solve for. 00:25:09.890 --> 00:25:14.000 It's actually the motion itself described by theta for x or y, 00:25:14.000 --> 00:25:15.190 whatever you do. 00:25:15.190 --> 00:25:18.940 There's at least three unknowns in this problem the way 00:25:18.940 --> 00:25:22.580 you see it in this free body diagram. 00:25:22.580 --> 00:25:25.100 So you've got to figure out ways around that. 00:25:25.100 --> 00:25:26.780 This instantaneous center of rotation 00:25:26.780 --> 00:25:30.060 gives you one possible way around that. 00:25:30.060 --> 00:25:33.780 Because about an instantaneous center, it's not moving. 00:25:33.780 --> 00:25:34.620 It's an axis. 00:25:34.620 --> 00:25:36.650 It's not moving. 00:25:36.650 --> 00:25:38.240 We have a little formula that says 00:25:38.240 --> 00:25:42.434 the time rate of change of angular momentum-- torque 00:25:42.434 --> 00:25:44.850 is related to the time rate of change of angular momentum. 00:25:44.850 --> 00:25:46.975 And you can have a messy formula or a not so messy. 00:25:46.975 --> 00:25:52.250 And it's not so messy when the axis of rotation is stationary. 00:25:52.250 --> 00:25:54.870 So at this instant in time, the axis is stationary. 00:25:54.870 --> 00:26:00.530 You can say that the torques about this point, 00:26:00.530 --> 00:26:04.410 the ICR, summation of the torques with respect 00:26:04.410 --> 00:26:13.150 to the ICR, is equal to d, is now a rigid body with respect 00:26:13.150 --> 00:26:19.820 to the ICR, dt. 00:26:19.820 --> 00:26:22.860 OK, in this problem, what are the torques? 00:26:22.860 --> 00:26:24.730 Where do the torques come from? 00:26:24.730 --> 00:26:28.629 Is there any torque caused by Nx? 00:26:28.629 --> 00:26:30.420 No, because there's no moment on it, right? 00:26:30.420 --> 00:26:31.900 It's pointing right at the center. 00:26:31.900 --> 00:26:34.320 Same thing-- no torque caused by this. 00:26:34.320 --> 00:26:36.640 You want to find equations that get rid of unknowns. 00:26:36.640 --> 00:26:38.787 So neither unknown appear in this equation. 00:26:38.787 --> 00:26:40.120 Where does the torque come from? 00:26:44.620 --> 00:26:45.770 Gravity, right? 00:26:45.770 --> 00:26:50.700 And it's going to be some Mg times a moment arm. 00:26:50.700 --> 00:26:53.740 And the moment arm is going to be like that. 00:26:53.740 --> 00:26:56.026 So it's an L over 2 sine theta. 00:27:00.140 --> 00:27:02.030 And the sine you'll have to figure out 00:27:02.030 --> 00:27:06.420 from an r cross and f. 00:27:06.420 --> 00:27:12.300 So you have an i cross j, gives you a k. 00:27:12.300 --> 00:27:13.635 But it's in the minus direction. 00:27:13.635 --> 00:27:17.470 So I think it'll come out minus. 00:27:17.470 --> 00:27:18.750 But I could be wrong. 00:27:18.750 --> 00:27:20.610 I did that on the fly. 00:27:23.870 --> 00:27:25.690 So we're not going to go further with this. 00:27:25.690 --> 00:27:30.440 But the instantaneous centers of rotation could be really handy. 00:27:30.440 --> 00:27:36.634 OK, a final example in this stuff-- 00:27:36.634 --> 00:27:37.675 how are we doing on time? 00:27:52.170 --> 00:27:56.670 Couple of carts, so the floor constrains the motion. 00:27:56.670 --> 00:28:03.305 And I've got a spring and a dashpot and an M1 and an M2. 00:28:06.230 --> 00:28:08.173 And I want to figure out how many-- yeah? 00:28:08.173 --> 00:28:09.960 AUDIENCE: I was just curious. 00:28:09.960 --> 00:28:14.109 Why is torque negative in your earlier solution? 00:28:14.109 --> 00:28:15.900 PROFESSOR: Let's just figure out r cross f. 00:28:15.900 --> 00:28:21.610 So my r-- you're saying, why is the torque negative? 00:28:21.610 --> 00:28:31.140 The r is L/2 in the i hat. 00:28:31.140 --> 00:28:37.790 And the gravity crossed with the force, which is Mg in the-- 00:28:37.790 --> 00:28:43.180 but it's minus Mg in the j hat. 00:28:43.180 --> 00:28:45.130 So i cross j is positive k. 00:28:45.130 --> 00:28:47.863 But the minus sign comes from there. 00:28:47.863 --> 00:28:51.560 AUDIENCE: What is that over to the left [INAUDIBLE]? 00:28:51.560 --> 00:28:54.190 PROFESSOR: Well, I'm working on the center of mass here. 00:28:54.190 --> 00:28:55.220 That's my equation. 00:28:55.220 --> 00:28:56.550 And this thing is L long. 00:28:56.550 --> 00:28:59.860 So half of the length must be L/2. 00:28:59.860 --> 00:29:05.517 And I'm interested in this side of the right triangle. 00:29:05.517 --> 00:29:07.892 AUDIENCE: Why are you interested in that side rather than 00:29:07.892 --> 00:29:09.530 the side that connects it to the-- 00:29:09.530 --> 00:29:13.670 PROFESSOR: Because this side crossed with that gives me 0. 00:29:13.670 --> 00:29:14.420 There's no moment. 00:29:14.420 --> 00:29:16.140 So this is a moment equation. 00:29:16.140 --> 00:29:18.930 I'm trying to compute moments. 00:29:18.930 --> 00:29:19.660 Yeah? 00:29:19.660 --> 00:29:22.366 AUDIENCE: Should the moment come from the ICR load? 00:29:22.366 --> 00:29:23.532 PROFESSOR: Oh, you're right. 00:29:23.532 --> 00:29:24.823 I don't know what I'm thinking. 00:29:24.823 --> 00:29:26.440 I could have messed this up. 00:29:26.440 --> 00:29:27.700 It's got to be about the ICR. 00:29:27.700 --> 00:29:29.910 So the force is down. 00:29:29.910 --> 00:29:33.050 Ooh, it's got to be this one. 00:29:33.050 --> 00:29:36.450 I messed up-- good catch. 00:29:36.450 --> 00:29:41.910 So that's a cosine, still L/2. 00:29:41.910 --> 00:29:42.890 You've got a theta. 00:29:42.890 --> 00:29:48.790 You have a-- this is also theta. 00:29:48.790 --> 00:29:50.920 And we're looking now for this side. 00:29:50.920 --> 00:29:55.130 Eh, it's still sine theta, right? 00:29:55.130 --> 00:29:57.609 Does the sign still work out the right way? 00:29:57.609 --> 00:29:58.525 AUDIENCE: [INAUDIBLE]. 00:30:02.890 --> 00:30:05.840 PROFESSOR: Good, OK, I've got to keep rolling. 00:30:05.840 --> 00:30:09.720 I've got something else really fun I want to talk about. 00:30:09.720 --> 00:30:11.426 So let's do this example quickly. 00:30:11.426 --> 00:30:13.925 This is mostly to get you to think about free body diagrams. 00:30:17.210 --> 00:30:21.080 This-- two rigid bodies. 00:30:21.080 --> 00:30:23.950 The degrees of freedom quickly here-- 6 times 2. 00:30:23.950 --> 00:30:28.909 There's no particles-- minus c, so 12 minus c. 00:30:28.909 --> 00:30:29.450 Now how many? 00:30:29.450 --> 00:30:30.866 What does your intuition tell you? 00:30:30.866 --> 00:30:32.830 How many independent coordinates is it 00:30:32.830 --> 00:30:36.330 going to take to solve this problem? 00:30:36.330 --> 00:30:38.340 Hold up your fingers. 00:30:38.340 --> 00:30:44.137 I see two, one, two, one. 00:30:44.137 --> 00:30:45.300 OK, one or two. 00:30:45.300 --> 00:30:51.240 Well, I think you can find 10 constraints in this problem. 00:30:51.240 --> 00:30:52.715 If you assume it doesn't roll over 00:30:52.715 --> 00:30:55.100 and you assume it doesn't move, you can find 10. 00:30:55.100 --> 00:30:57.880 You're going to need two coordinates. 00:30:57.880 --> 00:31:01.290 Because just because you've got a spring and a dashpot, 00:31:01.290 --> 00:31:02.950 they don't fix. 00:31:02.950 --> 00:31:05.050 They don't say there's any particular relation 00:31:05.050 --> 00:31:07.644 between the motion of this and the motion of that. 00:31:07.644 --> 00:31:09.560 You're going to need an independent coordinate 00:31:09.560 --> 00:31:16.220 to describe the-- whoops, 2 times 6. 00:31:16.220 --> 00:31:18.240 This is 12 minus 10. 00:31:18.240 --> 00:31:22.250 You don't need an independent coordinate 00:31:22.250 --> 00:31:24.730 to describe the motion of each of these masses. 00:31:24.730 --> 00:31:27.720 And I'd probably choose a coordinate that, let's say, 00:31:27.720 --> 00:31:29.560 goes from the center of mass of this one. 00:31:29.560 --> 00:31:34.430 I'll call it x1, center of mass of this one call it x2. 00:31:34.430 --> 00:31:36.930 And because I've been doing problems, vibration problems 00:31:36.930 --> 00:31:39.520 and stuff like this, for a long time, 00:31:39.520 --> 00:31:44.180 I'll tell you it's smart to start your coordinates at 0 00:31:44.180 --> 00:31:47.030 when you have zero spring forces. 00:31:47.030 --> 00:31:50.319 Or from the static equilibrium position-- 00:31:50.319 --> 00:31:51.610 that's the good place to start. 00:31:51.610 --> 00:31:55.100 So then your answer, if you're at the static equilibrium 00:31:55.100 --> 00:32:00.220 position, then any non-zero answers that come out of it 00:32:00.220 --> 00:32:02.990 are movement around that point. 00:32:02.990 --> 00:32:06.730 That's what mother nature does. 00:32:06.730 --> 00:32:09.480 A spring hanging on a mass hanging on a spring, 00:32:09.480 --> 00:32:11.350 it has a static equilibrium position. 00:32:11.350 --> 00:32:13.554 The vibration occurs around it. 00:32:13.554 --> 00:32:14.970 Your car is sitting on the ground. 00:32:14.970 --> 00:32:18.840 The vibration is around its static equilibrium position. 00:32:18.840 --> 00:32:27.770 So you usually choose these at the static equilibrium. 00:32:27.770 --> 00:32:40.540 And you'll find that that makes for the simplest 00:32:40.540 --> 00:32:42.080 equations of motion. 00:32:42.080 --> 00:32:46.200 They tend to not have constants on the right hand 00:32:46.200 --> 00:32:51.730 side that's caused by gravity or offsets or things like that. 00:32:51.730 --> 00:32:54.990 So this thing, actually a problem very similar to this 00:32:54.990 --> 00:32:56.680 is on your homework. 00:32:56.680 --> 00:32:58.996 The homework says, go find what these 10 are. 00:32:58.996 --> 00:33:00.370 So you can name them pretty fast. 00:33:00.370 --> 00:33:02.970 That's why I'm not doing it for you. 00:33:02.970 --> 00:33:08.150 What I do want to do quickly here 00:33:08.150 --> 00:33:21.000 is just talk about how you assign free body 00:33:21.000 --> 00:33:22.500 diagrams for this problem. 00:33:32.230 --> 00:33:34.620 Because people, one of the most easiest thing 00:33:34.620 --> 00:33:37.650 to get confused about is figuring out 00:33:37.650 --> 00:33:40.010 the directions of the forces that 00:33:40.010 --> 00:33:43.890 come from the springs in the dashpots, which direction 00:33:43.890 --> 00:33:46.720 acts on each. 00:33:46.720 --> 00:33:51.620 So I've made my x's so that x1 is 00:33:51.620 --> 00:33:55.090 0 when this thing is at its static equilibrium position. 00:33:55.090 --> 00:34:08.210 And I'm going to start by assuming that x1 and x1 dot are 00:34:08.210 --> 00:34:09.730 positive. 00:34:09.730 --> 00:34:13.040 You just establish positive motions. 00:34:13.040 --> 00:34:17.234 And you deduce the direction of the spring and dashpot forces. 00:34:30.679 --> 00:34:34.030 OK, and you do them one at a time. 00:34:34.030 --> 00:34:36.560 The problems we're doing here are linear. 00:34:36.560 --> 00:34:39.480 Spring force is equal to kx. 00:34:39.480 --> 00:34:42.066 Dashpot forces are Bx dot. 00:34:42.066 --> 00:34:43.190 So they're linear problems. 00:34:43.190 --> 00:34:45.870 So the superposition holds. 00:34:45.870 --> 00:34:48.380 You can just do these conceptually one at a time 00:34:48.380 --> 00:34:49.374 and figure them out. 00:34:49.374 --> 00:34:50.790 And then you add them all together 00:34:50.790 --> 00:34:51.914 to get the complete answer. 00:34:51.914 --> 00:34:59.240 So if x1 moves in the positive direction, 00:34:59.240 --> 00:35:03.380 what is the direction that that spring 00:35:03.380 --> 00:35:06.400 puts on this mass as a result? 00:35:06.400 --> 00:35:07.750 Which direction is it? 00:35:07.750 --> 00:35:11.420 And my coordinate system is positive. 00:35:11.420 --> 00:35:13.230 Here's x1. 00:35:13.230 --> 00:35:14.680 We'll do the same thing here. 00:35:14.680 --> 00:35:15.510 Here's x2. 00:35:15.510 --> 00:35:20.080 So if it moves in the positive direction out to here, 00:35:20.080 --> 00:35:22.520 what is the spring going to do? 00:35:22.520 --> 00:35:25.315 No other motion allowed in the system, just that one. 00:35:25.315 --> 00:35:27.080 It moves a little bit. 00:35:27.080 --> 00:35:28.960 What does the spring do on that mass? 00:35:28.960 --> 00:35:29.922 AUDIENCE: Push back. 00:35:29.922 --> 00:35:34.760 PROFESSOR: Push back kx. 00:35:34.760 --> 00:35:37.890 And I let the sign be indicated by the direction of the arrow. 00:35:37.890 --> 00:35:40.730 And I'll use that when I write out my equation of motion. 00:35:40.730 --> 00:35:44.360 OK, now I'm going to assume a positive bx1, so 00:35:44.360 --> 00:35:45.810 its velocity in that direction. 00:35:45.810 --> 00:35:47.408 What does the dashpot do? 00:35:47.408 --> 00:35:49.360 AUDIENCE: [INAUDIBLE]. 00:35:49.360 --> 00:35:51.040 PROFESSOR: Resist or [INAUDIBLE]? 00:35:51.040 --> 00:35:51.804 AUDIENCE: Resist. 00:35:51.804 --> 00:35:53.053 PROFESSOR: Pushes back, right? 00:35:53.053 --> 00:35:56.514 OK, so you have another force here, bx1 dot. 00:35:59.300 --> 00:36:02.270 Now, this is if you have two bodies. 00:36:02.270 --> 00:36:06.000 You have to have two free body diagrams, one for each. 00:36:06.000 --> 00:36:11.100 But I need to know what's the effect of x2 and x2 dot 00:36:11.100 --> 00:36:13.440 on this body. 00:36:13.440 --> 00:36:15.990 Well, let's just go do the same thing. 00:36:15.990 --> 00:36:17.730 Let's let x2 be positive. 00:36:17.730 --> 00:36:20.275 Now it's the only motion. 00:36:20.275 --> 00:36:26.690 I have a positive movement of x2 that stretches the spring. 00:36:26.690 --> 00:36:30.370 What is that spring force applied to this? 00:36:30.370 --> 00:36:31.530 What direction is it in? 00:36:36.620 --> 00:36:38.481 This body is moving in that way. 00:36:38.481 --> 00:36:39.773 AUDIENCE: Positive x direction. 00:36:39.773 --> 00:36:41.189 PROFESSOR: Which way is the spring 00:36:41.189 --> 00:36:42.387 going to pull on this thing? 00:36:42.387 --> 00:36:44.520 It's going to pull on it, right? 00:36:44.520 --> 00:36:50.950 So k, by amount kx2, and of positive velocity, 00:36:50.950 --> 00:36:53.850 to the right. 00:36:53.850 --> 00:36:56.040 It makes the dashpot open up. 00:36:56.040 --> 00:36:59.871 What direction is the force that dashpot puts on this? 00:36:59.871 --> 00:37:03.240 AUDIENCE: Positive x. 00:37:03.240 --> 00:37:06.160 PROFESSOR: bx2 dot. 00:37:06.160 --> 00:37:08.660 And now in this problem, except for gravity, 00:37:08.660 --> 00:37:12.040 I've got a normal force f and an Mg down. 00:37:12.040 --> 00:37:14.059 But all the action, all the motion, 00:37:14.059 --> 00:37:15.350 is in the horizontal direction. 00:37:15.350 --> 00:37:20.481 I can write out an equation of motion for the first rigid body 00:37:20.481 --> 00:37:20.980 here. 00:37:20.980 --> 00:37:26.020 And that's a sum of the forces in the x direction on body 00:37:26.020 --> 00:37:28.670 one-- I'll give this a 1 here-- in the x 00:37:28.670 --> 00:37:34.070 is equal to M1 x1 double dot. 00:37:34.070 --> 00:37:36.610 And now I can just write it out. 00:37:36.610 --> 00:37:45.730 It is k x2, because that one's positive, 00:37:45.730 --> 00:37:53.710 minus x1 plus b x2 dot minus x1 dot. 00:37:56.530 --> 00:38:00.200 And that has the signs right. 00:38:00.200 --> 00:38:04.250 And the whole key is just one at a time assume positive motions 00:38:04.250 --> 00:38:06.220 and deduce what happens, and then use 00:38:06.220 --> 00:38:10.330 the arrows, the direction of the arrows, to set the signs. 00:38:10.330 --> 00:38:13.520 And now there's your equation of motion. 00:38:13.520 --> 00:38:14.900 Now we can do the same thing, sum 00:38:14.900 --> 00:38:29.800 of the forces on 2 in the x 2 direction, M2 x2 double dot. 00:38:29.800 --> 00:38:32.440 And now we would do exactly the same thing. 00:38:32.440 --> 00:38:40.930 So positive motion of x1, what does it do over here? 00:38:43.899 --> 00:38:46.315 It gives me a force through the spring in which direction, 00:38:46.315 --> 00:38:47.476 positive or negative? 00:38:47.476 --> 00:38:48.316 AUDIENCE: Positive. 00:38:48.316 --> 00:38:49.690 PROFESSOR: Right, so now you just 00:38:49.690 --> 00:38:56.980 end up with a kx1, bx1 dot. 00:38:56.980 --> 00:39:03.710 And you'll find that kx2 bx2 dot. 00:39:03.710 --> 00:39:05.130 And you sum it up. 00:39:05.130 --> 00:39:12.080 And you'll end up-- this should switch around, k x1 minus x2 00:39:12.080 --> 00:39:16.320 plus b x1 dot minus x2 dot. 00:39:23.920 --> 00:39:25.720 I've got two equations of motion. 00:39:25.720 --> 00:39:26.530 And they're mixed. 00:39:26.530 --> 00:39:29.810 So each one has both coordinates in it. 00:39:29.810 --> 00:39:33.090 So this problem has two questions of motion, 00:39:33.090 --> 00:39:34.435 and they're coupled. 00:39:34.435 --> 00:39:35.435 They're not independent. 00:39:35.435 --> 00:39:37.870 You have to solve them together. 00:39:37.870 --> 00:39:41.400 OK, now we're going to move on to a subject which 00:39:41.400 --> 00:39:43.070 has come up in conversation. 00:39:43.070 --> 00:39:44.820 People have asked about this lots of time. 00:39:44.820 --> 00:39:50.470 And they say, what about the centrifugal force? 00:39:50.470 --> 00:39:53.040 And you sometimes use the term "fictitious force." 00:39:53.040 --> 00:39:59.150 How many of you use or heard the word used "fictitious force"? 00:39:59.150 --> 00:40:07.580 And how many of you heard us say that centripetal acceleration 00:40:07.580 --> 00:40:12.620 is not a force, it's an acceleration? 00:40:12.620 --> 00:40:14.645 And yet we love to talk about this concept 00:40:14.645 --> 00:40:19.840 of centrifugal force, which doesn't exist. 00:40:19.840 --> 00:40:21.020 But it trips us all up. 00:40:21.020 --> 00:40:22.680 Because it's handy to think about it. 00:40:22.680 --> 00:40:26.210 And so we're going to talk about fictitious forces now. 00:40:38.730 --> 00:40:40.160 They're handy. 00:40:40.160 --> 00:40:42.370 But they are dangerous. 00:40:42.370 --> 00:40:44.680 You really have to understand your fundamentals 00:40:44.680 --> 00:40:47.260 if you're going to use the concept of fictitious forces 00:40:47.260 --> 00:40:49.640 without getting yourself in trouble. 00:40:49.640 --> 00:40:52.220 OK, what is a fictitious force? 00:40:52.220 --> 00:40:57.800 Well, Newton's law, let's start there, Newton's second. 00:41:03.480 --> 00:41:11.030 Sum of the forces external on the body 00:41:11.030 --> 00:41:13.780 equals a mass times acceleration. 00:41:13.780 --> 00:41:15.070 It's a vector equation. 00:41:15.070 --> 00:41:19.360 So we can break it down into its components. 00:41:19.360 --> 00:41:57.935 So a fictitious force, you take the true acceleration times 00:41:57.935 --> 00:41:58.435 the mass. 00:41:58.435 --> 00:42:00.060 So you have a fictitious force. 00:42:00.060 --> 00:42:08.660 It's going to multiply the true acceleration times the mass 00:42:08.660 --> 00:42:25.540 and put it on the summation of forces side of the equation. 00:42:29.320 --> 00:42:30.320 That's really all it is. 00:42:30.320 --> 00:42:33.160 And you move it over to this side of the equation. 00:42:33.160 --> 00:42:35.620 And you think of it as a force. 00:42:35.620 --> 00:42:37.490 So what you've done is you've said 00:42:37.490 --> 00:42:44.220 that the summation of the forces minus Ma-- because to move it 00:42:44.220 --> 00:42:46.510 over here, you've got to subtract it from this side-- 00:42:46.510 --> 00:42:48.785 equals 0. 00:42:48.785 --> 00:42:50.160 And you're saying, I'm just going 00:42:50.160 --> 00:42:51.830 to think of this as a force. 00:42:51.830 --> 00:42:57.010 And it makes this whole equation be conveniently equal to 0. 00:42:57.010 --> 00:43:00.820 But now, that's kind of abstract. 00:43:00.820 --> 00:43:04.506 Let's see if we can figure out an example or two 00:43:04.506 --> 00:43:05.780 to illustrate this. 00:43:20.020 --> 00:43:22.030 I'm sure you've done this problem in physics. 00:43:22.030 --> 00:43:23.905 I'm going to pick a really elementary problem 00:43:23.905 --> 00:43:27.770 so you don't get hung up on the physics to start with. 00:43:27.770 --> 00:43:29.315 This is the elevator problem. 00:43:29.315 --> 00:43:31.220 You've got cables pulling it up. 00:43:37.730 --> 00:43:40.450 You've got some scales in here, and you're standing on it. 00:43:44.246 --> 00:43:45.370 Here's your center of mass. 00:43:45.370 --> 00:43:51.870 We'll call that A. And I need a coordinate system, 00:43:51.870 --> 00:43:54.130 my inertial system. 00:43:54.130 --> 00:43:57.690 Newton's law only applies in inertial systems. 00:43:57.690 --> 00:43:59.040 So here's my x. 00:43:59.040 --> 00:43:59.920 I called this z here. 00:43:59.920 --> 00:44:00.785 I'll call this y. 00:44:04.520 --> 00:44:06.710 So I need to write an equation of motion 00:44:06.710 --> 00:44:10.260 about this person riding up in the elevator. 00:44:12.976 --> 00:44:13.850 We take a look at it. 00:44:13.850 --> 00:44:15.016 How many degrees of freedom? 00:44:18.110 --> 00:44:20.824 Intuitively obvious-- it's probably how many? 00:44:20.824 --> 00:44:21.590 AUDIENCE: One. 00:44:21.590 --> 00:44:23.870 PROFESSOR: One-- all sorts of constraints. 00:44:23.870 --> 00:44:25.400 It's only going up. 00:44:25.400 --> 00:44:27.720 So we're going to say we need one degree of freedom. 00:44:31.640 --> 00:44:36.650 Free body diagrams-- well, this object 00:44:36.650 --> 00:44:40.430 has some force pushing up on it. 00:44:40.430 --> 00:44:42.040 This is the person. 00:44:42.040 --> 00:44:45.620 This is your mass. 00:44:45.620 --> 00:44:55.280 It has a force pulling down on it, mg. 00:44:55.280 --> 00:44:58.089 And that's it. 00:44:58.089 --> 00:44:59.630 Now this N is going to work out to be 00:44:59.630 --> 00:45:01.566 what the scales read, right? 00:45:01.566 --> 00:45:03.190 Because it's coming through the scales. 00:45:03.190 --> 00:45:04.917 So that's really what we're looking for. 00:45:11.130 --> 00:45:14.490 And the problem here is, define the weight on the scales. 00:45:14.490 --> 00:45:22.290 So we say that the summation of the forces in the y direction-- 00:45:22.290 --> 00:45:25.390 and in this problem, we would say 00:45:25.390 --> 00:45:28.960 it's the mass times the acceleration of point 00:45:28.960 --> 00:45:37.390 A with respect to O. And the sum of those 00:45:37.390 --> 00:45:45.530 forces-- you've got an N minus mg. 00:45:45.530 --> 00:45:47.125 And that's a pretty simple equation. 00:45:51.400 --> 00:45:53.630 Now I'm going to specify. 00:45:53.630 --> 00:46:02.000 It's given that acceleration of A with respect to O 00:46:02.000 --> 00:46:03.125 is 1/4 of g. 00:46:06.590 --> 00:46:08.790 So that's what the cables in the elevator are doing. 00:46:08.790 --> 00:46:10.510 It's making this thing move, and it's 00:46:10.510 --> 00:46:12.630 going up at 1/4 of g, the N acceleration. 00:46:12.630 --> 00:46:14.290 So it's getting faster and faster. 00:46:16.830 --> 00:46:30.400 OK, so if I solve this for N, I will get-- 00:46:30.400 --> 00:46:32.276 let me stop there for a second. 00:46:35.872 --> 00:46:37.330 It's normally what I would just do. 00:46:37.330 --> 00:46:39.730 But since we're talking about fictitious forces, 00:46:39.730 --> 00:46:42.830 I need to go through that step for a second. 00:46:42.830 --> 00:46:45.035 So now I say that, well, the summation 00:46:45.035 --> 00:46:54.300 of the forces in the y direction here minus M acceleration of A 00:46:54.300 --> 00:47:00.200 with respect to O, that total is equal to 0. 00:47:00.200 --> 00:47:13.090 And that then is N minus Mg minus M times g over 4. 00:47:18.740 --> 00:47:24.960 And now I have taken this upward acceleration. 00:47:24.960 --> 00:47:26.510 And I've treated it like a force. 00:47:26.510 --> 00:47:28.850 I've just moved it to this other side, 00:47:28.850 --> 00:47:30.890 set the whole thing equal to 0. 00:47:30.890 --> 00:47:32.655 This is the fictitious force. 00:47:39.224 --> 00:47:41.640 I'm going to say, OK, that's all the forces in the system, 00:47:41.640 --> 00:47:50.210 solve for N. And of course you get N is M times g plus g/4. 00:47:50.210 --> 00:47:54.290 And that's 5/4 mg. 00:47:54.290 --> 00:47:57.280 And so you read 25% heavier. 00:47:57.280 --> 00:47:58.530 It's a really trivial example. 00:47:58.530 --> 00:48:01.850 But the notion is that you think of this 00:48:01.850 --> 00:48:03.730 as a force that's been applied. 00:48:03.730 --> 00:48:05.950 It's the mass times acceleration with a minus 00:48:05.950 --> 00:48:06.920 sign in front of it. 00:48:06.920 --> 00:48:09.420 It'll always turn out like that. 00:48:09.420 --> 00:48:11.390 It's minus the mass times the acceleration. 00:48:11.390 --> 00:48:16.190 Now, the acceleration can come from Coriolis acceleration. 00:48:16.190 --> 00:48:19.740 It can come from centripetal acceleration. 00:48:19.740 --> 00:48:21.650 So if we do this in a rotating thing, 00:48:21.650 --> 00:48:25.850 this fictitious force might be the centrifugal force, 00:48:25.850 --> 00:48:29.590 which is minus M times the centripetal acceleration. 00:48:29.590 --> 00:48:31.748 And we'll do an example like that. 00:48:43.973 --> 00:48:48.190 All right, stop-- there we go. 00:48:48.190 --> 00:48:49.920 OK, trivial example-- let's see if we can 00:48:49.920 --> 00:48:50.870 find something a little harder. 00:48:50.870 --> 00:48:51.950 How are we doing on time? 00:48:51.950 --> 00:48:53.470 We're doing pretty good. 00:48:53.470 --> 00:49:01.270 So let's do an example where the notion is really 00:49:01.270 --> 00:49:02.800 quite powerful. 00:49:02.800 --> 00:49:05.990 Now, I showed you this last time. 00:49:05.990 --> 00:49:09.180 And we talked a lot about-- we did the time derivative is 00:49:09.180 --> 00:49:10.080 the angular momentum. 00:49:10.080 --> 00:49:13.190 And we computed the torques that this thing 00:49:13.190 --> 00:49:17.280 exerts around different axes with respect 00:49:17.280 --> 00:49:18.890 to the point of attachment to this. 00:49:18.890 --> 00:49:20.700 Well, this is an example which, if you're 00:49:20.700 --> 00:49:22.400 comfortable with fictitious forces, 00:49:22.400 --> 00:49:26.840 you can figure out those torques really rather quickly. 00:49:26.840 --> 00:49:30.520 So I'm going to-- these two problems are identical. 00:49:30.520 --> 00:49:33.880 This problem, or with a shaft running like that, 00:49:33.880 --> 00:49:35.980 are really exactly identical. 00:49:35.980 --> 00:49:39.180 But I'm going to pretend that my thing is made this way. 00:49:39.180 --> 00:49:41.440 Because it makes it easier to see where these torques 00:49:41.440 --> 00:49:43.840 are coming from. 00:49:43.840 --> 00:49:45.660 So here's my z. 00:49:45.660 --> 00:49:48.760 And I've got my rotation about the z-axis. 00:49:48.760 --> 00:49:52.942 Here's this mass, my coordinates. 00:49:52.942 --> 00:49:56.740 Here's my r hat. 00:49:56.740 --> 00:50:01.550 Here's my z k hat, is this distance here. 00:50:01.550 --> 00:50:10.050 And I'm going to set conditions in the problem, my rotation 00:50:10.050 --> 00:50:10.990 rate. 00:50:10.990 --> 00:50:17.470 It's also theta dot about the k z-axis. 00:50:17.470 --> 00:50:24.410 And omega dot, theta double dot-- also in the k. 00:50:24.410 --> 00:50:25.270 It's not restricted. 00:50:25.270 --> 00:50:28.220 So I'm allowing this thing to accelerate. 00:50:28.220 --> 00:50:32.810 But r dot equals r double dot is 0. 00:50:32.810 --> 00:50:34.430 So it's not changing a position. 00:50:34.430 --> 00:50:36.180 It's just fixed. 00:50:36.180 --> 00:50:39.880 And I need to figure out the forces on this thing 00:50:39.880 --> 00:50:43.150 and talk about how we might consider some of them 00:50:43.150 --> 00:50:45.870 as fictitious forces. 00:50:45.870 --> 00:50:48.070 So let's think about in the r direction-- 00:50:48.070 --> 00:50:54.660 summation of the forces here in this r hat direction. 00:50:54.660 --> 00:50:56.040 So it's one vector component. 00:50:56.040 --> 00:51:01.900 I don't have to carry along all of the other baggage. 00:51:01.900 --> 00:51:08.946 It's equal to-- and I'll call this B. 00:51:08.946 --> 00:51:10.900 This is going to be the point about which I 00:51:10.900 --> 00:51:11.950 care about things. 00:51:11.950 --> 00:51:19.380 There's a fixed coordinate system here, Oxyz. 00:51:19.380 --> 00:51:22.837 But then a rotating coordinate system-- we'll call this A. 00:51:22.837 --> 00:51:24.420 And it's going to have my coordinates. 00:51:24.420 --> 00:51:27.820 I'll use polar coordinates, r theta z. 00:51:27.820 --> 00:51:28.930 But this one rotates. 00:51:28.930 --> 00:51:36.040 But it also has its origin right there coincident with O. 00:51:36.040 --> 00:51:38.610 So the sum in the r direction, then, we 00:51:38.610 --> 00:51:45.310 have the mass times the acceleration of B 00:51:45.310 --> 00:51:50.670 with respect to O. And the acceleration of B with respect 00:51:50.670 --> 00:51:56.440 to O is the acceleration of-- we'll 00:51:56.440 --> 00:52:01.860 just write out the whole formula using cylindrical coordinates. 00:52:07.570 --> 00:52:14.650 r double dot minus r theta dot squared, this is in the r hat, 00:52:14.650 --> 00:52:24.170 plus r theta double dot plus 2r dot theta 00:52:24.170 --> 00:52:26.570 dot in the theta hat direction. 00:52:26.570 --> 00:52:31.100 That's my full expression for-- whoops, not quite full. 00:52:31.100 --> 00:52:34.430 Forgot a term, right? 00:52:34.430 --> 00:52:36.600 There we go. 00:52:36.600 --> 00:52:40.850 That's my full expression for acceleration 00:52:40.850 --> 00:52:45.970 using cylindrical coordinates, but with a moving, 00:52:45.970 --> 00:52:48.890 possibly translating and rotating reference frame. 00:52:48.890 --> 00:52:50.998 This piece here counts for what? 00:52:54.910 --> 00:52:57.355 AUDIENCE: [INAUDIBLE]. 00:52:57.355 --> 00:52:58.822 PROFESSOR: Acceleration of what? 00:52:58.822 --> 00:53:01.756 In general, why do we have that term? 00:53:01.756 --> 00:53:04.690 AUDIENCE: [INAUDIBLE]. 00:53:04.690 --> 00:53:07.310 PROFESSOR: It's the translational acceleration 00:53:07.310 --> 00:53:08.629 of the A frame. 00:53:08.629 --> 00:53:09.670 So we can account for it. 00:53:09.670 --> 00:53:11.770 In this case, what is that? 00:53:11.770 --> 00:53:14.460 OK, so this term is 0. 00:53:14.460 --> 00:53:18.660 But this formula you really need to know. 00:53:18.660 --> 00:53:21.190 Because it accounts for all of the accelerations 00:53:21.190 --> 00:53:23.000 in the problem. 00:53:23.000 --> 00:53:24.770 You know that, you just go in and start 00:53:24.770 --> 00:53:26.260 assigning, picking out things. 00:53:26.260 --> 00:53:27.144 What's this? 00:53:29.800 --> 00:53:30.720 0. 00:53:30.720 --> 00:53:32.440 Is this 0? 00:53:32.440 --> 00:53:34.960 No, how about this one? 00:53:34.960 --> 00:53:37.550 Not 0-- I'm allowing it to accelerate. 00:53:37.550 --> 00:53:39.860 This one-- definitely 0. 00:53:39.860 --> 00:53:41.090 And this one? 00:53:41.090 --> 00:53:41.590 0. 00:53:41.590 --> 00:53:43.990 So I don't have many terms left in this. 00:53:46.680 --> 00:53:59.721 So this then is the mass times minus r theta dot squared. 00:53:59.721 --> 00:54:12.450 And this is r hat plus r theta double dot 00:54:12.450 --> 00:54:15.420 in the theta hat direction. 00:54:15.420 --> 00:54:17.606 Now I said I was summing the forces in the r. 00:54:24.360 --> 00:54:26.370 I put all of them in at the moment. 00:54:29.430 --> 00:54:31.465 It may be smarter to do that. 00:54:31.465 --> 00:54:33.330 Let me reverse course here. 00:54:43.030 --> 00:54:47.710 This is the vector summation of all the forces for a moment. 00:54:47.710 --> 00:54:53.140 And then we'll take the r component next. 00:54:53.140 --> 00:54:55.670 So just the r component, the summation 00:54:55.670 --> 00:55:01.000 of the forces in the r, we look at this, 00:55:01.000 --> 00:55:07.830 and we say, oh, it's just that minus Mr theta 00:55:07.830 --> 00:55:11.390 dot squared r hat. 00:55:11.390 --> 00:55:13.160 So that's the r direction. 00:55:13.160 --> 00:55:14.805 But now we need a free body diagram. 00:55:18.670 --> 00:55:19.270 Yeah? 00:55:19.270 --> 00:55:23.570 AUDIENCE: Why did you take the 2r theta [INAUDIBLE]? 00:55:23.570 --> 00:55:29.660 PROFESSOR: Oh, because that's r dot 0. 00:55:33.360 --> 00:55:36.580 So in my r direction, this is my acceleration. 00:55:36.580 --> 00:55:38.970 That's the only acceleration in the r direction. 00:55:38.970 --> 00:55:41.790 It's what we know to be centripetal. 00:55:41.790 --> 00:55:43.570 And the minus tells us it's inward. 00:55:47.230 --> 00:55:49.560 Free body diagram-- need that. 00:55:52.810 --> 00:55:59.108 So looking-- a side view. 00:56:03.500 --> 00:56:04.430 Here's your bead. 00:56:07.580 --> 00:56:10.070 I'm going to draw it as an unknown here. 00:56:10.070 --> 00:56:13.800 There is an unknown force in the r direction 00:56:13.800 --> 00:56:18.480 that comes from this bar holding it. 00:56:18.480 --> 00:56:20.940 It's applying-- there's got to be a force that 00:56:20.940 --> 00:56:22.510 makes this go in a circle. 00:56:22.510 --> 00:56:24.070 And that bar is what provides it. 00:56:24.070 --> 00:56:25.403 It's the only thing [INAUDIBLE]. 00:56:25.403 --> 00:56:27.157 I'll just call this unknown and r. 00:56:27.157 --> 00:56:29.240 And I'm just drawing it in the positive direction. 00:56:29.240 --> 00:56:32.620 The way you can do this, if you're not sure the direction, 00:56:32.620 --> 00:56:33.500 draw it positive. 00:56:33.500 --> 00:56:36.600 And the sign that falls out tells you 00:56:36.600 --> 00:56:38.680 what the right answer is. 00:56:38.680 --> 00:56:42.740 There'll be some force in the z provided by the rod. 00:56:42.740 --> 00:56:43.760 There'll be Mg. 00:56:46.540 --> 00:56:50.260 And that should be it. 00:56:50.260 --> 00:56:57.550 If we did a top view, then there'd 00:56:57.550 --> 00:57:02.675 be an unknown in the theta direction. 00:57:07.810 --> 00:57:11.590 You'd also see the N in the r direction. 00:57:11.590 --> 00:57:12.480 And anything else? 00:57:12.480 --> 00:57:16.770 No, the Mg is down where you can't see it. 00:57:16.770 --> 00:57:18.680 So these are your two free body diagrams. 00:57:24.300 --> 00:57:32.680 So now, in this direction, this is the acceleration. 00:57:32.680 --> 00:57:34.935 And the external forces are just that. 00:57:37.790 --> 00:57:40.905 So the sum of the forces is Nr. 00:57:43.870 --> 00:57:48.180 Now I want to treat it, bring in this concept 00:57:48.180 --> 00:57:49.410 of a fictitious force. 00:57:54.780 --> 00:57:56.992 I'm going to move the acceleration 00:57:56.992 --> 00:57:57.950 term to the other side. 00:58:22.170 --> 00:58:24.640 Nr-- unknown positive. 00:58:24.640 --> 00:58:28.530 Now I'm going to move that acceleration 00:58:28.530 --> 00:58:32.560 term over here minus-- ah, but now it's 00:58:32.560 --> 00:58:33.680 minus the acceleration. 00:58:36.490 --> 00:58:38.800 So this term looks like that, has a minus sign. 00:58:38.800 --> 00:58:48.580 If you move it over here, that actually becomes plus, 00:58:48.580 --> 00:58:49.295 equals 0. 00:58:51.930 --> 00:58:55.100 Now this is your fictitious force. 00:59:03.719 --> 00:59:05.510 Sometimes people call them inertial forces. 00:59:08.340 --> 00:59:13.500 And it acts like it's pulling out on the object. 00:59:13.500 --> 00:59:20.490 So the force that the mass appears to apply to the rod 00:59:20.490 --> 00:59:24.170 is this centrifugal force pulling out. 00:59:24.170 --> 00:59:26.090 And of course now we can solve for Nr. 00:59:26.090 --> 00:59:28.900 And we find it's minus Mr theta dot squared, which 00:59:28.900 --> 00:59:29.870 we knew all along. 00:59:29.870 --> 00:59:33.750 It's the force applied to the mass by the arm 00:59:33.750 --> 00:59:34.640 as it spins around. 00:59:34.640 --> 00:59:37.650 It has to pull in on it to make it go in that circle. 00:59:40.170 --> 00:59:44.230 Now, we could do the same thing in the theta direction. 00:59:44.230 --> 00:59:48.020 The theta direction will have an Mr theta double dot. 00:59:48.020 --> 00:59:52.350 And when we look at the theta free body diagram, 00:59:52.350 --> 00:59:53.920 it's plus N theta. 00:59:53.920 --> 00:59:55.650 You could solve that, and you immediately 00:59:55.650 --> 01:00:02.152 come up with a solution for the force in the theta direction. 01:00:02.152 --> 01:00:03.735 I'm just going to write that one down. 01:00:11.460 --> 01:00:13.970 When you solve for this one, you get a minus sign. 01:00:13.970 --> 01:00:19.430 This one ends up plus, Mr theta double dot. 01:00:19.430 --> 01:00:22.660 I'm not going to go through the gyrations of getting to this. 01:00:22.660 --> 01:00:25.170 You can do that one. 01:00:25.170 --> 01:00:30.646 Now, I want to do one quick thing with this. 01:00:34.540 --> 01:00:40.400 Once you develop confidence in knowing 01:00:40.400 --> 01:00:45.752 when you can use a fictitious force and not get in trouble, 01:00:45.752 --> 01:00:47.377 this is the sort of thing you might do. 01:00:53.720 --> 01:00:55.220 Here's my system. 01:00:55.220 --> 01:00:56.960 It's rotating. 01:00:56.960 --> 01:01:02.370 And I want a quick estimate of, what's the torque? 01:01:02.370 --> 01:01:08.340 What's the bending moment about this point caused by the fact 01:01:08.340 --> 01:01:11.660 that it is the centripetal acceleration? 01:01:11.660 --> 01:01:14.180 Well, centripetal acceleration is 01:01:14.180 --> 01:01:17.400 equivalent to having this fictitious force outward 01:01:17.400 --> 01:01:25.820 on this of an amount Mr theta dot squared. 01:01:25.820 --> 01:01:31.790 And this is the moment arm z. 01:01:31.790 --> 01:01:35.330 what's the torque that that causes about this point? 01:01:42.680 --> 01:01:45.046 This is just levers now, forces and lever arms. 01:01:45.046 --> 01:01:46.002 AUDIENCE: 0. 01:01:46.002 --> 01:01:48.680 PROFESSOR: No, not 0. 01:01:48.680 --> 01:01:50.910 About this point here-- this is O. 01:01:50.910 --> 01:01:53.260 And I've got that centrifugal force pulling out 01:01:53.260 --> 01:01:54.680 on this fictitious force. 01:01:54.680 --> 01:01:55.632 Yeah? 01:01:55.632 --> 01:01:57.445 AUDIENCE: zMR of theta dot squared? 01:01:57.445 --> 01:02:01.993 PROFESSOR: Yeah, in the-- this direction, 01:02:01.993 --> 01:02:03.870 which is theta, right? 01:02:03.870 --> 01:02:08.710 So that's the moment the torque about this point caused 01:02:08.710 --> 01:02:16.860 by that-- torque at O is minus Mr theta dot squared 01:02:16.860 --> 01:02:20.430 times z, the moment arm. 01:02:20.430 --> 01:02:23.192 And it's not minus. 01:02:23.192 --> 01:02:24.150 It's in that direction. 01:02:24.150 --> 01:02:25.676 So it's plus. 01:02:25.676 --> 01:02:29.190 It's in the theta hat direction, the torque, 01:02:29.190 --> 01:02:31.940 that that force is applying about this point. 01:02:35.580 --> 01:02:37.830 We could have solved this problem the way 01:02:37.830 --> 01:02:41.300 we did in the last lecture, very carefully 01:02:41.300 --> 01:02:45.660 going through the dh dt's and following it all out. 01:02:45.660 --> 01:02:49.140 And we would have gotten that answer for the torque 01:02:49.140 --> 01:02:50.827 except for a minus sign. 01:02:54.490 --> 01:02:57.640 Now, why the difference? 01:02:57.640 --> 01:03:00.900 This is the torque that the centripetal force 01:03:00.900 --> 01:03:03.170 causes down here. 01:03:03.170 --> 01:03:08.050 When you do dh dt, you get the torque 01:03:08.050 --> 01:03:12.516 required to make what's happening happen. 01:03:15.780 --> 01:03:18.730 It's just the opposite. 01:03:18.730 --> 01:03:19.640 This is the torque. 01:03:19.640 --> 01:03:21.130 This is putting about that. 01:03:21.130 --> 01:03:23.730 There must be an equal and opposite torque 01:03:23.730 --> 01:03:27.820 that this system puts on this arm out here 01:03:27.820 --> 01:03:30.110 to keep it from flopping out. 01:03:30.110 --> 01:03:32.030 So this is applying a torque here. 01:03:32.030 --> 01:03:33.960 It must resist with a torque. 01:03:33.960 --> 01:03:43.200 And so when you do dh dt, you're going 01:03:43.200 --> 01:03:50.840 to end up with a minus Mr theta dot squared theta hat. 01:03:53.600 --> 01:03:58.210 So you've got to be careful what you mean. 01:03:58.210 --> 01:04:00.720 But from an engineering point of view, 01:04:00.720 --> 01:04:02.930 if I were just trying to calculate the bending moment 01:04:02.930 --> 01:04:06.030 down here and deciding whether or not this shaft sticking out 01:04:06.030 --> 01:04:07.950 here is going to break off or not, 01:04:07.950 --> 01:04:11.480 I could make a very quick estimate of the bending moment 01:04:11.480 --> 01:04:14.785 by knowing this centrifugal force times the moment arm. 01:04:14.785 --> 01:04:15.284 Yeah? 01:04:15.284 --> 01:04:21.190 AUDIENCE: Is that supposed to be a z in the dh dt expression? 01:04:21.190 --> 01:04:24.190 Does there have to be a z in that expression? 01:04:24.190 --> 01:04:25.508 PROFESSOR: A v, a velocity? 01:04:25.508 --> 01:04:26.464 AUDIENCE: A z. 01:04:26.464 --> 01:04:27.898 PROFESSOR: A z, yeah, right. 01:04:27.898 --> 01:04:32.708 I'm just getting a little speedy in writing the equation. 01:04:32.708 --> 01:04:34.640 Good catch. 01:04:34.640 --> 01:04:41.110 All right, so now, we also have a theta direction thing here, 01:04:41.110 --> 01:04:42.840 right? 01:04:42.840 --> 01:04:46.820 We can think of a fictitious force in the theta direction. 01:04:46.820 --> 01:04:49.650 And that was the Mr theta double dot term. 01:04:49.650 --> 01:04:55.980 Does it generate torques that you could quickly compute? 01:04:55.980 --> 01:04:58.570 So as this thing is trying to accelerate 01:04:58.570 --> 01:05:02.380 in the positive acceleration, omega dot, theta double dot, 01:05:02.380 --> 01:05:04.730 it's trying to accelerate into the board. 01:05:04.730 --> 01:05:07.490 The bar is having to push that thing into the board. 01:05:07.490 --> 01:05:10.580 There's a force in the positive theta hat direction. 01:05:10.580 --> 01:05:15.630 But the fictitious inertial force 01:05:15.630 --> 01:05:22.190 is the mass pushing back on the rod, pushing this way, 01:05:22.190 --> 01:05:25.740 Mr theta double dot pushing this way. 01:05:25.740 --> 01:05:28.170 What moments does that cause about this point? 01:05:33.742 --> 01:05:34.658 AUDIENCE: [INAUDIBLE]. 01:05:40.125 --> 01:05:42.380 PROFESSOR: So there's actually two moment arms. 01:05:42.380 --> 01:05:46.730 If you have a force this way now, the force in, 01:05:46.730 --> 01:05:48.550 I'll call it, the theta direction 01:05:48.550 --> 01:05:53.240 is Mr theta double dot. 01:05:53.240 --> 01:05:56.436 Now, this is my fictitious force. 01:05:56.436 --> 01:05:58.010 This is this fictitious force. 01:05:58.010 --> 01:06:00.900 It's in the minus theta hat direction. 01:06:00.900 --> 01:06:04.330 It's pushing back. 01:06:04.330 --> 01:06:08.090 It's in the minus theta hat direction, Mr theta double dot. 01:06:08.090 --> 01:06:15.920 And a torque is some R cross with the force. 01:06:15.920 --> 01:06:23.922 And this R is that, RBA. 01:06:23.922 --> 01:06:31.930 And it's composed of r R hat and z k hat here. 01:06:31.930 --> 01:06:37.300 So we have to do a little r r hat 01:06:37.300 --> 01:06:51.950 plus z k hat crossed with minus Mr theta double dot theta hat. 01:06:51.950 --> 01:06:55.460 So you get an r hat cross theta gives you a k. 01:06:55.460 --> 01:06:58.085 A k cross theta hat gives you an r. 01:06:58.085 --> 01:07:01.650 You're going to get two terms out of this. 01:07:01.650 --> 01:07:07.130 One is going to look like Mr squared theta double dot. 01:07:07.130 --> 01:07:09.400 And that's a torque. 01:07:09.400 --> 01:07:13.150 And that is the one that it takes to-- that's 01:07:13.150 --> 01:07:16.680 the torque about this axis. 01:07:16.680 --> 01:07:19.780 There's a force times this moment arm. 01:07:19.780 --> 01:07:21.260 That's a torque about this axis. 01:07:21.260 --> 01:07:23.730 That's the actual torque it takes to speed this thing up. 01:07:26.640 --> 01:07:30.860 And then there's a torque about this axis, which is 01:07:30.860 --> 01:07:34.320 this length times this force. 01:07:34.320 --> 01:07:35.830 And that's the other term. 01:07:35.830 --> 01:07:40.800 That'll give you the term in the R direction. 01:07:40.800 --> 01:07:41.750 It'll be a twist. 01:07:41.750 --> 01:07:44.380 It'll be trying to twist this bar like that. 01:07:44.380 --> 01:07:46.570 Because there's a force in this direction. 01:07:46.570 --> 01:07:47.780 Yeah. 01:07:47.780 --> 01:07:49.700 AUDIENCE: It seems like these forces 01:07:49.700 --> 01:07:52.580 are coming from the systems. 01:07:52.580 --> 01:07:54.980 They aren't even fictitious. 01:07:54.980 --> 01:08:02.080 PROFESSOR: So she's asking, it's like the forces are real. 01:08:02.080 --> 01:08:06.510 But the force-- I mean, this is why 01:08:06.510 --> 01:08:11.030 fictitious forces, this concept of them, is so dangerous. 01:08:11.030 --> 01:08:14.870 It's because it's really tempting 01:08:14.870 --> 01:08:17.620 to start thinking of them like real forces. 01:08:17.620 --> 01:08:22.500 So any time you get stuck, you go back to the basics. 01:08:22.500 --> 01:08:26.830 And you say, Newton's law, F equals ma, 01:08:26.830 --> 01:08:30.605 you see torques is dh dt plus that v cross p term 01:08:30.605 --> 01:08:31.630 if you need it. 01:08:31.630 --> 01:08:37.100 And you work it out carefully using all the vector math. 01:08:37.100 --> 01:08:40.529 And then you are dealing-- and then 01:08:40.529 --> 01:08:47.779 Coriolis, centripetal, Euler terms are only accelerations. 01:08:47.779 --> 01:08:49.779 You treat them as pure accelerations. 01:08:49.779 --> 01:08:50.850 That's all they are. 01:08:53.580 --> 01:09:01.790 Now, to cause accelerations, you need to apply forces. 01:09:01.790 --> 01:09:05.160 And we give these forces names. 01:09:05.160 --> 01:09:07.350 Because it's helpful conceptually 01:09:07.350 --> 01:09:11.080 to think about these things as forces sometimes. 01:09:11.080 --> 01:09:14.445 But they are not real forces. 01:09:17.490 --> 01:09:19.795 But it's quick and easy. 01:09:19.795 --> 01:09:21.819 If you get too comfortable with them, 01:09:21.819 --> 01:09:24.760 they are great assets to your intuition. 01:09:24.760 --> 01:09:26.870 So I know immediately just looking 01:09:26.870 --> 01:09:30.420 at this thing, as soon as I see a machine that 01:09:30.420 --> 01:09:35.470 has a part that rotates like this, I say, rotational motion. 01:09:35.470 --> 01:09:41.920 That's central, circular motion, must be the equivalent 01:09:41.920 --> 01:09:45.290 of a centrifugal force. 01:09:45.290 --> 01:09:48.000 This mass is going to pull in that direction. 01:09:48.000 --> 01:09:49.770 Because it's going around and around. 01:09:49.770 --> 01:09:51.960 And it's going to cause a moment about this thing. 01:09:51.960 --> 01:09:54.326 I know for sure that's got to be there. 01:09:54.326 --> 01:09:56.080 And if I'm going to do it rigorously, 01:09:56.080 --> 01:09:57.960 I call it centripetal acceleration. 01:09:57.960 --> 01:09:59.040 And I compute. 01:09:59.040 --> 01:10:03.390 It's really the force that this bar puts on that mass 01:10:03.390 --> 01:10:07.010 to make it go in a circle. 01:10:07.010 --> 01:10:09.200 But it's handy to think of it as a force 01:10:09.200 --> 01:10:12.760 if what I'm interested in is the force. 01:10:12.760 --> 01:10:17.210 The other even better-- let's go back to the old really simple 01:10:17.210 --> 01:10:18.670 demonstration. 01:10:18.670 --> 01:10:21.920 This thing going around constant speed, 01:10:21.920 --> 01:10:23.930 there's definitely a tension in the string. 01:10:23.930 --> 01:10:25.100 And I am pulling. 01:10:25.100 --> 01:10:27.690 The tension is inward, right? 01:10:27.690 --> 01:10:31.230 That tension causes an acceleration 01:10:31.230 --> 01:10:33.780 of r theta dot squared. 01:10:33.780 --> 01:10:37.440 And that acceleration is inward. 01:10:37.440 --> 01:10:39.220 But it's easy. 01:10:39.220 --> 01:10:42.860 If you ask me, come on, Kim, quick, 01:10:42.860 --> 01:10:45.760 tell me, what's the tension in the string, 01:10:45.760 --> 01:10:49.000 I just say, that's easy-- Mr theta dot squared, 01:10:49.000 --> 01:10:50.050 centrifugal force. 01:10:59.230 --> 01:11:00.010 They're handy. 01:11:00.010 --> 01:11:04.550 But any time you get stuck, go back and be really strict 01:11:04.550 --> 01:11:06.810 and say, what's acceleration? 01:11:06.810 --> 01:11:09.650 And what's real external forces? 01:11:09.650 --> 01:11:13.330 Gravity is an external force. 01:11:13.330 --> 01:11:16.580 The force that the bar puts on the mass, that's a real force. 01:11:16.580 --> 01:11:20.040 But these other things are just accelerations. 01:11:20.040 --> 01:11:23.030 All right, OK, hey, perfect. 01:11:23.030 --> 01:11:25.400 I didn't get to the thing I really wanted to do. 01:11:25.400 --> 01:11:26.400 So I'll do it next time. 01:11:26.400 --> 01:11:29.027 And I'm going to show you this thing. 01:11:29.027 --> 01:11:30.860 And I want you to fill out your muddy cards. 01:11:30.860 --> 01:11:32.960 I'm giving you a couple minutes. 01:11:32.960 --> 01:11:35.755 So this is a mechanical shaker. 01:11:35.755 --> 01:11:37.300 You see it moving already. 01:11:37.300 --> 01:11:42.720 And we're going to talk about that next time.