WEBVTT 00:00:00.100 --> 00:00:02.450 The following content is provided under a Creative 00:00:02.450 --> 00:00:03.830 Commons license. 00:00:03.830 --> 00:00:06.070 Your support will help MIT OpenCourseWare 00:00:06.070 --> 00:00:10.170 continue to offer high quality educational resources for free. 00:00:10.170 --> 00:00:12.710 To make a donation or to view additional materials 00:00:12.710 --> 00:00:16.620 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.620 --> 00:00:17.270 at ocw.mit.edu. 00:00:22.010 --> 00:00:26.250 PROFESSOR: OK, let's talk about the important thoughts 00:00:26.250 --> 00:00:27.625 for the week, important concepts. 00:00:30.442 --> 00:00:31.150 What do you have? 00:00:33.928 --> 00:00:35.317 AUDIENCE: Energy method. 00:00:35.317 --> 00:00:36.706 PROFESSOR: Energy method. 00:00:36.706 --> 00:00:40.745 I'm going to say this work energy theorem. 00:00:54.820 --> 00:01:01.053 And from that, these turn out to be our potential energies. 00:01:16.481 --> 00:01:20.610 And this minus the work done by the conservative forces 00:01:20.610 --> 00:01:23.890 we call V2 minus V1. 00:01:23.890 --> 00:01:26.870 That's the change in the potential energy of the system. 00:01:26.870 --> 00:01:28.380 So that's the work energy theorem. 00:01:28.380 --> 00:01:30.250 That's an important one for the week. 00:01:30.250 --> 00:01:30.750 What else? 00:01:41.420 --> 00:01:43.370 AUDIENCE: Different types of problems. 00:01:43.370 --> 00:01:45.870 PROFESSOR: Yeah, we did several example problem 00:01:45.870 --> 00:01:47.120 particularly emphasizing what? 00:01:52.669 --> 00:01:54.335 AUDIENCE: With respect to moving points? 00:01:54.335 --> 00:01:57.202 PROFESSOR: Yeah, or points that look like they're moving. 00:01:57.202 --> 00:02:00.020 But you can use special considerations 00:02:00.020 --> 00:02:03.360 in calling them instantaneous centers of rotation and things 00:02:03.360 --> 00:02:07.560 like that, so summation of the torques with respect to points 00:02:07.560 --> 00:02:18.030 sometimes moving defined with respect to ICRs, Instantaneous 00:02:18.030 --> 00:02:24.250 Centers of Rotation, and also moving points. 00:02:27.365 --> 00:02:29.410 We definitely did some of that. 00:02:29.410 --> 00:02:30.320 How about a third? 00:02:45.950 --> 00:02:48.820 I'd say one more thing that we got started on 00:02:48.820 --> 00:03:07.710 is expressions-- 1/2 i, 1/2 omega dot h plus 1/2 mv, v dot 00:03:07.710 --> 00:03:09.730 v, those kind of terms for this. 00:03:09.730 --> 00:03:14.229 And for potential energy, if it's purely mechanical devices, 00:03:14.229 --> 00:03:15.770 we have two kinds of potential energy 00:03:15.770 --> 00:03:17.020 that we'll be dealing with. 00:03:17.020 --> 00:03:24.000 And they are associated with springs and gravity. 00:03:24.000 --> 00:03:25.970 So that's the two we'll deal with. 00:03:25.970 --> 00:03:27.836 Are there others? 00:03:27.836 --> 00:03:30.210 What's an example of a different kind of potential energy 00:03:30.210 --> 00:03:31.643 that you might run into? 00:03:31.643 --> 00:03:32.550 AUDIENCE: Magnetic. 00:03:32.550 --> 00:03:33.985 PROFESSOR: Magnetic fields, yeah. 00:03:33.985 --> 00:03:36.850 We can do work pulling something out of a magnetic field. 00:03:36.850 --> 00:03:37.420 What else? 00:03:37.420 --> 00:03:38.100 AUDIENCE: Electric fields. 00:03:38.100 --> 00:03:39.570 PROFESSOR: Electric fields, sure. 00:03:39.570 --> 00:03:41.230 But they do have their complications. 00:03:41.230 --> 00:03:45.370 So that's pretty good for the week. 00:03:47.890 --> 00:03:52.990 Now, we're going to talk about this problem. 00:03:52.990 --> 00:03:54.770 I'll give him a chance to shoot it. 00:03:54.770 --> 00:03:56.210 This is a good diagram of it. 00:03:56.210 --> 00:04:00.570 You guys had a homework problem that instead of a uniform rod, 00:04:00.570 --> 00:04:06.820 you had a concentrated mass on the end of a massless rod. 00:04:06.820 --> 00:04:09.690 So this is almost identical to that. 00:04:09.690 --> 00:04:15.530 But we now have actual mass, distributed mass, in the stick. 00:04:15.530 --> 00:04:20.579 I've got a little demo here, which will work, 00:04:20.579 --> 00:04:21.760 works after a fashion. 00:04:25.440 --> 00:04:27.030 Put a little tension on these springs, 00:04:27.030 --> 00:04:29.740 so now I've got my cart on a spring. 00:04:29.740 --> 00:04:31.390 The damping is natural in the system, 00:04:31.390 --> 00:04:34.154 comes from the wheels and everything else. 00:04:34.154 --> 00:04:36.070 I've got to get this kind of close to the edge 00:04:36.070 --> 00:04:36.880 so it'll work. 00:04:40.600 --> 00:04:44.020 This thing has two natural frequencies actually. 00:04:44.020 --> 00:04:46.790 They're kind of hard to isolate them. 00:04:46.790 --> 00:04:51.300 There's one-- no, can't do it. 00:04:51.300 --> 00:04:54.590 In one natural mode, the thing swings forward 00:04:54.590 --> 00:04:56.540 and moves forward together. 00:04:56.540 --> 00:04:59.135 And in the other natural mode, this'll go aft, that way, 00:04:59.135 --> 00:05:00.760 while the other one's going frontwards. 00:05:00.760 --> 00:05:03.220 So let's see if I can-- that's higher frequency. 00:05:03.220 --> 00:05:07.480 That's the motion of what's called the second mode, 00:05:07.480 --> 00:05:08.300 opposite. 00:05:08.300 --> 00:05:11.720 And the first mode is lower frequency than that. 00:05:25.490 --> 00:05:28.010 It's hard to get a perfect start to it. 00:05:28.010 --> 00:05:31.010 So this is the kind of motion we're talking about. 00:05:31.010 --> 00:05:33.530 And we're going to have you help me solve. 00:05:33.530 --> 00:05:37.910 We want to obtain equations of motion for this system. 00:05:37.910 --> 00:05:41.890 And let's quickly run through the exercise of-- you 00:05:41.890 --> 00:05:43.814 can do this intuitively first, it's so simple. 00:05:43.814 --> 00:05:44.980 Plus you've already done it. 00:05:44.980 --> 00:05:48.240 How many independent coordinates does it take to do it? 00:05:48.240 --> 00:05:57.150 Two, and how many rigid bodies are there in this problem? 00:05:57.150 --> 00:05:59.140 How many rigid bodies in the problem? 00:06:02.932 --> 00:06:03.806 AUDIENCE: Two. 00:06:03.806 --> 00:06:05.430 PROFESSOR: There's gotta be two, right? 00:06:05.430 --> 00:06:07.250 There's a block that moves back and forth, 00:06:07.250 --> 00:06:08.970 and there's this rod that swings. 00:06:08.970 --> 00:06:11.380 So the number of degrees of freedom, we'll call it d, 00:06:11.380 --> 00:06:15.260 is 6 times the number of bodies. 00:06:15.260 --> 00:06:18.270 n is the number bodies minus the number of constraints. 00:06:18.270 --> 00:06:21.330 So this is 6 times 2 minus the constraints. 00:06:21.330 --> 00:06:23.870 So we've got 12 minus c. 00:06:23.870 --> 00:06:25.860 And you know the answers. 00:06:25.860 --> 00:06:29.150 The answer, how many independent coordinates to completely 00:06:29.150 --> 00:06:30.237 describe this motion? 00:06:30.237 --> 00:06:30.820 AUDIENCE: Two. 00:06:30.820 --> 00:06:31.819 PROFESSOR: Just the two. 00:06:31.819 --> 00:06:34.135 So you know the answer is two. 00:06:34.135 --> 00:06:36.380 So that means you've got to come up with 10, 00:06:36.380 --> 00:06:37.940 a list of 10 of these things. 00:06:37.940 --> 00:06:42.250 So let's see if you can write down quickly your own list. 00:06:42.250 --> 00:06:47.370 How would you eliminate through coming up with constraints? 00:06:47.370 --> 00:06:49.420 Write down the 10 constraints. 00:06:49.420 --> 00:06:52.720 And I recommend you do it by taking one body at a time. 00:06:52.720 --> 00:06:56.824 So constraints on mass one, the first block, what do you got? 00:06:56.824 --> 00:06:59.610 AUDIENCE: y, y dot, and y double dot equal 0. 00:06:59.610 --> 00:07:02.200 PROFESSOR: OK, so that says no motion in the y direction. 00:07:02.200 --> 00:07:03.815 Because we're on rollers. 00:07:03.815 --> 00:07:06.670 It's fixed that way, so no motion in the y. 00:07:06.670 --> 00:07:07.629 What else? 00:07:07.629 --> 00:07:08.670 AUDIENCE: No motion in z. 00:07:08.670 --> 00:07:09.030 PROFESSOR: No motion in z. 00:07:09.030 --> 00:07:11.430 We're telling it it can't come in and out of the board. 00:07:11.430 --> 00:07:15.190 OK, any other translational constraints on that one? 00:07:15.190 --> 00:07:21.685 So no y or z translation. 00:07:24.260 --> 00:07:26.470 And so that gives us two constraints there. 00:07:26.470 --> 00:07:28.030 What else on that body? 00:07:28.030 --> 00:07:29.620 AUDIENCE: No rotations. 00:07:29.620 --> 00:07:31.370 PROFESSOR: No rotations in what direction? 00:07:31.370 --> 00:07:32.210 AUDIENCE: x, y, or z. 00:07:32.210 --> 00:07:33.251 PROFESSOR: In x, y, or z. 00:07:33.251 --> 00:07:39.980 The thing can only translate, so no x, y, z rotations. 00:07:42.920 --> 00:07:43.540 That's three. 00:07:43.540 --> 00:07:45.510 So that's a total of five. 00:07:45.510 --> 00:07:46.620 But we have one left. 00:07:46.620 --> 00:07:48.600 It's got to be able to move in the x. 00:07:48.600 --> 00:07:51.710 And how about M2? 00:07:51.710 --> 00:07:55.786 What are the constraints on the second one? 00:07:55.786 --> 00:07:58.030 AUDIENCE: [INAUDIBLE] 00:07:58.030 --> 00:07:59.400 PROFESSOR: So one at a time. 00:07:59.400 --> 00:08:00.275 AUDIENCE: [INAUDIBLE] 00:08:02.950 --> 00:08:04.840 PROFESSOR: So omega which? 00:08:04.840 --> 00:08:06.010 AUDIENCE: x and y. 00:08:06.010 --> 00:08:09.986 PROFESSOR: No rotations about which axes? 00:08:09.986 --> 00:08:10.820 AUDIENCE: x and y. 00:08:10.820 --> 00:08:13.683 PROFESSOR: So you're saying no rotations of this about the x. 00:08:13.683 --> 00:08:15.340 So it can't roll over. 00:08:15.340 --> 00:08:17.960 About the y, it can't turn. 00:08:17.960 --> 00:08:20.260 And about the z. 00:08:20.260 --> 00:08:22.756 But we're doing this one. 00:08:22.756 --> 00:08:24.380 So it's no spinning about-- let's do it 00:08:24.380 --> 00:08:25.550 in its body coordinates. 00:08:25.550 --> 00:08:30.070 x1-- no spinning about x1. 00:08:30.070 --> 00:08:33.120 No spinning about y1. 00:08:33.120 --> 00:08:35.679 But can it rotate around z1? 00:08:35.679 --> 00:08:47.310 OK, so for the second mass, we have no x1, y1 rotations. 00:08:47.310 --> 00:08:48.100 That's two. 00:08:48.100 --> 00:08:50.180 And what about translations on the second body? 00:08:53.916 --> 00:08:54.850 Hmm? 00:08:54.850 --> 00:08:56.730 AUDIENCE: No translations x, y, and z. 00:08:56.730 --> 00:08:58.380 PROFESSOR: No translations x, y, and z. 00:08:58.380 --> 00:09:01.320 So z, pretty obvious-- we're not letting it do this. 00:09:01.320 --> 00:09:07.136 But why can you argue there's no translation in x and y? 00:09:07.136 --> 00:09:10.608 AUDIENCE: Because it's strictly rotation. 00:09:10.608 --> 00:09:13.856 PROFESSOR: You say it's strictly rotation. 00:09:13.856 --> 00:09:17.994 AUDIENCE: For it to translate [INAUDIBLE]. 00:09:17.994 --> 00:09:19.785 PROFESSOR: OK, so you're saying translation 00:09:19.785 --> 00:09:22.680 is strictly described as parallel motion 00:09:22.680 --> 00:09:24.980 of all points on it. 00:09:24.980 --> 00:09:27.915 But where is it constrained in x and y? 00:09:27.915 --> 00:09:30.532 AUDIENCE: [INAUDIBLE] 00:09:30.532 --> 00:09:31.990 PROFESSOR: It's pinned to the cart. 00:09:31.990 --> 00:09:34.240 And if you fix x, which you're allowed to do, 00:09:34.240 --> 00:09:36.712 a test you're allowed to do, you have set this 00:09:36.712 --> 00:09:37.920 as an independent coordinate. 00:09:37.920 --> 00:09:41.230 If you fix it, this point is fixed. 00:09:41.230 --> 00:09:44.780 And you've constrained it in x and y. 00:09:44.780 --> 00:09:48.100 And if any point in the object is constrained in x and y, 00:09:48.100 --> 00:09:54.590 then all points are constrained in x and y pure translation. 00:09:54.590 --> 00:10:02.035 So we have in this one no x, y, and z translations. 00:10:04.610 --> 00:10:05.520 So that's three. 00:10:05.520 --> 00:10:06.880 You add them up, you get 10. 00:10:06.880 --> 00:10:08.590 OK, good. 00:10:08.590 --> 00:10:12.720 Now, the next thing I want you to do, 00:10:12.720 --> 00:10:16.320 another key step in doing problems like this, 00:10:16.320 --> 00:10:17.930 is our free body diagrams. 00:10:17.930 --> 00:10:20.850 And how many do you need? 00:10:20.850 --> 00:10:27.525 As many as you have rigid bodies. 00:10:27.525 --> 00:10:30.400 You need one for every rigid body. 00:10:30.400 --> 00:10:32.030 So we got two rigid bodies. 00:10:32.030 --> 00:10:41.770 Draw for me your free body diagram for the cart 00:10:41.770 --> 00:10:44.265 and for the rod. 00:11:16.930 --> 00:11:20.830 OK, what group wants to volunteer here, 00:11:20.830 --> 00:11:23.319 tell me what I ought to do? 00:11:23.319 --> 00:11:24.610 How about you guys in the back? 00:11:24.610 --> 00:11:26.330 You're quite today. 00:11:26.330 --> 00:11:29.460 So pick this first one. 00:11:29.460 --> 00:11:33.880 AUDIENCE: You have the spring force kx to the left. 00:11:33.880 --> 00:11:35.002 PROFESSOR: kx. 00:11:35.002 --> 00:11:36.970 AUDIENCE: Also bx dot. 00:11:36.970 --> 00:11:38.710 PROFESSOR: bx dot. 00:11:38.710 --> 00:11:40.570 AUDIENCE: Normal force up. 00:11:40.570 --> 00:11:41.487 PROFESSOR: N. 00:11:41.487 --> 00:11:46.844 AUDIENCE: You have M1g down, and then the force of the rod 00:11:46.844 --> 00:11:48.305 is down and to the right. 00:11:48.305 --> 00:11:52.155 PROFESSOR: Oh, so you've made-- ahh, colinear with the rod? 00:11:52.155 --> 00:11:52.780 AUDIENCE: Yeah. 00:11:52.780 --> 00:11:54.490 PROFESSOR: OK, so you're saying we've 00:11:54.490 --> 00:11:59.000 got a vector force like that, some F, right? 00:11:59.000 --> 00:12:02.240 OK, how about a little help on this. 00:12:02.240 --> 00:12:06.994 Anybody have a different result here for this one? 00:12:06.994 --> 00:12:10.930 AUDIENCE: I broke up my normal into two [INAUDIBLE]. 00:12:10.930 --> 00:12:13.950 PROFESSOR: So you have two pieces here? 00:12:13.950 --> 00:12:17.712 So in what direction do you draw them? 00:12:17.712 --> 00:12:20.060 AUDIENCE: Reaction force of the first failed reaction? 00:12:20.060 --> 00:12:22.750 PROFESSOR: I guess which coordinate system? 00:12:22.750 --> 00:12:27.260 We need two coordinate systems to define this problem. 00:12:27.260 --> 00:12:29.140 AUDIENCE: Just in the regular x, y. 00:12:29.140 --> 00:12:34.100 PROFESSOR: So I've got an inertial frame here. 00:12:34.100 --> 00:12:37.720 And that gives me my X, capital X, motion for this guy. 00:12:37.720 --> 00:12:40.000 But we need, because if we want to be able to moments 00:12:40.000 --> 00:12:43.550 of inertia and things like that, a coordinate system attached 00:12:43.550 --> 00:12:45.440 to this rigid body, right? 00:12:45.440 --> 00:12:48.990 So I've called it little x1, y1, and it moves with the body. 00:12:48.990 --> 00:12:52.130 OK, so you've picked your two reaction 00:12:52.130 --> 00:12:57.094 forces aligned with these two or aligned in this system? 00:12:57.094 --> 00:12:59.360 AUDIENCE: With the inertial. 00:12:59.360 --> 00:13:01.300 PROFESSOR: OK, you've done it this way. 00:13:01.300 --> 00:13:03.010 So you're coming up with a pair of forces 00:13:03.010 --> 00:13:07.198 that have a piece like this and a piece like that? 00:13:07.198 --> 00:13:11.011 AUDIENCE: No, I'm talking about for the wheels. 00:13:11.011 --> 00:13:12.260 PROFESSOR: Oh, for the wheels. 00:13:12.260 --> 00:13:13.930 Oh, OK. 00:13:13.930 --> 00:13:15.750 You came up with individual ones here? 00:13:15.750 --> 00:13:17.530 Yeah, OK. 00:13:17.530 --> 00:13:20.390 I'll call that N1 and N2. 00:13:20.390 --> 00:13:23.520 And that's probably an improvement. 00:13:23.520 --> 00:13:26.210 Because that's what keeps it from rolling over. 00:13:26.210 --> 00:13:28.780 OK, that's true. 00:13:28.780 --> 00:13:29.990 And what else? 00:13:29.990 --> 00:13:32.892 So you've got a reaction force here like that. 00:13:32.892 --> 00:13:33.850 Can we improve on that? 00:13:36.670 --> 00:13:39.960 AUDIENCE: You can break it up into tangential and radial. 00:13:39.960 --> 00:13:41.470 PROFESSOR: Yeah, but I understood. 00:13:41.470 --> 00:13:44.570 The intention here was it was drawn colinear 00:13:44.570 --> 00:13:47.290 with the axis of the thing. 00:13:47.290 --> 00:13:50.060 And that suggested just that one part. 00:13:50.060 --> 00:13:53.630 You could break that into a horizontal in the master frame, 00:13:53.630 --> 00:13:54.130 yeah. 00:13:54.130 --> 00:13:58.090 But does anybody have something different about that, 00:13:58.090 --> 00:13:59.520 another way to do this? 00:13:59.520 --> 00:14:01.450 Do you agree with this? 00:14:01.450 --> 00:14:03.510 Is this correct? 00:14:03.510 --> 00:14:04.010 Pardon? 00:14:07.080 --> 00:14:09.160 You're treating it like a string. 00:14:09.160 --> 00:14:09.966 Is it a string? 00:14:12.840 --> 00:14:15.935 Can that shaft transmit shear forces? 00:14:20.111 --> 00:14:20.610 Hmm? 00:14:25.330 --> 00:14:27.730 If I grab this thing and go left and right on it, 00:14:27.730 --> 00:14:30.020 it's putting forces in that direction perpendicular 00:14:30.020 --> 00:14:32.650 to this. 00:14:32.650 --> 00:14:35.150 So is it conceivable that there are 00:14:35.150 --> 00:14:37.915 forces on this axle caused by that 00:14:37.915 --> 00:14:42.120 that are in that direction as well as that direction? 00:14:42.120 --> 00:14:44.000 What do you think? 00:14:44.000 --> 00:14:45.800 Because I can't make an argument for sure 00:14:45.800 --> 00:14:48.169 that would eliminate either one of those. 00:14:48.169 --> 00:14:49.960 Pretty sure there's certainly a radial one, 00:14:49.960 --> 00:14:52.585 because that's centripetal, and there's got to be some of that. 00:14:52.585 --> 00:14:56.100 But might there be some in the other direction? 00:14:56.100 --> 00:15:01.270 OK, so an improvement on this, a necessary improvement, I think, 00:15:01.270 --> 00:15:05.370 is that rather than just one, we'd better have two. 00:15:05.370 --> 00:15:08.360 And we're going to use my system of naming 00:15:08.360 --> 00:15:11.110 so that the answers I've got here will work out. 00:15:11.110 --> 00:15:16.640 I'm going to call that one F2 and this one F1. 00:15:16.640 --> 00:15:20.900 So now the rod is placing on the cart two reaction forces 00:15:20.900 --> 00:15:22.180 which you don't know. 00:15:22.180 --> 00:15:26.160 Now, is that the complete thing for the cart? 00:15:26.160 --> 00:15:28.390 OK, now let's do this one. 00:15:28.390 --> 00:15:33.210 Another group, tell me how to do the second one, free body 00:15:33.210 --> 00:15:36.165 diagram. 00:15:36.165 --> 00:15:39.116 AUDIENCE: So you have F1 and F2, but opposite. 00:15:39.116 --> 00:15:40.240 PROFESSOR: Ahh, good point. 00:15:40.240 --> 00:15:46.950 You've got an F2 this way and an F1 like that, 00:15:46.950 --> 00:15:48.200 is what you're saying? 00:15:48.200 --> 00:15:50.330 AUDIENCE: Yeah, and an Mg comes down. 00:15:50.330 --> 00:15:53.863 PROFESSOR: And an Mg, yeah, M2g. 00:15:53.863 --> 00:15:55.530 OK, good. 00:15:55.530 --> 00:16:00.340 Real important point-- Newton's third law. 00:16:00.340 --> 00:16:02.810 If you draw F1 and F2 here, those on that side 00:16:02.810 --> 00:16:07.300 have to be equal and opposite in order for all this 00:16:07.300 --> 00:16:08.890 to work out without problems. 00:16:08.890 --> 00:16:11.110 OK, those look pretty good to me. 00:16:14.010 --> 00:16:15.401 Now let's move on. 00:16:15.401 --> 00:16:16.650 This is a complicated problem. 00:16:16.650 --> 00:16:20.480 And to get a lot with it in an hour is a challenge. 00:16:20.480 --> 00:16:22.490 And you've done a bunch of this problem already. 00:16:22.490 --> 00:16:24.615 You did the one with the particle on it. 00:16:24.615 --> 00:16:27.050 So there's not a great deal of difference here. 00:16:27.050 --> 00:16:29.690 So I'm emphasizing a few of the nuances. 00:16:29.690 --> 00:16:32.020 What I now want to do is to talk about, 00:16:32.020 --> 00:16:36.870 how do we get equations of motion for this problem? 00:16:36.870 --> 00:16:38.956 How many will there be to start with? 00:16:38.956 --> 00:16:39.789 AUDIENCE: Two. 00:16:39.789 --> 00:16:41.580 PROFESSOR: Got to get two, two rigid bodies 00:16:41.580 --> 00:16:43.831 and two degrees of freedom. 00:16:43.831 --> 00:16:46.330 And you're going to end up-- it's the two degrees of freedom 00:16:46.330 --> 00:16:48.705 that tell you you're going to get two equations of motion 00:16:48.705 --> 00:16:49.260 at the end. 00:16:49.260 --> 00:16:53.210 OK, how many-- and I've written three up there, 00:16:53.210 --> 00:16:55.730 but it's not necessarily true. 00:16:55.730 --> 00:16:58.490 You've been taught to this point two different methods 00:16:58.490 --> 00:17:02.990 for getting at these equations of motion. 00:17:02.990 --> 00:17:05.079 Somebody describe one for me. 00:17:05.079 --> 00:17:08.400 How would you personally go about this problem? 00:17:08.400 --> 00:17:12.730 Just pick a method and in words describe it. 00:17:12.730 --> 00:17:14.821 And think of what you did in the last homework. 00:17:14.821 --> 00:17:19.640 AUDIENCE: Sum of external force equals [INAUDIBLE]. 00:17:19.640 --> 00:17:22.894 PROFESSOR: So sum of external forces is-- 00:17:22.894 --> 00:17:23.819 AUDIENCE: [INAUDIBLE] 00:17:23.819 --> 00:17:28.460 PROFESSOR: OK, so on method one here, I'll call it, 00:17:28.460 --> 00:17:33.191 you're going to say sum of the forces on what? 00:17:33.191 --> 00:17:34.232 AUDIENCE: The rigid body. 00:17:34.232 --> 00:17:36.145 PROFESSOR: So you've got two bodies here. 00:17:36.145 --> 00:17:38.011 So pick one. 00:17:38.011 --> 00:17:39.454 Huh? 00:17:39.454 --> 00:17:42.650 All right, so I'm going to put this on M1 for sure. 00:17:42.650 --> 00:17:44.360 That one is translating. 00:17:44.360 --> 00:17:46.260 And you're going to get an equation that 00:17:46.260 --> 00:17:49.040 involves its x double dot. 00:17:49.040 --> 00:17:50.870 It's got to be able to move back and forth. 00:17:50.870 --> 00:17:53.200 You're going to need an equation like this. 00:17:53.200 --> 00:17:55.270 So the sum of the external forces on that mass 00:17:55.270 --> 00:18:04.900 has got to be equal to M1 times-- right? 00:18:04.900 --> 00:18:11.390 And we can write an equation for that in terms of this free body 00:18:11.390 --> 00:18:12.790 diagram. 00:18:12.790 --> 00:18:15.310 In the i direction, you've got this. 00:18:15.310 --> 00:18:16.750 You've got this. 00:18:16.750 --> 00:18:21.330 You have the normal forces gravity don't contribute. 00:18:21.330 --> 00:18:28.700 And components of this and this have to be added in. 00:18:28.700 --> 00:18:32.650 You have to do an F2, probably cosine theta, and an F1 sine 00:18:32.650 --> 00:18:33.960 theta, and add them in. 00:18:33.960 --> 00:18:35.460 Because they're the external forces. 00:18:35.460 --> 00:18:37.710 So you've got all these summation of forces 00:18:37.710 --> 00:18:39.390 on that side. 00:18:39.390 --> 00:18:41.560 That gives you one equation. 00:18:41.560 --> 00:18:46.530 This will give you an equation for the motion 00:18:46.530 --> 00:18:47.900 of the first mass. 00:18:47.900 --> 00:18:49.950 And it'll have some unknowns in it. 00:18:49.950 --> 00:18:50.700 And what are they? 00:18:53.510 --> 00:18:56.205 Describe them. 00:18:56.205 --> 00:18:57.580 So you're going to get a function 00:18:57.580 --> 00:19:00.640 over here of a couple unknowns. 00:19:00.640 --> 00:19:03.491 This is going to be a function of what that you don't know? 00:19:03.491 --> 00:19:04.892 AUDIENCE: [INAUDIBLE] 00:19:04.892 --> 00:19:08.840 PROFESSOR: F1 and F2, and then also x 00:19:08.840 --> 00:19:12.366 double dot and thetas and theta dots and so forth, 00:19:12.366 --> 00:19:13.740 all the variables in the problem. 00:19:13.740 --> 00:19:15.114 But these are two unknowns you've 00:19:15.114 --> 00:19:16.800 got to ultimately get rid of. 00:19:16.800 --> 00:19:18.870 OK, what else are you going to do in method one? 00:19:22.132 --> 00:19:24.090 You've got to deal with the second mass, right? 00:19:24.090 --> 00:19:25.300 What would you do then? 00:19:25.300 --> 00:19:26.175 AUDIENCE: [INAUDIBLE] 00:19:31.420 --> 00:19:34.360 PROFESSOR: So some of the force is now on M2, for sure. 00:19:34.360 --> 00:19:37.120 And that's going to give you a couple equations. 00:19:37.120 --> 00:19:40.020 What else do you do in your method? 00:19:40.020 --> 00:19:42.241 AUDIENCE: [INAUDIBLE] 00:19:42.241 --> 00:19:43.990 PROFESSOR: [INAUDIBLE] torques about what? 00:19:47.490 --> 00:19:50.232 AUDIENCE: [INAUDIBLE] 00:19:50.232 --> 00:19:51.440 PROFESSOR: Have I named them? 00:19:51.440 --> 00:19:53.820 Look at that diagram there. 00:19:53.820 --> 00:19:56.460 So you're going to do your torques, sum 00:19:56.460 --> 00:20:00.020 of the torques, about A, OK? 00:20:00.020 --> 00:20:06.690 And if you sum torques about A, what are they? 00:20:06.690 --> 00:20:10.470 I'll let you guys pursue that a little bit, figure out, 00:20:10.470 --> 00:20:17.890 what are the torques about this point in this problem? 00:20:17.890 --> 00:20:20.040 OK, what do you get? 00:20:20.040 --> 00:20:21.670 Sum of the torques about A? 00:20:26.090 --> 00:20:29.430 z is in which-- which positive z? 00:20:29.430 --> 00:20:31.960 x, y, positive z coming out of the board. 00:20:31.960 --> 00:20:36.060 What produces torque about this point in this problem? 00:20:36.060 --> 00:20:36.940 Just what? 00:20:36.940 --> 00:20:37.690 AUDIENCE: Gravity. 00:20:37.690 --> 00:20:38.690 PROFESSOR: Just gravity. 00:20:38.690 --> 00:20:40.822 And is it positive or negative? 00:20:40.822 --> 00:20:44.390 And it's a component of gravity perpendicular to this. 00:20:44.390 --> 00:20:49.340 So it's probably sine theta M2g sine theta in the minus. 00:20:49.340 --> 00:20:57.220 So it's minus M2g L over 2-- you need the moment arm-- sine 00:20:57.220 --> 00:21:01.030 theta in the k hat direction. 00:21:01.030 --> 00:21:17.840 And that's going to be equal to the angular dH, dT about A. 00:21:17.840 --> 00:21:20.310 Sum of the torques, time rate of change of the angular 00:21:20.310 --> 00:21:21.960 momentum about A-- because that's what 00:21:21.960 --> 00:21:23.580 you've chosen to work about. 00:21:23.580 --> 00:21:27.010 And you always have this term you have to check on. 00:21:30.460 --> 00:21:32.770 OK, so that's the method. 00:21:32.770 --> 00:21:34.530 This is method one. 00:21:34.530 --> 00:21:37.730 I'll call this the tau with respect to A method. 00:21:40.940 --> 00:21:42.560 You have to write this equation. 00:21:42.560 --> 00:21:47.700 You have two unknowns that pop up in it. 00:21:47.700 --> 00:21:49.120 You go to the second mass. 00:21:49.120 --> 00:21:51.180 You write the F equals ma for that, 00:21:51.180 --> 00:21:54.150 and you're going to get two equations. 00:21:54.150 --> 00:21:56.520 If you do it in the local coordinate system, 00:21:56.520 --> 00:21:59.590 in the rotating one, you get mass times acceleration 00:21:59.590 --> 00:22:01.210 in the i direction. 00:22:01.210 --> 00:22:04.970 You get F1 and a component of gravity. 00:22:04.970 --> 00:22:08.610 And in the j direction, m2 acceleration in j, 00:22:08.610 --> 00:22:11.730 you get F2 and a piece that's acceleration of gravity. 00:22:16.610 --> 00:22:18.600 You don't know this yet. 00:22:18.600 --> 00:22:20.310 But you can find it. 00:22:20.310 --> 00:22:24.520 You know how to find those velocities now. 00:22:24.520 --> 00:22:25.650 You've done it before. 00:22:25.650 --> 00:22:30.440 So the velocity of G with respect to O, 00:22:30.440 --> 00:22:32.230 probably going to have an X double 00:22:32.230 --> 00:22:38.930 dot I hat plus something that will be like L/2 theta dot, 00:22:38.930 --> 00:22:40.995 and in what direction? 00:22:40.995 --> 00:22:41.495 Hmm? 00:22:44.080 --> 00:22:49.130 It's rotating like this-- j hat, little j. 00:22:49.130 --> 00:22:50.350 So there's your velocity. 00:22:50.350 --> 00:22:54.030 You can take the time derivative of that to get an acceleration, 00:22:54.030 --> 00:22:56.370 X double dot-- that's not true. 00:22:56.370 --> 00:22:58.170 Down here, you get an X double dot. 00:22:58.170 --> 00:22:59.580 In this one, you get two terms. 00:22:59.580 --> 00:23:01.580 Because that's a rotating piece. 00:23:01.580 --> 00:23:06.200 You get a centripetal term and an Euler term 00:23:06.200 --> 00:23:07.410 that come out of that. 00:23:07.410 --> 00:23:10.660 So then now you know velocities and accelerations. 00:23:10.660 --> 00:23:13.320 You can put the accelerations on this side. 00:23:13.320 --> 00:23:15.060 You can solve this first equation 00:23:15.060 --> 00:23:18.645 for F1, second equation for F2. 00:23:18.645 --> 00:23:21.340 You take those and put them in here. 00:23:21.340 --> 00:23:25.110 And you finally have one of your equations of motion. 00:23:25.110 --> 00:23:32.590 Down here, the external torque only involves gravity. 00:23:32.590 --> 00:23:36.150 So do your F1 and F2 appear-- do you have to mess with those 00:23:36.150 --> 00:23:43.450 in this final equation or not? 00:23:43.450 --> 00:23:47.800 This is the reason you use point A. This equation does not 00:23:47.800 --> 00:23:49.220 involve F1 or F2. 00:23:49.220 --> 00:23:51.860 There's no moments created by F1 and F2. 00:23:51.860 --> 00:23:55.120 The stuff on the right hand side is only kinematics and mass 00:23:55.120 --> 00:23:57.890 moments of inertia. 00:23:57.890 --> 00:24:00.260 This thing will give you an i with respect-- 00:24:00.260 --> 00:24:03.260 you can think of this as it's going to give you 00:24:03.260 --> 00:24:07.450 an i with respect to G theta double dot 00:24:07.450 --> 00:24:10.210 plus some other terms. 00:24:10.210 --> 00:24:12.145 And I want to talk about this. 00:24:12.145 --> 00:24:13.410 Does this term go to 0? 00:24:17.200 --> 00:24:20.210 Is the velocity of A 0 in this problem? 00:24:20.210 --> 00:24:20.960 No way. 00:24:20.960 --> 00:24:23.800 And in what direction is it? 00:24:23.800 --> 00:24:27.450 Horizontal, right, capital I hat. 00:24:27.450 --> 00:24:29.570 P, the momentum of the second thing, 00:24:29.570 --> 00:24:33.170 has these velocity components. 00:24:33.170 --> 00:24:36.330 The P is equal to m times that. 00:24:36.330 --> 00:24:38.520 And you have velocity components in the I, 00:24:38.520 --> 00:24:39.960 which is the same as vA. 00:24:39.960 --> 00:24:43.440 But you have another velocity component that's not the same. 00:24:43.440 --> 00:24:45.890 So is that cross product going to be 0? 00:24:45.890 --> 00:24:46.930 No way. 00:24:46.930 --> 00:24:49.140 You have to deal with that term. 00:24:49.140 --> 00:24:53.780 It happens that you can express H with respect 00:24:53.780 --> 00:25:02.440 to A as H with respect to G plus rG A cross P. 00:25:02.440 --> 00:25:07.440 And you know P, because you just found v. It's mv. 00:25:07.440 --> 00:25:14.140 And r, rG A, is just L over 2 j, little j, 00:25:14.140 --> 00:25:16.700 which is from here to here. 00:25:16.700 --> 00:25:21.896 It's L over 2 little-- no i, excuse me. 00:25:21.896 --> 00:25:23.920 So that's r, L over 2 i. 00:25:23.920 --> 00:25:27.239 So you can crank out this cross product. 00:25:27.239 --> 00:25:29.530 Then you have to take the time derivative of this right 00:25:29.530 --> 00:25:31.560 here, this whole thing. 00:25:31.560 --> 00:25:38.340 Every time when you take the time derivative of this part, 00:25:38.340 --> 00:25:41.000 you will get a P. So you get multiple pieces. 00:25:41.000 --> 00:25:43.150 But one of them will be minus that. 00:25:45.920 --> 00:25:48.940 This thing creates a piece that's exactly minus that, 00:25:48.940 --> 00:25:50.300 and they'll cancel. 00:25:50.300 --> 00:25:54.290 But you've got to go through the cranking it out 00:25:54.290 --> 00:25:55.930 till you get to that point. 00:25:55.930 --> 00:25:58.050 OK, that's one method of doing it. 00:25:58.050 --> 00:26:01.370 Let's talk briefly about a second method. 00:26:01.370 --> 00:26:03.620 What is the other way to go about doing this problem? 00:26:07.564 --> 00:26:09.540 AUDIENCE: [INAUDIBLE] 00:26:09.540 --> 00:26:12.330 PROFESSOR: That's method three. 00:26:12.330 --> 00:26:15.452 We're going to hurry so I can get to it. 00:26:15.452 --> 00:26:17.785 What's a second way to do this problem, a close relative 00:26:17.785 --> 00:26:18.750 of the first way? 00:26:21.956 --> 00:26:24.630 But I want you to have all these different ways 00:26:24.630 --> 00:26:26.320 of doing it in your tool kit. 00:26:26.320 --> 00:26:29.490 Because some problems are easy, more easily done. 00:26:29.490 --> 00:26:33.050 There's another way than this. 00:26:33.050 --> 00:26:37.090 This is perfectly appropriate for this type of problem. 00:26:37.090 --> 00:26:44.890 If the problem we were doing involved this, 00:26:44.890 --> 00:26:46.520 would you do this method? 00:26:46.520 --> 00:26:48.460 What would you do? 00:26:48.460 --> 00:26:49.400 AUDIENCE: [INAUDIBLE] 00:26:49.400 --> 00:26:50.900 PROFESSOR: No, can't use energy yet. 00:26:50.900 --> 00:26:51.920 That's next week. 00:26:51.920 --> 00:26:52.860 AUDIENCE: [INAUDIBLE] 00:26:52.860 --> 00:26:54.405 PROFESSOR: Ahh, right. 00:26:54.405 --> 00:26:57.270 So the second method is basically 00:26:57.270 --> 00:26:59.900 this is the same, this piece. 00:26:59.900 --> 00:27:04.200 But the second part-- and this is the same, 00:27:04.200 --> 00:27:05.740 sum of the forces about M2. 00:27:05.740 --> 00:27:07.600 You do both of those forces. 00:27:07.600 --> 00:27:11.460 You do sum forces for sure, M1 and M2. 00:27:11.460 --> 00:27:17.838 But the third step is you do sum of torques about g. 00:27:17.838 --> 00:27:26.600 If you sum torques about g, which is here, 00:27:26.600 --> 00:27:29.720 what appears in the sum torques? 00:27:29.720 --> 00:27:31.343 Does Mg contribute? 00:27:31.343 --> 00:27:35.021 No, what are the external torques about g? 00:27:35.021 --> 00:27:36.452 AUDIENCE: [INAUDIBLE] 00:27:36.452 --> 00:27:43.550 PROFESSOR: F2, L/2, positive F2 L/2 k. 00:27:43.550 --> 00:27:47.390 So that equation, summing torques about g, 00:27:47.390 --> 00:27:51.030 will give you an F2 L/2 k. 00:27:51.030 --> 00:27:53.310 And now you've got that unknown popping up 00:27:53.310 --> 00:27:54.830 in your torque equation. 00:27:54.830 --> 00:27:58.020 And you know that it also pops up in these two equations. 00:27:58.020 --> 00:28:01.140 Because we solved them. 00:28:01.140 --> 00:28:07.030 So now why does this method have a slight advantage 00:28:07.030 --> 00:28:09.162 over this method? 00:28:09.162 --> 00:28:10.940 AUDIENCE: [INAUDIBLE] 00:28:10.940 --> 00:28:13.641 PROFESSOR: Well, you don't get F2 or F1 popping up 00:28:13.641 --> 00:28:14.640 in your torque equation. 00:28:14.640 --> 00:28:16.680 You immediately get to the answer for that one. 00:28:16.680 --> 00:28:18.013 So it's a little more efficient. 00:28:24.490 --> 00:28:32.800 So is there a way to get directly 00:28:32.800 --> 00:28:36.870 at the two equations we're after? 00:28:36.870 --> 00:28:41.950 The second equation we get from this torque thing. 00:28:41.950 --> 00:28:45.880 The first equation we get from resolving this. 00:28:45.880 --> 00:28:50.510 This is a mass times-- M1 X double dot. 00:28:50.510 --> 00:28:52.680 And you get this function of F1 and F2. 00:28:52.680 --> 00:28:54.920 And you have to substitute in for these F1's 00:28:54.920 --> 00:28:58.060 and F2's until you finally get that equation. 00:28:58.060 --> 00:29:00.570 So that's kind of tedious. 00:29:00.570 --> 00:29:05.230 So is there a way to get directly at this without having 00:29:05.230 --> 00:29:07.700 to mess with F1 and F2 at all? 00:29:07.700 --> 00:29:10.670 And that's what is the key to what you're talking about. 00:29:10.670 --> 00:29:11.980 Let's look at something here. 00:29:11.980 --> 00:29:20.360 Let's do the vector sum of the forces on M1. 00:29:24.270 --> 00:29:29.170 We get an F. Let's look at our free body diagram, 00:29:29.170 --> 00:29:47.920 F2j plus F1i minus kx minus bx dot and a minus M1g. 00:29:47.920 --> 00:29:49.920 Well, those are-- well, we'll put them in there, 00:29:49.920 --> 00:29:52.600 at least throw them in there. 00:29:52.600 --> 00:30:04.370 This is in my capital J plus N1 plus N2 capital J hat here. 00:30:04.370 --> 00:30:07.320 Is that all the forces, all the vector forces in that problem? 00:30:12.630 --> 00:30:13.860 Did I miss anything? 00:30:13.860 --> 00:30:16.400 I've got spring, the dashpot, the forces 00:30:16.400 --> 00:30:19.614 caused by the rod, gravity, and the normal forces. 00:30:19.614 --> 00:30:32.490 OK, and we could figure out the component of this, 00:30:32.490 --> 00:30:37.150 the sum of the forces on M1 in the capital X direction. 00:30:37.150 --> 00:30:39.200 We could figure that out from-- this 00:30:39.200 --> 00:30:41.450 has component in that direction. 00:30:41.450 --> 00:30:43.160 It just involves thetas. 00:30:43.160 --> 00:30:44.195 These go away, then. 00:30:44.195 --> 00:30:46.570 You don't have to deal with this stuff in the x equation. 00:30:49.950 --> 00:30:52.217 But there's our sum of the forces on the first body. 00:30:52.217 --> 00:30:54.300 Let's do the sum of the forces on the second body. 00:30:57.870 --> 00:30:58.930 What are they? 00:31:04.056 --> 00:31:05.460 AUDIENCE: Gravity. 00:31:05.460 --> 00:31:10.800 PROFESSOR: OK, so you've got a minus M2g on it. 00:31:10.800 --> 00:31:12.120 And what else? 00:31:15.992 --> 00:31:17.882 AUDIENCE: [INAUDIBLE] 00:31:17.882 --> 00:31:19.590 PROFESSOR: But with opposite sign, right? 00:31:19.590 --> 00:31:25.710 F2 little j minus F1 little i, and anything else? 00:31:28.330 --> 00:31:29.570 Nothing, right? 00:31:29.570 --> 00:31:30.700 Let's add the two together. 00:31:33.570 --> 00:31:39.020 Oh, and this one has got to be equal to the mass, M1, 00:31:39.020 --> 00:31:47.720 times the acceleration, the x component I plus M1 00:31:47.720 --> 00:31:53.380 acceleration in the-- I can't write over here-- ay 00:31:53.380 --> 00:31:58.250 in the capital J hat direction. 00:31:58.250 --> 00:31:59.940 That gives us mass times acceleration, 00:31:59.940 --> 00:32:02.790 is what all the sums of the forces is equal to. 00:32:02.790 --> 00:32:09.180 The same thing here-- this is equal to the little mass, M2, 00:32:09.180 --> 00:32:11.250 times its acceleration. 00:32:11.250 --> 00:32:19.500 And it'll have components in the-- this is mass 2. 00:32:19.500 --> 00:32:23.680 And it'll have components we can put in the I hat direction, 00:32:23.680 --> 00:32:33.720 and plus M2 acceleration 2-- maybe I'll do it like this. 00:32:37.380 --> 00:32:40.430 a2y, and this is in the j direction. 00:32:40.430 --> 00:32:45.090 And those two things we can add together. 00:32:45.090 --> 00:32:53.148 And so if I add them together, and I put M1 plus M2 times-- 00:32:53.148 --> 00:32:57.151 I'll just call it the total acceleration. 00:32:57.151 --> 00:32:58.850 Well, I don't want to do that. 00:32:58.850 --> 00:33:00.780 I don't want to say this. 00:33:07.362 --> 00:33:09.320 I'm hurrying because we're running out of time. 00:33:09.320 --> 00:33:17.850 M1 times its acceleration times M2 plus its acceleration 00:33:17.850 --> 00:33:20.740 is equal to-- and now we sum up everything on the right hand 00:33:20.740 --> 00:33:22.610 side. 00:33:22.610 --> 00:33:25.440 And the key thing that happens here 00:33:25.440 --> 00:33:26.780 is what happens at F1 and F2? 00:33:26.780 --> 00:33:27.610 AUDIENCE: Drop out. 00:33:27.610 --> 00:33:29.766 PROFESSOR: They completely drop out. 00:33:33.080 --> 00:33:40.630 So on the right hand side, it's not a function of F1 and F2. 00:33:47.050 --> 00:33:51.880 The equating of motion that we were trying to get here 00:33:51.880 --> 00:33:54.270 was just in the I direction. 00:33:54.270 --> 00:33:56.810 That's all we need for that big cart. 00:33:56.810 --> 00:33:59.460 So if we go down here and extract 00:33:59.460 --> 00:34:06.830 the I component of this total sum, it'll be an equation. 00:34:06.830 --> 00:34:08.710 It'll involve M1 and M2. 00:34:08.710 --> 00:34:15.760 And it'll have cosines and sines and thetas and things like 00:34:15.760 --> 00:34:17.860 that, but no F1's, no F2's. 00:34:17.860 --> 00:34:24.090 You will get directly to this, the final result that you got 00:34:24.090 --> 00:34:26.250 here where you had to go in and substitute 00:34:26.250 --> 00:34:29.376 in for F1's and F2's. 00:34:29.376 --> 00:34:30.709 So it'll get it to you directly. 00:34:33.790 --> 00:34:36.420 What do you need to solve it, though? 00:34:36.420 --> 00:34:38.540 You need to know the acceleration of mass 1. 00:34:41.070 --> 00:34:41.900 You know that. 00:34:41.900 --> 00:34:43.340 That's trivial. 00:34:43.340 --> 00:34:47.530 You need to know the acceleration of mass 2. 00:34:47.530 --> 00:34:52.400 And that you did by taking the time 00:34:52.400 --> 00:34:53.580 derivatives of these terms. 00:34:53.580 --> 00:34:56.210 You know the acceleration of that mass. 00:34:56.210 --> 00:34:58.870 So you need that expression. 00:34:58.870 --> 00:35:02.190 And you plug that in here. 00:35:02.190 --> 00:35:04.770 And everything else in the right hand side you know. 00:35:04.770 --> 00:35:11.022 It's kx dot, bx dot, minus Mg, M1g minus M2g, and so forth. 00:35:11.022 --> 00:35:12.480 All the rest of the stuff is known. 00:35:12.480 --> 00:35:12.770 Yeah. 00:35:12.770 --> 00:35:15.170 AUDIENCE: Does this give you three equations of motion? 00:35:15.170 --> 00:35:17.003 PROFESSOR: Well, it gives you two equations. 00:35:18.980 --> 00:35:22.120 At this point, you'd break this into capital I and capital 00:35:22.120 --> 00:35:26.280 J. The one in the capital J direction is equal to 0. 00:35:26.280 --> 00:35:27.770 There's no motion in that direction 00:35:27.770 --> 00:35:30.490 for the main mass, the cart. 00:35:30.490 --> 00:35:36.990 There's only motion in the inertial frame x direction. 00:35:36.990 --> 00:35:39.417 So that's all. 00:35:39.417 --> 00:35:40.500 This is a vector equation. 00:35:40.500 --> 00:35:46.535 It has two pieces to it, I and J. The J is a static equation. 00:35:46.535 --> 00:35:48.310 The I is your dynamic one. 00:35:48.310 --> 00:35:52.250 And it's this one that you would have found in this problem, 00:35:52.250 --> 00:35:55.630 but without ever having to solve for F1 and F2. 00:35:55.630 --> 00:36:00.700 But you do have to go out and find the acceleration 00:36:00.700 --> 00:36:02.910 of both masses. 00:36:02.910 --> 00:36:06.050 Mass times acceleration, what this is really saying-- this 00:36:06.050 --> 00:36:08.380 is what you asked about the system. 00:36:08.380 --> 00:36:09.940 This is sort of the system approach. 00:36:09.940 --> 00:36:15.230 If you think of this whole thing as a system, 00:36:15.230 --> 00:36:20.080 draw a box around it, the sum of the external forces 00:36:20.080 --> 00:36:24.910 on the system is equal to the mass, 00:36:24.910 --> 00:36:30.650 the total mass, times actually the acceleration of the center 00:36:30.650 --> 00:36:32.620 of mass of the system. 00:36:32.620 --> 00:36:35.250 But this acceleration times the center 00:36:35.250 --> 00:36:38.480 of mass of the system, you can break this into two pieces. 00:36:38.480 --> 00:36:45.040 It is this term plus this term. 00:36:45.040 --> 00:36:49.080 And then that's the total system. 00:36:49.080 --> 00:36:52.170 And all you have to put on this side 00:36:52.170 --> 00:36:56.170 is the summation of the external forces. 00:36:56.170 --> 00:37:00.060 But the external forces are not-- the internal forces 00:37:00.060 --> 00:37:00.919 don't count. 00:37:00.919 --> 00:37:02.210 They're not part of the system. 00:37:02.210 --> 00:37:04.100 So you only have to put the external forces 00:37:04.100 --> 00:37:10.040 on it-- Mg's springs, dashpots, normal forces, outside forces, 00:37:10.040 --> 00:37:13.820 the only ones that go over here. 00:37:13.820 --> 00:37:16.450 The reason we don't usually think of doing this 00:37:16.450 --> 00:37:18.470 is because when you think about the system, 00:37:18.470 --> 00:37:23.240 you write it as the acceleration of the center of mass. 00:37:23.240 --> 00:37:25.270 But you have to realize that you can 00:37:25.270 --> 00:37:29.680 get that expression as the sum of the parts, the sum 00:37:29.680 --> 00:37:32.970 of each bit times its acceleration, 00:37:32.970 --> 00:37:35.110 and sum them all up, is the same answer 00:37:35.110 --> 00:37:37.900 as the sum of the masses times the acceleration 00:37:37.900 --> 00:37:41.240 of the center of mass of the whole system. 00:37:41.240 --> 00:37:44.275 So we've still got a couple of minutes. 00:37:44.275 --> 00:37:46.400 Last hour, I ran over, and we couldn't get to this. 00:37:46.400 --> 00:37:50.010 So there is, in a way, yet another thing you can do. 00:37:50.010 --> 00:37:52.770 And that's think of the thing as a whole system. 00:37:52.770 --> 00:37:54.520 And sometimes that'll give you an equation 00:37:54.520 --> 00:37:56.540 without having to use F1, without having 00:37:56.540 --> 00:37:59.110 to solve for internal forces-- pretty cool. 00:38:04.170 --> 00:38:09.980 All right, so questions, thoughts? 00:38:09.980 --> 00:38:10.650 Yeah. 00:38:10.650 --> 00:38:13.548 AUDIENCE: Is it safe to say that if we 00:38:13.548 --> 00:38:17.090 have two more rigid bodies, we should probably consider it 00:38:17.090 --> 00:38:19.830 as a whole system [INAUDIBLE]? 00:38:19.830 --> 00:38:22.930 PROFESSOR: Well, I can't generalize on that. 00:38:22.930 --> 00:38:25.980 I haven't done enough problems like that myself to even 00:38:25.980 --> 00:38:30.330 have enough experience to know just how often is 00:38:30.330 --> 00:38:31.455 that going to help you out. 00:38:34.780 --> 00:38:37.500 This one, there's a pretty obvious equation 00:38:37.500 --> 00:38:40.950 that you're going to write in this direction on that mass. 00:38:40.950 --> 00:38:45.360 It's constrained to a single motion and a single coordinate 00:38:45.360 --> 00:38:46.930 you're assigning to it. 00:38:46.930 --> 00:38:48.870 And probably in cases like that where 00:38:48.870 --> 00:38:51.730 you can isolate one of the rigid bodies, 00:38:51.730 --> 00:38:54.436 there might be some advantage in doing something like this. 00:38:54.436 --> 00:38:56.102 AUDIENCE: --more difficult to figure out 00:38:56.102 --> 00:38:58.440 what the accelerations for both rigid bodies was, 00:38:58.440 --> 00:39:01.210 then it would be better [INAUDIBLE]. 00:39:01.210 --> 00:39:04.890 PROFESSOR: Yeah, this one, all of these methods 00:39:04.890 --> 00:39:08.880 you're going to end up having to compute the velocities 00:39:08.880 --> 00:39:11.720 and accelerations of each piece. 00:39:11.720 --> 00:39:13.400 And if you've got that information, 00:39:13.400 --> 00:39:17.770 you may as well use it if there's 00:39:17.770 --> 00:39:24.020 something to be gained by doing a clever step like this. 00:39:24.020 --> 00:39:24.827 Yeah. 00:39:24.827 --> 00:39:27.326 AUDIENCE: What confuses me is that method one, method three, 00:39:27.326 --> 00:39:29.410 there's no-- so you said there's going 00:39:29.410 --> 00:39:30.535 to be two things in motion. 00:39:30.535 --> 00:39:31.680 But why is it-- 00:39:31.680 --> 00:39:33.410 PROFESSOR: So this, this method three, 00:39:33.410 --> 00:39:37.940 this shows you a different way to get that equation. 00:39:37.940 --> 00:39:39.900 You still need this. 00:39:39.900 --> 00:39:42.260 AUDIENCE: Oh, this is just for the first-- 00:39:42.260 --> 00:39:45.160 PROFESSOR: This is a clever way to get just this first equation 00:39:45.160 --> 00:39:46.810 down. 00:39:46.810 --> 00:39:49.240 You still need a second equation. 00:39:49.240 --> 00:39:50.890 And you still are going to have to-- 00:39:50.890 --> 00:39:53.780 and then you don't want to have to mess with F1 and F2. 00:39:53.780 --> 00:39:55.950 So we've just found a way to never have 00:39:55.950 --> 00:39:59.010 to mess with F1 and F2 in at least this problem. 00:39:59.010 --> 00:40:00.790 And that is get the first equation 00:40:00.790 --> 00:40:04.370 by using this trick, the system equation. 00:40:04.370 --> 00:40:07.500 And it's a sum forces expression. 00:40:07.500 --> 00:40:11.790 And then the second equation you need is this moment equation. 00:40:11.790 --> 00:40:15.070 And you have two choices, about G or about A. 00:40:15.070 --> 00:40:17.400 And which one can you do it without having 00:40:17.400 --> 00:40:19.760 to find F1 and F2? 00:40:19.760 --> 00:40:22.784 I'm doing it about A. So there's a way 00:40:22.784 --> 00:40:24.200 that you can do this whole problem 00:40:24.200 --> 00:40:28.130 and never have to address what F1 and F2 are. 00:40:28.130 --> 00:40:32.490 And that's kind of a combination of these two pieces here. 00:40:32.490 --> 00:40:34.740 The traditional way, the textbook way, you open it up, 00:40:34.740 --> 00:40:36.510 the textbook will teach you this one. 00:40:36.510 --> 00:40:39.776 And it'll teach you about A and teach you about G. 00:40:39.776 --> 00:40:44.560 But you don't often run into-- there's sometimes 00:40:44.560 --> 00:40:49.890 a direct way of doing it without finding the F1 and F2. 00:40:52.704 --> 00:40:54.120 Unfortunately, there are thousands 00:40:54.120 --> 00:40:55.161 of problems in the world. 00:40:55.161 --> 00:41:01.400 And it's a little hard to say in advance what method to use 00:41:01.400 --> 00:41:03.310 and what's going to work the easiest. 00:41:03.310 --> 00:41:05.990 But there's some insight. 00:41:05.990 --> 00:41:07.820 There's some generalizations that you 00:41:07.820 --> 00:41:13.020 can get from looking at things in these different ways. 00:41:13.020 --> 00:41:18.140 If you have forces at a pivot point, 00:41:18.140 --> 00:41:20.360 taking the moments of the pivot point, 00:41:20.360 --> 00:41:23.070 unknowns of the pivot point, taking the moments 00:41:23.070 --> 00:41:25.290 about that point will get rid of them for you. 00:41:25.290 --> 00:41:27.400 That's the one generalization that you can usually 00:41:27.400 --> 00:41:28.170 take to the bank. 00:41:31.720 --> 00:41:35.310 But beyond that-- and if the thing's unconstrained. 00:41:35.310 --> 00:41:38.960 So if it's fixed axis rotation, then you 00:41:38.960 --> 00:41:42.100 want to use about the axis of fixed axis. 00:41:42.100 --> 00:41:44.190 This is a moving axis. 00:41:44.190 --> 00:41:48.999 But it is still rotating about a fixed point that's moving. 00:41:48.999 --> 00:41:50.790 There's still some advantage of doing this. 00:41:54.270 --> 00:41:59.400 If it's the eraser problem, this thing, 00:41:59.400 --> 00:42:02.330 where you have no fixed points of rotation-- 00:42:02.330 --> 00:42:06.070 if you take a body, and you do this with it, 00:42:06.070 --> 00:42:08.270 where is it spinning about? 00:42:08.270 --> 00:42:10.295 Where is its axis of rotation? 00:42:10.295 --> 00:42:11.170 AUDIENCE: [INAUDIBLE] 00:42:11.170 --> 00:42:12.890 PROFESSOR: Always, right? 00:42:12.890 --> 00:42:15.130 That you can take to the bank. 00:42:15.130 --> 00:42:19.780 If there's no forced pivot, bodies on their own 00:42:19.780 --> 00:42:24.000 only rotate about their centers of gravity, centers of mass. 00:42:24.000 --> 00:42:26.530 And then when you have free bodies that 00:42:26.530 --> 00:42:30.250 don't have constraints that are forcing them to move 00:42:30.250 --> 00:42:32.800 with respect to something, if there's no constraints, 00:42:32.800 --> 00:42:36.350 then you want to take the torques about G for sure. 00:42:40.190 --> 00:42:47.140 Good, all right, see you guys next Tuesday.