WEBVTT 00:00:00.120 --> 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 --> 00:00:06.060 Your support will help MIT OpenCourseWare 00:00:06.060 --> 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 --> 00:00:17.325 at ocw.mit.edu. 00:00:22.630 --> 00:00:23.800 PROFESSOR: Right. 00:00:23.800 --> 00:00:24.960 And we've done some review. 00:00:36.090 --> 00:00:37.130 Something else. 00:00:37.130 --> 00:00:38.046 AUDIENCE: [INAUDIBLE]. 00:00:42.880 --> 00:00:50.500 PROFESSOR: Yeah, symmetry, review principal axes ideas. 00:00:55.910 --> 00:00:56.420 What else? 00:01:07.310 --> 00:01:09.140 I think we spent quite a bit of time 00:01:09.140 --> 00:01:13.825 on generalized forces and ways for computing them. 00:01:39.270 --> 00:01:41.920 All right. 00:01:41.920 --> 00:01:46.069 And I'm trying to begin to summarize what we've 00:01:46.069 --> 00:01:47.610 done up to this point because we have 00:01:47.610 --> 00:01:49.730 a quiz coming next Tuesday. 00:01:49.730 --> 00:01:52.830 So some of yesterday's lecture was 00:01:52.830 --> 00:01:54.690 intended to kind of start pulling together, 00:01:54.690 --> 00:01:58.060 comparing direct versus Lagrange, advantages, 00:01:58.060 --> 00:01:59.900 disadvantages. 00:01:59.900 --> 00:02:03.360 And on Tuesday next, we'll do more examples. 00:02:03.360 --> 00:02:05.550 OK, so the problem you're going to work on today-- 00:02:05.550 --> 00:02:07.090 and you really are going to work in groups. 00:02:07.090 --> 00:02:08.298 And how many do we have here? 00:02:08.298 --> 00:02:13.380 We've got four-- about enough for three groups, let's say. 00:02:13.380 --> 00:02:17.170 We have a double pendulum. 00:02:17.170 --> 00:02:20.440 It's made out of two-- now I have two pieces of chalk. 00:02:24.440 --> 00:02:28.110 It's two slender rods, but an approximation-- this 00:02:28.110 --> 00:02:29.110 is a double pendulum. 00:02:29.110 --> 00:02:33.100 They have many-- that one and this one have a lot in common. 00:02:33.100 --> 00:02:36.755 Takes how many coordinates to completely describe the motion? 00:02:39.670 --> 00:02:41.450 One for each rod, right? 00:02:41.450 --> 00:02:44.360 And the coordinates up there that are drawn in that diagram 00:02:44.360 --> 00:02:47.640 is the angle of the first one with the vertical 00:02:47.640 --> 00:02:49.640 and the angle of the second one with a vertical. 00:02:49.640 --> 00:02:51.570 And with that, you can completely 00:02:51.570 --> 00:02:54.190 describe any allowable motion of the system. 00:02:54.190 --> 00:02:56.755 So a double pendulum has a property 00:02:56.755 --> 00:03:01.100 that it's got two natural frequencies and two mode 00:03:01.100 --> 00:03:01.600 shapes. 00:03:01.600 --> 00:03:04.160 That's the shape of the first mode. 00:03:04.160 --> 00:03:07.260 Both masses go in the same direction. 00:03:07.260 --> 00:03:10.990 And the second mode looks like that. 00:03:10.990 --> 00:03:14.080 The two masses go in opposite directions. 00:03:14.080 --> 00:03:17.532 Not with equal amplitudes, but opposite directions. 00:03:17.532 --> 00:03:18.990 And it's a different frequency too. 00:03:18.990 --> 00:03:20.030 It's higher. 00:03:20.030 --> 00:03:22.670 So it has two natural frequencies, two mode shapes. 00:03:22.670 --> 00:03:25.400 And if you had-- in this case it shows 00:03:25.400 --> 00:03:29.610 a force acting on the system, pushing this thing 00:03:29.610 --> 00:03:30.960 back and forth. 00:03:30.960 --> 00:03:37.800 And we're interested in the generalized forces, 00:03:37.800 --> 00:03:40.140 or thinking in terms of the work done 00:03:40.140 --> 00:03:41.880 by the non-conservative forces. 00:03:41.880 --> 00:03:49.750 So the exercise to do in groups-- like four or five 00:03:49.750 --> 00:03:50.970 of you are a group. 00:03:50.970 --> 00:03:52.270 You five are a group. 00:03:52.270 --> 00:03:57.840 And you two, four, five are a group there. 00:03:57.840 --> 00:03:59.080 Work this out. 00:03:59.080 --> 00:04:00.650 Find this vector. 00:04:00.650 --> 00:04:03.400 This is the position vector to the point 00:04:03.400 --> 00:04:06.370 of application of the force. 00:04:06.370 --> 00:04:08.520 And we want you to do that. 00:04:08.520 --> 00:04:10.170 Actually, I did something last time, 00:04:10.170 --> 00:04:12.780 and I'm almost forgetting to do it. 00:04:12.780 --> 00:04:14.880 I want to remind you of something. 00:04:14.880 --> 00:04:17.600 Professor Gossard, each week for the recitations-- 00:04:17.600 --> 00:04:21.430 he teaches Thursday ones-- does a write-up of the recitation. 00:04:21.430 --> 00:04:22.660 And they're posted. 00:04:22.660 --> 00:04:25.990 So this week the solution to this problem is posted. 00:04:25.990 --> 00:04:29.510 But also, a little quick review of the important stuff. 00:04:29.510 --> 00:04:30.980 So this would be posted. 00:04:30.980 --> 00:04:34.850 And basically I've redrawn it up here on the board. 00:04:34.850 --> 00:04:37.820 So this is this kinematic method for obtaining 00:04:37.820 --> 00:04:39.430 generalized forces. 00:04:39.430 --> 00:04:45.120 So a body with N forces on it. 00:04:45.120 --> 00:04:48.480 Here is the ith one. 00:04:48.480 --> 00:04:52.590 At each point of application of the force, 00:04:52.590 --> 00:04:56.850 there is a total virtual displacement 00:04:56.850 --> 00:04:58.820 that basically comes from the sum 00:04:58.820 --> 00:05:01.240 of all the individual generalized 00:05:01.240 --> 00:05:03.480 coordinate virtual motions. 00:05:03.480 --> 00:05:06.100 You sum them up, you get the total. 00:05:06.100 --> 00:05:11.960 And so therefore, at the point of application of each force, 00:05:11.960 --> 00:05:16.270 there is a total amount of virtual work that's done. 00:05:16.270 --> 00:05:18.780 So the total non-conservative work 00:05:18.780 --> 00:05:23.870 is the sum of the virtual work done 00:05:23.870 --> 00:05:27.710 at the point of application of every force summed over 00:05:27.710 --> 00:05:29.880 all the forces. 00:05:29.880 --> 00:05:32.820 And remember, inside of here, the total displacement 00:05:32.820 --> 00:05:36.450 at every location is a summation over of all the generalized 00:05:36.450 --> 00:05:37.300 coordinates. 00:05:37.300 --> 00:05:41.030 So it really ends up as a double sum, this dot product, 00:05:41.030 --> 00:05:43.960 the forces times delta ri. 00:05:43.960 --> 00:05:48.440 It's a double summation over the forces 00:05:48.440 --> 00:05:53.410 that are applied and over the generalized coordinates. 00:05:53.410 --> 00:05:56.850 But at the end of the day, what you end up doing, 00:05:56.850 --> 00:05:58.450 the calculation you want to do, is 00:05:58.450 --> 00:06:03.630 you need to know the generalized force-- the force associated 00:06:03.630 --> 00:06:05.970 with each generalized coordinate. 00:06:05.970 --> 00:06:10.240 And that's the summation of the contribution 00:06:10.240 --> 00:06:18.080 of the virtual work caused by that coordinate's motion dotted 00:06:18.080 --> 00:06:21.540 into the force at every location. 00:06:21.540 --> 00:06:24.730 So at every location, there's a little contribution 00:06:24.730 --> 00:06:27.500 to the total work done. 00:06:27.500 --> 00:06:32.190 And therefore, the total generalized force is this sum. 00:06:35.115 --> 00:06:40.570 So today's problem is easy in the sense that i is 1. 00:06:40.570 --> 00:06:43.140 You only have to deal with one force. 00:06:43.140 --> 00:06:44.980 But this is a little messier problem 00:06:44.980 --> 00:06:47.660 in that in order to do this, you need 00:06:47.660 --> 00:06:49.840 to be able-- this is the r sub i in this case. 00:06:49.840 --> 00:06:55.010 Here's point P. I want you to first, just in groups, 00:06:55.010 --> 00:07:00.180 work out what this vector is, rP, in the system as drawn. 00:07:00.180 --> 00:07:03.530 Here's the inertial system xy, generalized 00:07:03.530 --> 00:07:05.610 coordinates theta 1, theta 2. 00:07:05.610 --> 00:07:10.840 And this bar is L1 long, and this bar is L2 long. 00:07:10.840 --> 00:07:14.280 And come up with an answer in a form like this, something 00:07:14.280 --> 00:07:18.181 in the i hat direction plus something in the j hat 00:07:18.181 --> 00:07:18.680 direction. 00:07:18.680 --> 00:07:20.810 And then as your group finishes-- this is group 00:07:20.810 --> 00:07:22.830 one here-- come put your answer up. 00:07:22.830 --> 00:07:24.390 Second group, put your answer up. 00:07:24.390 --> 00:07:25.730 Third group, put your answer up. 00:07:25.730 --> 00:07:26.820 And we'll move on. 00:07:26.820 --> 00:07:29.628 This will go pretty fast. 00:07:29.628 --> 00:07:31.083 All right, looks like everybody's 00:07:31.083 --> 00:07:31.550 pretty much in agreement. 00:07:31.550 --> 00:07:32.508 Not much to talk about. 00:07:32.508 --> 00:07:33.900 So let's do the next piece. 00:07:33.900 --> 00:07:39.310 So the next part is to compute your variations, 00:07:39.310 --> 00:07:43.270 this part of the calculation, for each of the coordinates. 00:07:43.270 --> 00:07:53.500 So do rP with respect to theta 1 and rP with respect to theta 2 00:07:53.500 --> 00:07:56.280 and write down your answers. 00:08:00.770 --> 00:08:06.375 So the last step here is get q theta 1 q theta 2. 00:08:12.810 --> 00:08:15.530 So move on to getting the two generalized forces now. 00:08:20.590 --> 00:08:21.506 AUDIENCE: [INAUDIBLE]. 00:08:25.720 --> 00:08:26.595 PROFESSOR: You could. 00:08:26.595 --> 00:08:28.840 And then if you did, then this would 00:08:28.840 --> 00:08:32.419 have to be a summation over two components. 00:08:32.419 --> 00:08:33.780 Remember, this is a sum. 00:08:33.780 --> 00:08:35.690 It just happens in this example, we only 00:08:35.690 --> 00:08:38.171 have i goes from 1 to 1. 00:08:38.171 --> 00:08:42.400 But if there were two forces, then you would do this twice. 00:08:42.400 --> 00:08:47.480 You'd have an r1 with respect to theta 1 and an r2 with respect 00:08:47.480 --> 00:08:49.460 to theta 1. 00:08:49.460 --> 00:08:53.520 And you'd do this product twice, add the two pieces together, 00:08:53.520 --> 00:08:57.220 to get the total-- the total generalized force is 00:08:57.220 --> 00:09:00.950 the sum of the bits that come from all 00:09:00.950 --> 00:09:03.368 of the individual forces. 00:09:03.368 --> 00:09:05.813 AUDIENCE: So if [INAUDIBLE] just doing theta 1, 00:09:05.813 --> 00:09:08.747 but [INAUDIBLE] two forces, you just do one force, 00:09:08.747 --> 00:09:10.710 and then you do [INAUDIBLE] force-- 00:09:10.710 --> 00:09:12.450 PROFESSOR: Yeah, so now here let's 00:09:12.450 --> 00:09:17.550 have a B. We'll call it just B force. 00:09:17.550 --> 00:09:22.760 And it has to Bx components and By components. 00:09:22.760 --> 00:09:28.172 You would now also have to compute what would be rB. 00:09:28.172 --> 00:09:29.350 You would find it. 00:09:29.350 --> 00:09:32.980 Then you'd have rB would be this vector, which 00:09:32.980 --> 00:09:37.910 is just that piece plus this piece, right? 00:09:37.910 --> 00:09:42.250 So you'd have just those-- rB would be this plus this, 00:09:42.250 --> 00:09:44.370 and then you would do the derivative of rB 00:09:44.370 --> 00:09:45.830 with respect to theta 1. 00:09:45.830 --> 00:09:46.930 And you'd get something. 00:09:46.930 --> 00:09:50.152 And derivative of rB with respect to theta 2. 00:09:50.152 --> 00:09:54.232 AUDIENCE: So then you'd just do the rB equals [INAUDIBLE] 00:09:54.232 --> 00:09:58.224 for one force, and then you do the rPe [INAUDIBLE]? 00:09:58.224 --> 00:09:58.890 PROFESSOR: Yeah. 00:09:58.890 --> 00:10:01.650 So you have two-- so this thing would end up 00:10:01.650 --> 00:10:03.870 being the summation of two contributions, right? 00:10:03.870 --> 00:10:16.710 This would look like an FB dot rB with respect to-- 00:10:16.710 --> 00:10:19.400 and this is just for just one of them. 00:10:19.400 --> 00:10:21.250 So we'll do the theta 1. 00:10:21.250 --> 00:10:27.165 With respect to theta 1 plus F-- I'm 00:10:27.165 --> 00:10:35.350 just calling this one F-- dot derivative rP with respect 00:10:35.350 --> 00:10:37.780 to theta 1. 00:10:37.780 --> 00:10:42.920 So you get two contributions from the two forces 00:10:42.920 --> 00:10:45.110 give you the total q. 00:10:45.110 --> 00:10:47.070 This would be q theta 1. 00:10:47.070 --> 00:10:49.920 And then you'd do it again for the theta 2. 00:10:49.920 --> 00:10:51.715 And you'd have two possible pieces. 00:10:58.882 --> 00:11:02.900 AUDIENCE: So you try to find r's for every force [INAUDIBLE]. 00:11:02.900 --> 00:11:05.420 PROFESSOR: At every point of application of a force, 00:11:05.420 --> 00:11:07.820 you define an r. 00:11:07.820 --> 00:11:12.875 You eventually need to do this summation for every force. 00:11:17.320 --> 00:11:17.820 All right. 00:11:17.820 --> 00:11:20.230 Everybody's got the same answers. 00:11:20.230 --> 00:11:21.790 So now I'm going to try to-- this 00:11:21.790 --> 00:11:24.670 is an indirect way of answering your question. 00:11:24.670 --> 00:11:28.910 I'm now going to try to convince you that you're wrong. 00:11:28.910 --> 00:11:33.172 And I want you to tell me why my logic is flawed. 00:11:33.172 --> 00:11:36.120 AUDIENCE: No. 00:11:36.120 --> 00:11:37.750 PROFESSOR: OK? 00:11:37.750 --> 00:11:44.520 So I look at these answers here, and so this 00:11:44.520 --> 00:11:49.940 says that if-- this line right here, if I caused 00:11:49.940 --> 00:11:57.190 a little delta theta 1 here, that swings this arm 00:11:57.190 --> 00:11:59.360 through that amount. 00:11:59.360 --> 00:12:06.540 And it moves a distance in the x direction, L1 cosine theta 00:12:06.540 --> 00:12:07.470 delta theta 1. 00:12:07.470 --> 00:12:10.110 And it moves over that little bit right here, right? 00:12:10.110 --> 00:12:10.730 OK. 00:12:10.730 --> 00:12:14.560 And you're telling me that this point over here 00:12:14.560 --> 00:12:16.900 moves that amount. 00:12:16.900 --> 00:12:21.030 The point of application of this force 00:12:21.030 --> 00:12:27.980 due to delta x, delta theta 1, is this delta theta 1. 00:12:27.980 --> 00:12:30.770 There's no L2 involved. 00:12:30.770 --> 00:12:37.600 So to me, it seems like here's this pendulum. 00:12:37.600 --> 00:12:44.150 And if I move this first piece by a little bit, delta theta 1, 00:12:44.150 --> 00:12:46.280 this piece also moves. 00:12:46.280 --> 00:12:48.370 And so if you just think of a straight piece, 00:12:48.370 --> 00:12:51.200 delta theta 1 times L1 moves a little bit. 00:12:51.200 --> 00:12:55.050 But down here it moves like twice as much. 00:12:55.050 --> 00:12:59.390 So I think that when you move this little delta theta 00:12:59.390 --> 00:13:03.880 one, that you ought to get even more motion down here. 00:13:03.880 --> 00:13:08.330 And yet your answers say that that's not true. 00:13:08.330 --> 00:13:13.096 Tell me why-- what's flawed in my reasoning? 00:13:13.096 --> 00:13:15.576 AUDIENCE: So when you-- I'll try. 00:13:15.576 --> 00:13:19.874 When you [INAUDIBLE] theta 2, [INAUDIBLE]. 00:13:19.874 --> 00:13:21.770 And now when you move by theta 1, 00:13:21.770 --> 00:13:25.090 [INAUDIBLE] all points move the same theta. 00:13:25.090 --> 00:13:26.232 So even though [INAUDIBLE]. 00:13:31.050 --> 00:13:31.790 PROFESSOR: OK. 00:13:31.790 --> 00:13:35.220 So you're saying that we argue that when 00:13:35.220 --> 00:13:39.890 you have one virtual displacement, or in this case 00:13:39.890 --> 00:13:46.630 a rotation delta theta 1, we freeze all other generalized 00:13:46.630 --> 00:13:47.530 coordinates. 00:13:47.530 --> 00:13:51.970 That means we freeze delta theta 2 and don't let it change. 00:13:51.970 --> 00:13:55.420 Delta theta 2 is measured with respect to the vertical, right? 00:13:55.420 --> 00:13:59.090 So if that moves over delta theta 1, 00:13:59.090 --> 00:14:01.500 this whole thing would-- this angle here 00:14:01.500 --> 00:14:04.980 we go from delta theta 2 to delta theta 2 plus delta theta 00:14:04.980 --> 00:14:08.630 1 if my argument's correct. 00:14:08.630 --> 00:14:12.160 But for my argument to be correct, 00:14:12.160 --> 00:14:14.730 that would have to swing by delta theta 1 over here 00:14:14.730 --> 00:14:15.530 as well. 00:14:15.530 --> 00:14:19.990 But the fact that we freeze theta 2, in fact, 00:14:19.990 --> 00:14:24.720 the angle between these two things has to change. 00:14:24.720 --> 00:14:26.340 This has to open up a little bit. 00:14:26.340 --> 00:14:27.740 So when this is moving over, this 00:14:27.740 --> 00:14:30.670 is dropping down so that this point, this whole body 00:14:30.670 --> 00:14:31.620 only translates. 00:14:31.620 --> 00:14:34.890 So that whoever described it as just pure translation 00:14:34.890 --> 00:14:38.310 of the second body, that's another way of saying it. 00:14:38.310 --> 00:14:41.570 So you've got to remember what it means, 00:14:41.570 --> 00:14:43.940 freezing all the other generalized coordinates 00:14:43.940 --> 00:14:46.920 and only allowing the one you picked to move. 00:14:46.920 --> 00:14:48.000 Good. 00:14:48.000 --> 00:14:48.660 All right. 00:14:48.660 --> 00:14:50.520 Nice work. 00:14:50.520 --> 00:14:52.190 We've got some time left. 00:14:52.190 --> 00:14:55.370 So do you have any questions from problem 00:14:55.370 --> 00:14:57.950 sets, lectures, just-- 00:14:57.950 --> 00:14:59.450 AUDIENCE: [INAUDIBLE]. 00:14:59.450 --> 00:15:01.575 PROFESSOR: Yeah. 00:15:01.575 --> 00:15:02.950 AUDIENCE: When you're calculating 00:15:02.950 --> 00:15:04.450 kinetic energy of a rotating body, 00:15:04.450 --> 00:15:05.890 there's an omega [INAUDIBLE]. 00:15:09.610 --> 00:15:11.750 In the homework, the last homework, 00:15:11.750 --> 00:15:13.690 I didn't know which omega could take, 00:15:13.690 --> 00:15:16.200 whether to take it from the wheel or from the axis. 00:15:16.200 --> 00:15:18.417 PROFESSOR: Oh, this is the rotating thing with the-- 00:15:18.417 --> 00:15:19.000 AUDIENCE: Yes. 00:15:19.000 --> 00:15:19.700 PROFESSOR: Ah, OK. 00:15:19.700 --> 00:15:21.158 AUDIENCE: There's two omegas there, 00:15:21.158 --> 00:15:23.870 and I didn't know which omega [INAUDIBLE]. 00:15:23.870 --> 00:15:27.890 PROFESSOR: So should we just talk about that problem? 00:15:27.890 --> 00:15:32.830 Are other people-- this is the one where you had basically 00:15:32.830 --> 00:15:36.120 the rod, and on the end of the rod 00:15:36.120 --> 00:15:39.350 was the wheel, which was rotating. 00:15:39.350 --> 00:15:44.940 So this was going around at some omega 1 in the z, right? 00:15:44.940 --> 00:15:47.750 And this was rotating, and I think 00:15:47.750 --> 00:15:55.550 this was like z, maybe x and y. 00:15:55.550 --> 00:15:56.800 So it was rotating about that. 00:15:56.800 --> 00:15:58.632 AUDIENCE: [INAUDIBLE] it had a negative. 00:15:58.632 --> 00:16:00.010 It was a negative [INAUDIBLE]. 00:16:00.010 --> 00:16:01.301 PROFESSOR: Oh, it was a-- yeah. 00:16:01.301 --> 00:16:06.350 Actually, I think I have it here with its correct definition. 00:16:06.350 --> 00:16:08.440 So rather than try to-- 00:16:08.440 --> 00:16:11.821 AUDIENCE: Yeah, because there was the positive [INAUDIBLE], 00:16:11.821 --> 00:16:13.742 and then there's the [INAUDIBLE]. 00:16:13.742 --> 00:16:14.658 PROFESSOR: Here it is. 00:16:14.658 --> 00:16:16.860 Let's get the picture right. 00:16:29.940 --> 00:16:37.930 And there was an x, a y, and a z. 00:16:37.930 --> 00:16:41.205 And this was cap omega. 00:16:44.170 --> 00:16:47.670 And this was omega 1. 00:16:47.670 --> 00:16:51.210 And that's minus, in minus direction, 00:16:51.210 --> 00:16:53.000 for this system, right? 00:16:53.000 --> 00:16:55.560 OK. 00:16:55.560 --> 00:17:05.670 So then we're trying to compute T for this system. 00:17:09.210 --> 00:17:12.829 1/2, and what'd we call for the masses here? 00:17:12.829 --> 00:17:17.710 This was m1, and this was m2, I guess. 00:17:17.710 --> 00:17:19.140 Yeah. 00:17:19.140 --> 00:17:25.280 So this T for 2-- T1 you don't have a problem with, right? 00:17:25.280 --> 00:17:28.820 T1 is just a shaft, just a rod. 00:17:28.820 --> 00:17:34.100 It's pivoting about its end 1/2i about-- if this 00:17:34.100 --> 00:17:36.930 is-- I call it o. 00:17:36.930 --> 00:17:40.510 1/2i, i with respect to o, mL squared 00:17:40.510 --> 00:17:44.760 over 3 times that squared. 00:17:44.760 --> 00:17:45.370 You're done. 00:17:45.370 --> 00:17:47.200 That's the energy of this part. 00:17:47.200 --> 00:17:49.310 This is the problematic one, right? 00:17:49.310 --> 00:17:50.320 The piece of it. 00:17:50.320 --> 00:17:55.830 So 1/2, I would say mass 2. 00:17:55.830 --> 00:17:57.150 And this does have a point. 00:17:57.150 --> 00:18:00.110 They called this point B here. 00:18:00.110 --> 00:18:03.410 Velocity of B dot velocity of B. 00:18:03.410 --> 00:18:06.600 So that's the 1/2mv squared piece of it. 00:18:06.600 --> 00:18:21.240 Plus 1/2 omega dot h with respect to G 00:18:21.240 --> 00:18:26.270 for-- this is mass 2. 00:18:26.270 --> 00:18:28.460 So you have a contribution to the kinetic energy 00:18:28.460 --> 00:18:33.700 of this that comes from its center of gravity translating, 00:18:33.700 --> 00:18:38.320 and a contribution from the rotor rotating with respect 00:18:38.320 --> 00:18:42.260 to its center of mass here, G here. 00:18:45.610 --> 00:18:47.170 And that'd be the expression you use. 00:18:47.170 --> 00:18:56.420 And then the problem is, what omega is the trick here, right? 00:18:56.420 --> 00:18:57.050 OK. 00:18:57.050 --> 00:19:00.000 This is the one that had the concept question I talked 00:19:00.000 --> 00:19:01.730 about in class because it came up, 00:19:01.730 --> 00:19:05.630 and I hadn't thought about this, about the body 00:19:05.630 --> 00:19:10.660 coordinates attached to this are actually rotating with this. 00:19:10.660 --> 00:19:15.326 And that gets messy trying to figure out 00:19:15.326 --> 00:19:16.700 what are the-- how do you break-- 00:19:16.700 --> 00:19:19.810 you're supposed to express the angular 00:19:19.810 --> 00:19:21.800 rotation in the body coordinates. 00:19:27.500 --> 00:19:30.960 But this angular rotation, this one 00:19:30.960 --> 00:19:36.960 actually is lined up with this axis. 00:19:36.960 --> 00:19:39.250 And you could have body coordinates on here. 00:19:39.250 --> 00:19:45.420 You could have a-- if this is a y, 00:19:45.420 --> 00:19:55.310 you could have a body coordinate x and z that rotate with it. 00:19:55.310 --> 00:20:02.450 But this Component would just be still along this axis, right? 00:20:02.450 --> 00:20:10.170 So the concept question [INAUDIBLE] 00:20:10.170 --> 00:20:15.700 was basically, are these body coordinates, x, y, z, 00:20:15.700 --> 00:20:18.770 attached to the rod? 00:20:18.770 --> 00:20:23.040 Principal axes for this body. 00:20:23.040 --> 00:20:28.650 And the complication here is that this body is rotating. 00:20:28.650 --> 00:20:34.570 And so this x and this z don't rotate with that body. 00:20:34.570 --> 00:20:39.620 And so they're not body fixed coordinates for that body. 00:20:39.620 --> 00:20:42.980 But in fact, they're still principal coordinates 00:20:42.980 --> 00:20:44.070 for that body. 00:20:44.070 --> 00:20:45.350 Why is that true? 00:20:48.284 --> 00:20:51.710 AUDIENCE: Because you have [INAUDIBLE]. 00:20:51.710 --> 00:20:55.601 And you can take whatever you want for the [INAUDIBLE]. 00:20:55.601 --> 00:20:56.600 PROFESSOR: You're close. 00:20:56.600 --> 00:21:01.930 This body is axially symmetric. 00:21:01.930 --> 00:21:05.330 And so at any instant in time, you 00:21:05.330 --> 00:21:07.900 could say that-- you could define 00:21:07.900 --> 00:21:10.640 a principal coordinate on this body that 00:21:10.640 --> 00:21:12.240 is lined up with these. 00:21:16.520 --> 00:21:19.484 So for axially symmetric bodies, you could do that. 00:21:19.484 --> 00:21:20.150 You can do that. 00:21:20.150 --> 00:21:21.858 You can just say, at an incident in time, 00:21:21.858 --> 00:21:24.620 let's compute the kinetic energy. 00:21:24.620 --> 00:21:30.140 And we have the rotation of that body 00:21:30.140 --> 00:21:34.970 defined in the instantaneous principal coordinates 00:21:34.970 --> 00:21:36.290 of the body. 00:21:36.290 --> 00:21:43.540 So now omega for the second body is 00:21:43.540 --> 00:21:58.050 this in the k minus omega 1 in the j of the x, y, z system 00:21:58.050 --> 00:22:01.680 rotating with the rod. 00:22:01.680 --> 00:22:06.870 And then to complete this, what's H? 00:22:14.130 --> 00:22:30.650 So H with respect to G is Ixx Iyy Izz times-- 00:22:30.650 --> 00:22:38.570 and now this one's rotation is 0 minus omega 1 and cap omega. 00:22:42.570 --> 00:22:46.610 And I multiply that times this with the definition 00:22:46.610 --> 00:22:49.640 that the first one gives me the I, second one J. 00:22:49.640 --> 00:22:52.990 And so I end up with two pieces. 00:22:52.990 --> 00:22:54.730 The first one is zero. 00:22:54.730 --> 00:22:57.150 I get an Iyy. 00:22:57.150 --> 00:22:59.240 And remember, these are defined with respect 00:22:59.240 --> 00:23:03.130 to center of mass of the disk. 00:23:03.130 --> 00:23:17.200 Iyy minus Iyy omega 1 j plus Izz cap omega k. 00:23:17.200 --> 00:23:20.340 And that'd be the angular momentum for the disk. 00:23:23.120 --> 00:23:23.890 Right? 00:23:23.890 --> 00:23:34.400 And then the kinetic energy, the 1/2 omega dot H, 00:23:34.400 --> 00:23:41.110 you now just have another omega 1/2, and you have a vector, 00:23:41.110 --> 00:23:49.530 0 minus omega 1 J cap omega k dotted 00:23:49.530 --> 00:24:06.080 with the H vector, which is your 0 Iyy omega 1 00:24:06.080 --> 00:24:10.540 j and Izz cap omega k. 00:24:10.540 --> 00:24:13.359 AUDIENCE: I have a question. [INAUDIBLE]. 00:24:13.359 --> 00:24:14.025 PROFESSOR: Yeah. 00:24:14.025 --> 00:24:18.610 And if that one's negative, then it'll 00:24:18.610 --> 00:24:20.826 happily fix this one because it ought 00:24:20.826 --> 00:24:23.300 to come out looking like omega squareds, right? 00:24:23.300 --> 00:24:32.500 And so out of this, you get Iyy omega 1 squared plus Izz cap 00:24:32.500 --> 00:24:33.425 omega squared. 00:24:48.370 --> 00:24:54.440 And that's the second piece, and you add it to this piece. 00:24:54.440 --> 00:24:57.860 AUDIENCE: So is there ever a time where the omega's inside 00:24:57.860 --> 00:24:58.360 [INAUDIBLE]. 00:25:19.200 --> 00:25:37.820 PROFESSOR: So let me-- we have a total-- so let's just 00:25:37.820 --> 00:25:48.846 talk-- we have H with respect to G. 00:25:48.846 --> 00:25:50.990 Or a slightly simpler example. 00:26:28.770 --> 00:26:32.490 So H for this system with respect to the center 00:26:32.490 --> 00:26:39.720 is H with respect to G plus rG with respect 00:26:39.720 --> 00:26:44.740 to the center cross PG, right? 00:26:44.740 --> 00:26:47.720 I can write the angular momentum that way. 00:26:47.720 --> 00:26:52.950 And first piece we know. 00:26:52.950 --> 00:27:01.060 The second piece, rG, is in a r hat direction, if you will. 00:27:01.060 --> 00:27:06.300 The P is going around the circle. 00:27:06.300 --> 00:27:09.630 And so the cross product of those two 00:27:09.630 --> 00:27:13.280 gives me a component, what's the direction of the result. 00:27:13.280 --> 00:27:21.110 So this is r cross theta hat. 00:27:21.110 --> 00:27:22.850 You get a k. 00:27:22.850 --> 00:27:26.150 This ends up being-- this gives you a k component. 00:27:26.150 --> 00:27:31.590 And this one we write out. 00:27:31.590 --> 00:27:48.350 So this is minus Iyy omega 1 j plus Izz cap omega k 00:27:48.350 --> 00:27:54.080 plus some stuff that's in the k. 00:27:54.080 --> 00:27:56.400 Agreed? 00:27:56.400 --> 00:28:02.680 And I want to know the torques required to make 00:28:02.680 --> 00:28:05.750 this system do what it's doing. 00:28:05.750 --> 00:28:08.770 How do I get the torques required 00:28:08.770 --> 00:28:11.620 to make the system go around? 00:28:15.412 --> 00:28:16.644 AUDIENCE: [INAUDIBLE]. 00:28:16.644 --> 00:28:17.310 PROFESSOR: Yeah. 00:28:17.310 --> 00:28:24.760 Sum of the external torques equals dH. 00:28:24.760 --> 00:28:28.540 Now, with respect-- remember, we would ordinarily usually write 00:28:28.540 --> 00:28:33.130 that as an A. We'll call it H with respect to O dt 00:28:33.130 --> 00:28:36.910 plus velocity-- usually we call it A-- with respect to O, 00:28:36.910 --> 00:28:39.800 cross P. Right? 00:28:39.800 --> 00:28:41.590 In this system, what's the velocity 00:28:41.590 --> 00:28:43.140 of this piece, this term? 00:28:43.140 --> 00:28:43.640 Zero. 00:28:43.640 --> 00:28:45.690 So we don't have to worry about it. 00:28:45.690 --> 00:28:47.740 So the sum of the external torques 00:28:47.740 --> 00:28:52.320 should just be the time derivative of this angular 00:28:52.320 --> 00:28:55.500 momentum vector. 00:28:55.500 --> 00:28:57.400 Right? 00:28:57.400 --> 00:29:12.810 And dH then O dt is equal to the partial derivative with respect 00:29:12.810 --> 00:29:16.705 to t of H. And I'll write it like this. 00:29:16.705 --> 00:29:18.365 This is in the rotating frame. 00:29:22.700 --> 00:29:27.040 So this is the piece from inside the frame plus, 00:29:27.040 --> 00:29:29.900 this is just the derivative of a rotating vector. 00:29:29.900 --> 00:29:40.060 Omega cross H. And the issue here is, what's the omega? 00:29:40.060 --> 00:29:42.040 That's really what this question comes down to. 00:29:53.340 --> 00:29:55.450 When we did this problem, we found 00:29:55.450 --> 00:30:01.250 that H came from taking the omega, the vector, 00:30:01.250 --> 00:30:04.950 multiplying it by this, and we got these two pieces. 00:30:04.950 --> 00:30:08.930 And so it started off with components in the j and k 00:30:08.930 --> 00:30:11.920 direction, and it came out with components in the j and k 00:30:11.920 --> 00:30:15.510 because this was diagonal. 00:30:15.510 --> 00:30:17.260 So we have principal axes. 00:30:17.260 --> 00:30:19.860 We came out with these two pieces. 00:30:19.860 --> 00:30:22.360 So the angular momentum has a j component and a k component. 00:30:25.190 --> 00:30:27.350 What is actually-- now, this vector, 00:30:27.350 --> 00:30:29.740 this angular momentum vector, the reason 00:30:29.740 --> 00:30:34.480 we have this second term is because of the change 00:30:34.480 --> 00:30:36.610 of direction of a vector. 00:30:36.610 --> 00:30:40.270 This piece comes from taking the time derivatives of the unit 00:30:40.270 --> 00:30:41.880 vectors in the problem. 00:30:41.880 --> 00:30:44.190 The time derivatives of the other stuff in the problem 00:30:44.190 --> 00:30:46.110 has been taken care of by this. 00:30:46.110 --> 00:30:48.880 So this only deals with the time derivatives of the unit 00:30:48.880 --> 00:30:51.150 vectors. 00:30:51.150 --> 00:31:00.190 So what is the actual rotation rate 00:31:00.190 --> 00:31:05.070 in this problem, the rate at which things 00:31:05.070 --> 00:31:06.650 are changing direction? 00:31:06.650 --> 00:31:10.240 Which one's changing direction? 00:31:10.240 --> 00:31:11.406 AUDIENCE: [INAUDIBLE]. 00:31:11.406 --> 00:31:13.155 PROFESSOR: Is this one changing direction? 00:31:13.155 --> 00:31:13.580 No. 00:31:13.580 --> 00:31:15.450 So actually, it doesn't give you any derivative, right? 00:31:15.450 --> 00:31:16.810 Is this one changing direction? 00:31:16.810 --> 00:31:17.310 Yeah. 00:31:21.980 --> 00:31:33.410 And at what rotation rate is it changing direction? 00:31:33.410 --> 00:31:35.549 AUDIENCE: [INAUDIBLE]. 00:31:35.549 --> 00:31:36.590 PROFESSOR: Capital omega. 00:31:36.590 --> 00:31:44.810 But why not-- but it's not changing little omega, is it? 00:31:44.810 --> 00:31:49.180 The math, the vectors works that all out for you. 00:31:49.180 --> 00:31:53.920 Because this is in the j direction, and this 00:31:53.920 --> 00:31:55.030 is the same. 00:31:55.030 --> 00:31:59.540 You just use the same total rotation vector here. 00:31:59.540 --> 00:32:02.440 And this will have in it the zero, 00:32:02.440 --> 00:32:09.430 the omega minus omega 1 j, the plus cap omega k. 00:32:09.430 --> 00:32:13.450 And the parts that you don't care-- this 00:32:13.450 --> 00:32:16.290 crossed with itself goes to zero. 00:32:16.290 --> 00:32:21.390 So even though you're just going ahead and leaving this in here, 00:32:21.390 --> 00:32:26.400 it doesn't result in anything because the cross product 00:32:26.400 --> 00:32:28.940 would-- cross product with itself gives you nothing. 00:32:28.940 --> 00:32:32.100 So the only piece that actually contributes nonzero 00:32:32.100 --> 00:32:35.510 contribution to this answer is this crossed 00:32:35.510 --> 00:32:39.490 with the pieces in here. 00:32:39.490 --> 00:32:43.940 So when you're taking derivatives 00:32:43.940 --> 00:32:47.640 of rotating vectors, remember the original formula. 00:32:47.640 --> 00:32:50.290 And the original formula is the derivative 00:32:50.290 --> 00:32:53.330 of the vector in the rotating frame-- 00:32:53.330 --> 00:32:57.090 is this magnitude getting more or less-- 00:32:57.090 --> 00:33:00.869 plus just the rotation rate crossed 00:33:00.869 --> 00:33:01.910 with the original vector. 00:33:04.640 --> 00:33:07.670 And the pieces that are common just fall out. 00:33:07.670 --> 00:33:12.600 The piece of this that is due to its own rotation 00:33:12.600 --> 00:33:13.780 doesn't enter into it. 00:33:13.780 --> 00:33:17.840 This rotation doesn't contribute to that d 00:33:17.840 --> 00:33:19.205 by dt of the rotation. 00:33:21.792 --> 00:33:22.750 Just works out for you. 00:33:22.750 --> 00:33:23.290 Yeah. 00:33:23.290 --> 00:33:26.194 AUDIENCE: What would happen if that just [INAUDIBLE] 00:33:26.194 --> 00:33:29.098 rotates like this, but also rotates on its own axis 00:33:29.098 --> 00:33:31.520 as it does it? 00:33:31.520 --> 00:33:34.092 PROFESSOR: You mean if this one is going in the k? 00:33:34.092 --> 00:33:35.940 AUDIENCE: Yeah. 00:33:35.940 --> 00:33:37.682 PROFESSOR: Then-- 00:33:37.682 --> 00:33:42.056 AUDIENCE: [INAUDIBLE] you didn't have the big omega term 00:33:42.056 --> 00:33:45.570 because the actual thing isn't-- the big omega would be taken 00:33:45.570 --> 00:33:50.380 into account when looking at the first term of kinetic energy. 00:33:50.380 --> 00:33:55.720 PROFESSOR: If this one also had some k rotation in addition 00:33:55.720 --> 00:34:02.500 relative to here, we'll call it omega 2 in the k direction, 00:34:02.500 --> 00:34:05.840 then what's the total rotation rate in the k 00:34:05.840 --> 00:34:07.422 direction for the disk? 00:34:07.422 --> 00:34:09.030 AUDIENCE: [INAUDIBLE]. 00:34:09.030 --> 00:34:12.440 PROFESSOR: You add cap omega, 2 omega 2, 00:34:12.440 --> 00:34:17.489 and you'd have its total k directed rotation, right? 00:34:17.489 --> 00:34:20.350 And if you did that, you'd end up in here 00:34:20.350 --> 00:34:26.330 with an omega plus omega-- cap omega plus omega 2 k. 00:34:29.480 --> 00:34:36.550 But when you came over to do the time derivative, 00:34:36.550 --> 00:34:38.917 you just leave all those things in there. 00:34:38.917 --> 00:34:40.500 And the ones that you don't care about 00:34:40.500 --> 00:34:42.110 will just cancel out because they're 00:34:42.110 --> 00:34:45.498 cross products with themselves. 00:34:45.498 --> 00:34:49.929 AUDIENCE: [INAUDIBLE] omega also appears [INAUDIBLE]. 00:34:49.929 --> 00:34:51.900 PROFESSOR: Oh yeah. 00:34:51.900 --> 00:34:54.800 Well, the cap omega appears in this piece. 00:34:58.040 --> 00:35:00.060 But this goes to k. 00:35:00.060 --> 00:35:01.784 We did a particular problem. 00:35:01.784 --> 00:35:03.450 I'm not trying to prove this in general, 00:35:03.450 --> 00:35:05.241 although there's probably a way to do that. 00:35:05.241 --> 00:35:10.110 This problem, this term is in the k direction. 00:35:10.110 --> 00:35:15.270 And therefore, k hat does not change direction with time. 00:35:15.270 --> 00:35:19.980 Therefore, the dH dt of this piece goes to zero. 00:35:19.980 --> 00:35:22.370 So we only had to deal with this part 00:35:22.370 --> 00:35:25.080 in taking the time derivative that dealt 00:35:25.080 --> 00:35:28.300 with changes in direction. 00:35:28.300 --> 00:35:29.830 It's only the changes in direction 00:35:29.830 --> 00:35:32.140 that the omega cross something matters, right? 00:35:36.790 --> 00:35:37.700 Good question. 00:35:37.700 --> 00:35:40.630 Something else? 00:35:40.630 --> 00:35:41.435 Recent problems. 00:35:48.440 --> 00:35:51.466 Somebody in the last class asked about, 00:35:51.466 --> 00:35:55.275 had a question about this thing. 00:35:55.275 --> 00:35:57.830 That bring anything to memory? 00:35:57.830 --> 00:35:59.690 AUDIENCE: [INAUDIBLE]. 00:35:59.690 --> 00:36:01.780 PROFESSOR: I see some grimaces. 00:36:01.780 --> 00:36:05.230 What troubled you about that problem? 00:36:05.230 --> 00:36:08.460 What troubled the last group about that problem was the 00:36:08.460 --> 00:36:09.210 posted solution. 00:36:11.890 --> 00:36:14.300 They didn't understand the posted solution. 00:36:14.300 --> 00:36:18.390 So I ran through-- I showed why the posted 00:36:18.390 --> 00:36:20.390 solution works the way it does. 00:36:20.390 --> 00:36:21.890 AUDIENCE: Well, one confusion that I 00:36:21.890 --> 00:36:26.043 had had doing this problem was that the spring for the wheel 00:36:26.043 --> 00:36:31.610 mass, when that exerts displacements, Newton's 00:36:31.610 --> 00:36:34.120 third being what it is, should it also 00:36:34.120 --> 00:36:36.165 exert some force on the large base mass 00:36:36.165 --> 00:36:38.680 as well, so you should take that into account 00:36:38.680 --> 00:36:40.910 for the total displacemenet caused by springs 00:36:40.910 --> 00:36:44.452 for big mass, the big-- 00:36:44.452 --> 00:36:46.660 PROFESSOR: Are you talking about not computing-- just 00:36:46.660 --> 00:36:49.215 giving equations of motion, or computing generalized forces, 00:36:49.215 --> 00:36:50.470 or what's the context? 00:36:50.470 --> 00:36:51.605 AUDIENCE: Yes, when you're computing 00:36:51.605 --> 00:36:52.670 your equations of motion, you have 00:36:52.670 --> 00:36:54.390 to take into account from the larger 00:36:54.390 --> 00:36:56.200 mass of the displacements in the spring, 00:36:56.200 --> 00:36:58.040 it's attached to the wall, [INAUDIBLE]. 00:36:58.040 --> 00:37:00.286 PROFESSOR: As well as the-- yeah. 00:37:00.286 --> 00:37:04.040 So if you're doing this by the direct method, 00:37:04.040 --> 00:37:06.820 you have to figure that out. 00:37:06.820 --> 00:37:08.370 So here's a way of doing it. 00:37:08.370 --> 00:37:11.050 This is a way that I think the-- the way the student 00:37:11.050 --> 00:37:12.830 just read off his paper. 00:37:12.830 --> 00:37:17.440 So I said, let's talk about the forces in the x direction 00:37:17.440 --> 00:37:19.690 on the main mass. 00:37:19.690 --> 00:37:23.410 Well, that's mass 1 times its acceleration 00:37:23.410 --> 00:37:25.940 must be equal to all the external forces. 00:37:25.940 --> 00:37:28.663 This is minus kx minus bx dot. 00:37:28.663 --> 00:37:34.590 This is minus the x component of the spring force 00:37:34.590 --> 00:37:37.840 minus the x component of the friction 00:37:37.840 --> 00:37:43.500 force from the wheel minus the x component of any normal force 00:37:43.500 --> 00:37:45.620 from the wheel. 00:37:45.620 --> 00:37:47.895 And I do the same thing for the second mass, M2. 00:37:50.570 --> 00:37:53.490 The total forces on it must be equal to its mass 00:37:53.490 --> 00:37:54.640 times its acceleration. 00:37:54.640 --> 00:37:56.390 Well, its acceleration I'm doing this just 00:37:56.390 --> 00:37:57.990 to get in the x direction. 00:37:57.990 --> 00:38:02.480 Its acceleration in the x is the main mass, x double dot, 00:38:02.480 --> 00:38:04.750 which it's sitting on. 00:38:04.750 --> 00:38:12.420 And the x component of the relative coordinate 00:38:12.420 --> 00:38:13.605 x1 double dot. 00:38:13.605 --> 00:38:16.320 This is the total acceleration in the x direction 00:38:16.320 --> 00:38:18.030 of the second mass. 00:38:18.030 --> 00:38:20.440 And the total forces on that second mass 00:38:20.440 --> 00:38:23.450 are-- signs are changed. 00:38:23.450 --> 00:38:26.460 Internal force, internal force, internal force, 00:38:26.460 --> 00:38:29.820 plus the force that was applied. 00:38:29.820 --> 00:38:31.520 Actually, here it is written out. 00:38:31.520 --> 00:38:34.910 Here is that external force that was applied to that wheel. 00:38:34.910 --> 00:38:39.690 If you take these two equations, you just add them together. 00:38:39.690 --> 00:38:43.920 All of the internal forces drop out. 00:38:43.920 --> 00:38:46.480 Plus fx minus fx. 00:38:46.480 --> 00:38:47.610 All of those drop out. 00:38:47.610 --> 00:38:50.750 And you end up with an expression that says, 00:38:50.750 --> 00:38:53.620 this is a system. 00:38:53.620 --> 00:38:58.330 The total external forces on the system in the x direction 00:38:58.330 --> 00:39:03.710 must be equal to the total mass of the system 00:39:03.710 --> 00:39:07.620 times the acceleration of its center of mass. 00:39:07.620 --> 00:39:08.120 Right? 00:39:08.120 --> 00:39:10.050 That's what Newton said. 00:39:10.050 --> 00:39:14.290 Well, the acceleration of the center of mass 00:39:14.290 --> 00:39:18.030 is the sum of the individual pieces 00:39:18.030 --> 00:39:21.040 times their individual accelerations. 00:39:21.040 --> 00:39:26.790 The sum of the total mass times the acceleration 00:39:26.790 --> 00:39:28.960 of the center of mass is equal to the sum 00:39:28.960 --> 00:39:32.430 of the individual masses times their individual accelerations. 00:39:32.430 --> 00:39:34.700 Mass one, its acceleration. 00:39:34.700 --> 00:39:39.090 Mass two, its acceleration due to its local coordinate 00:39:39.090 --> 00:39:43.540 plus its acceleration due to the main mass acceleration. 00:39:43.540 --> 00:39:46.060 And all of that's got to be equal to the external forces, 00:39:46.060 --> 00:39:53.100 which are now only minus kx, minus bx dot, and plus F. 00:39:53.100 --> 00:39:55.820 And this term here, this stuff, all that, 00:39:55.820 --> 00:40:00.260 including this, these three M X double dot terms, 00:40:00.260 --> 00:40:01.820 all added together as the same thing 00:40:01.820 --> 00:40:05.020 is the total mass times the acceleration of the center 00:40:05.020 --> 00:40:06.250 of mass, which is somewhere. 00:40:08.900 --> 00:40:13.510 But that's a completely legitimate equation 00:40:13.510 --> 00:40:14.750 for this system. 00:40:14.750 --> 00:40:17.470 And it involves both masses, both coordinates. 00:40:20.590 --> 00:40:23.670 So this one here would just be some x1 double dot times 00:40:23.670 --> 00:40:25.754 a cosine phi or a sine phi or something like that. 00:40:25.754 --> 00:40:27.378 AUDIENCE: So for this problem, is there 00:40:27.378 --> 00:40:28.652 only one equation of motion? 00:40:28.652 --> 00:40:29.360 PROFESSOR: Oh no. 00:40:29.360 --> 00:40:30.670 There still has to be two. 00:40:30.670 --> 00:40:35.900 But I've just shown you a way of getting at one of them 00:40:35.900 --> 00:40:40.730 without ever solving for the internal forces. 00:40:40.730 --> 00:40:44.370 By just realizing that if I separate them, 00:40:44.370 --> 00:40:47.190 that Newton's third law, all the internal ones 00:40:47.190 --> 00:40:50.007 will cancel when I add the two pieces together. 00:40:50.007 --> 00:40:51.590 I still have to get a second equation. 00:40:51.590 --> 00:40:52.548 Where would you get it? 00:41:02.080 --> 00:41:05.496 AUDIENCE: Sum of the forces in x1? 00:41:05.496 --> 00:41:07.470 AUDIENCE: They're rotating [INAUDIBLE]. 00:41:07.470 --> 00:41:08.390 PROFESSOR: Yeah, you want to do something 00:41:08.390 --> 00:41:10.306 where you don't have to solve-- you don't want 00:41:10.306 --> 00:41:13.170 to have to solve for any of these internal forces 00:41:13.170 --> 00:41:15.060 that you don't know, right? 00:41:15.060 --> 00:41:19.040 So the spring force you can know because it 00:41:19.040 --> 00:41:22.420 is a minus k times the coordinate that you 00:41:22.420 --> 00:41:25.630 have to use, minus kx1. 00:41:25.630 --> 00:41:27.330 So I would go in here, and I'd say 00:41:27.330 --> 00:41:32.280 maybe the-- what's the sum of the torques about that point, 00:41:32.280 --> 00:41:33.325 perhaps. 00:41:33.325 --> 00:41:34.700 AUDIENCE: There's no slip, right? 00:41:34.700 --> 00:41:35.658 PROFESSOR: And no slip. 00:41:35.658 --> 00:41:37.340 So these things-- that avoids having 00:41:37.340 --> 00:41:39.490 to deal with any of these forces. 00:41:39.490 --> 00:41:41.550 The real force acting on it, real force, 00:41:41.550 --> 00:41:42.720 you have to deal with that. 00:41:42.720 --> 00:41:46.760 And the minus kx1 force you need. 00:41:46.760 --> 00:41:50.810 It comes into your equation of motion. 00:41:50.810 --> 00:41:53.370 If you do that particular approach, 00:41:53.370 --> 00:41:57.640 it has one problematic-- it's not terrible 00:41:57.640 --> 00:41:59.890 but you have to remember to deal with it-- 00:41:59.890 --> 00:42:06.250 is that the sum of the x torques about that point we'll call A 00:42:06.250 --> 00:42:19.100 is d HA dt plus vAo cross P, right? 00:42:19.100 --> 00:42:21.370 Or I gave you an equation yesterday. 00:42:21.370 --> 00:42:23.500 This is easier to do usually if you 00:42:23.500 --> 00:42:39.940 do d HG dt plus rG with respect to A cross the mass 00:42:39.940 --> 00:42:45.270 times the acceleration of g with respect to o. 00:42:45.270 --> 00:42:49.810 This would be the second mass, M2. 00:42:49.810 --> 00:42:51.540 You can derive. 00:42:51.540 --> 00:42:53.200 It's about three lines. 00:42:53.200 --> 00:42:57.390 You can prove that this statement's the same as that. 00:42:57.390 --> 00:43:02.280 This one, turns out this one's less work. 00:43:02.280 --> 00:43:05.850 And the reason that this one's more work is this term always 00:43:05.850 --> 00:43:07.520 cancels out. 00:43:07.520 --> 00:43:10.310 Cancels out with a piece of it is generated 00:43:10.310 --> 00:43:11.815 when you do this calculation. 00:43:11.815 --> 00:43:13.440 But you know how messy this stuff gets. 00:43:13.440 --> 00:43:15.000 You've got all these unit vectors running around 00:43:15.000 --> 00:43:16.590 and cosine thetas and sine thetas. 00:43:16.590 --> 00:43:19.050 It's a lot of work to get this piece. 00:43:19.050 --> 00:43:21.340 And it's twice the work because you 00:43:21.340 --> 00:43:24.794 have to go in and find the piece that cancels it in here. 00:43:24.794 --> 00:43:27.210 So if you actually-- I sat down about a week ago and said, 00:43:27.210 --> 00:43:29.970 there's got to be a way to get rid of this thing. 00:43:29.970 --> 00:43:32.160 And there is, and it comes out like this. 00:43:32.160 --> 00:43:33.940 Now there's no pieces that cancel. 00:43:33.940 --> 00:43:38.030 You cancelled out in the proof, and get this piece to go away. 00:43:38.030 --> 00:43:39.650 And you're left with this. 00:43:39.650 --> 00:43:44.750 So this is the distance from your point 00:43:44.750 --> 00:43:46.160 to the center of mass. 00:43:46.160 --> 00:43:48.794 In this case, it's just r. 00:43:48.794 --> 00:43:50.460 And the only thing you have to calculate 00:43:50.460 --> 00:43:52.080 is the acceleration of the center 00:43:52.080 --> 00:43:57.750 of mass, which is capital X double dot plus X1 double dot. 00:43:57.750 --> 00:43:59.480 You know that. 00:43:59.480 --> 00:44:02.342 So this is vastly easier to do. 00:44:02.342 --> 00:44:04.772 AUDIENCE: So I understand how to use that if I'm 00:44:04.772 --> 00:44:06.230 using different [INAUDIBLE]. 00:44:10.604 --> 00:44:13.034 PROFESSOR: Center of mass, I didn't get the last bit. 00:44:13.034 --> 00:44:13.950 AUDIENCE: [INAUDIBLE]. 00:44:18.224 --> 00:44:19.390 PROFESSOR: That's the other. 00:44:19.390 --> 00:44:20.889 Because you're doing it around point 00:44:20.889 --> 00:44:23.260 A is what gets you into this mess. 00:44:23.260 --> 00:44:25.830 If you're doing it-- as soon as you work with respect-- 00:44:25.830 --> 00:44:28.520 if you're working with G, the center of mass, 00:44:28.520 --> 00:44:34.200 then this is vG cross vG, which is always zero. 00:44:34.200 --> 00:44:36.800 This has a vG in it. 00:44:36.800 --> 00:44:40.320 So that's the reason we like to work around center of the mass, 00:44:40.320 --> 00:44:42.260 but we don't do centers of mass when 00:44:42.260 --> 00:44:45.490 it causes us to have a whole bunch of external things 00:44:45.490 --> 00:44:46.280 we don't know. 00:44:50.740 --> 00:44:52.840 Thanks. 00:44:52.840 --> 00:44:53.440 Yes. 00:44:53.440 --> 00:44:54.802 AUDIENCE: [INAUDIBLE].