WEBVTT 00:00:00.070 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. 00:00:03.800 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.580 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.580 --> 00:00:17.260 at ocw.mit.edu. 00:00:37.410 --> 00:00:40.110 PROFESSOR: OK, let's get started. 00:00:40.110 --> 00:00:42.260 Can we get them up? 00:00:42.260 --> 00:00:43.690 So this is our thing spinning. 00:00:43.690 --> 00:00:46.180 What are the units of the generalized force 00:00:46.180 --> 00:00:48.060 in this problem, which is going to be related 00:00:48.060 --> 00:00:50.130 to the torque at the bottom. 00:00:50.130 --> 00:00:54.000 So most people said Newton meters, 00:00:54.000 --> 00:00:56.340 which is the units of torque. 00:00:56.340 --> 00:01:00.550 And that would be correct. 00:01:00.550 --> 00:01:03.290 So you're going to get some i theta double dot 00:01:03.290 --> 00:01:05.940 kind of equation of motion with this, 00:01:05.940 --> 00:01:08.380 as units of moment or torque. 00:01:08.380 --> 00:01:12.440 And any external non-conservative force on it 00:01:12.440 --> 00:01:14.660 would in this case have units of torque. 00:01:14.660 --> 00:01:17.440 OK, next. 00:01:17.440 --> 00:01:19.920 So we have a pendulum, kind of an odd shape. 00:01:19.920 --> 00:01:21.590 That is, does it have symmetries? 00:01:24.290 --> 00:01:27.665 Name a symmetry that this thing has. 00:01:27.665 --> 00:01:29.640 AUDIENCE: [INAUDIBLE]. 00:01:29.640 --> 00:01:32.390 PROFESSOR: You mean axial then, or a plane, or what? 00:01:32.390 --> 00:01:35.030 AUDIENCE: [INAUDIBLE]. 00:01:35.030 --> 00:01:36.850 PROFESSOR: Axis or a plane, OK. 00:01:36.850 --> 00:01:38.880 So this thing has symmetries. 00:01:38.880 --> 00:01:40.930 You could convince yourself pretty quickly 00:01:40.930 --> 00:01:42.820 that the principal axes with one big 00:01:42.820 --> 00:01:44.590 break down the center of it. 00:01:44.590 --> 00:01:47.650 And the other two would be perpendicular to that. 00:01:47.650 --> 00:01:52.800 So is it appropriate to use the parallel axis theorem to find 00:01:52.800 --> 00:01:57.150 an equation of motion of this thing for when it's pinned 00:01:57.150 --> 00:01:58.470 at the top? 00:01:58.470 --> 00:02:01.790 And most people said yes if you said no. 00:02:01.790 --> 00:02:03.380 I think parallel axis theorem works 00:02:03.380 --> 00:02:05.440 just great in this problem. 00:02:05.440 --> 00:02:11.670 It's planar motion, and the, for example, kinetic energy 00:02:11.670 --> 00:02:20.245 you can write as 1/2i about o up there, the hinge point. 00:02:20.245 --> 00:02:24.160 i with respect to that point theta double dot. 00:02:24.160 --> 00:02:26.220 And to do i about that point, you'd 00:02:26.220 --> 00:02:27.740 use the parallel axis theorem. 00:02:27.740 --> 00:02:34.970 So i about g plus the distance l squared times m. 00:02:34.970 --> 00:02:35.820 This one. 00:02:35.820 --> 00:02:39.300 Do you expect a centripetal acceleration term 00:02:39.300 --> 00:02:42.730 to show up in your equations of motion? 00:02:42.730 --> 00:02:45.175 And some said yes, some said no. 00:02:48.010 --> 00:02:50.010 So the equation of motion for this, 00:02:50.010 --> 00:02:54.580 I would probably write in terms of some rotation 00:02:54.580 --> 00:02:58.890 theta of the big disk. 00:02:58.890 --> 00:03:02.055 And that will cause the mass on the string to move up and down. 00:03:05.280 --> 00:03:12.770 But for the rotating parts of the system, 00:03:12.770 --> 00:03:20.810 are there-- I'll think of a clear way to word this. 00:03:20.810 --> 00:03:25.630 Let's ignore the little mass going up and down for a moment. 00:03:25.630 --> 00:03:29.370 The axis of rotation of that double disk and hub 00:03:29.370 --> 00:03:32.270 and the bigger disk, does the axis of rotation 00:03:32.270 --> 00:03:35.636 go through the center of mass of that disk? 00:03:35.636 --> 00:03:36.552 AUDIENCE: [INAUDIBLE]. 00:03:36.552 --> 00:03:37.135 PROFESSOR: OK. 00:03:37.135 --> 00:03:39.713 Is it statically balanced? 00:03:47.220 --> 00:03:50.150 We haven't talked about balancing in a while. 00:03:50.150 --> 00:03:52.940 Good review question for thinking about a quiz 00:03:52.940 --> 00:03:55.260 next Tuesday. 00:03:55.260 --> 00:04:00.840 So what does it take for something 00:04:00.840 --> 00:04:07.400 which is rotating to be considered statically balanced? 00:04:07.400 --> 00:04:10.694 The axis of rotation must what? 00:04:10.694 --> 00:04:11.630 AUDIENCE: [INAUDIBLE]. 00:04:11.630 --> 00:04:13.754 PROFESSOR: He says pass through the center of mass. 00:04:13.754 --> 00:04:16.910 Anybody else have a suggestion, different suggestion? 00:04:16.910 --> 00:04:21.370 If the axis of rotation passes through the center of mass, 00:04:21.370 --> 00:04:24.105 is it statically balanced? 00:04:24.105 --> 00:04:27.210 AUDIENCE: It has to be a principal axis? 00:04:27.210 --> 00:04:29.210 PROFESSOR: Does it have to be a principal axis? 00:04:29.210 --> 00:04:30.850 What do you think? 00:04:30.850 --> 00:04:33.600 Is that axis passing through g have to be a principal 00:04:33.600 --> 00:04:36.945 axis in order for it to be statically balanced? 00:04:42.200 --> 00:04:42.865 Let's see. 00:04:49.270 --> 00:04:53.680 I don't have a little-- no, it's too big. 00:04:59.720 --> 00:05:01.780 This is just a wheel. 00:05:01.780 --> 00:05:05.270 And this is my axle going through the center of mass. 00:05:05.270 --> 00:05:07.400 And it goes through the center of mass, 00:05:07.400 --> 00:05:15.650 and it'll have no tendency to swing down to a low point 00:05:15.650 --> 00:05:19.710 because the weight, its weight, is acting right 00:05:19.710 --> 00:05:20.820 on its center of mass. 00:05:20.820 --> 00:05:23.885 And with respect to this axle, there's no moment arm. 00:05:23.885 --> 00:05:25.740 So there's no torque caused by gravity 00:05:25.740 --> 00:05:31.320 that could cause this to put its center of mass below the axle. 00:05:31.320 --> 00:05:33.380 So it doesn't matter, even if I had 00:05:33.380 --> 00:05:35.830 the axis going through here. 00:05:35.830 --> 00:05:39.850 As long as that axis passes through g, 00:05:39.850 --> 00:05:42.750 there is no tendency for this thing to swing to a low side 00:05:42.750 --> 00:05:47.630 because the mass mg is acting right on the axle. 00:05:47.630 --> 00:05:48.690 No moment arm. 00:05:48.690 --> 00:05:53.660 So the only condition for static balance 00:05:53.660 --> 00:05:56.820 for a rotor of any kind, something rotating, 00:05:56.820 --> 00:06:00.320 to be statically balanced is for the axis of rotation 00:06:00.320 --> 00:06:01.680 to pass through the mass center. 00:06:05.360 --> 00:06:11.060 So the rotating part of this system, the axis of rotation 00:06:11.060 --> 00:06:15.770 passes through the mass center, that big disk. 00:06:15.770 --> 00:06:18.875 Does the little mass, m, rotate? 00:06:22.420 --> 00:06:28.450 So it can't be statically or dynamically in balance. 00:06:28.450 --> 00:06:30.390 It's not a rotating system. 00:06:30.390 --> 00:06:32.570 It's just a little translating mass 00:06:32.570 --> 00:06:34.540 that happens to be pulled up and down 00:06:34.540 --> 00:06:36.770 by the action of this thing. 00:06:36.770 --> 00:06:39.610 So the question was, do you expect 00:06:39.610 --> 00:06:42.040 a centripetal acceleration term to show up? 00:06:42.040 --> 00:06:43.720 The answer's no. 00:06:43.720 --> 00:06:44.840 So let's get back to that. 00:06:44.840 --> 00:06:48.730 If you have a static imbalance-- let's say my axis of rotation 00:06:48.730 --> 00:06:51.570 is here. 00:06:51.570 --> 00:06:54.190 This is definitely statically imbalanced now. 00:06:54.190 --> 00:06:57.640 Its mass center is down here. 00:06:57.640 --> 00:07:02.080 The force of gravity pulls it down until it hangs straight. 00:07:02.080 --> 00:07:04.520 That's how you can just test to see if you're passing 00:07:04.520 --> 00:07:06.280 through the mass center. 00:07:06.280 --> 00:07:12.980 So if you have an axis that does not pass through g, 00:07:12.980 --> 00:07:14.760 then you're statically imbalanced. 00:07:14.760 --> 00:07:18.700 And if you rotate about that axis, 00:07:18.700 --> 00:07:25.250 is there a net centripetal force required as it goes around? 00:07:25.250 --> 00:07:26.510 Why? 00:07:26.510 --> 00:07:28.810 AUDIENCE: Because the center of mass is on the outside. 00:07:28.810 --> 00:07:31.620 PROFESSOR: So the center of mass is some distance away. 00:07:31.620 --> 00:07:33.820 And you're swinging it around and around 00:07:33.820 --> 00:07:37.150 and make that center of mass go in a circle. 00:07:37.150 --> 00:07:39.809 You're applying a centripetal acceleration 00:07:39.809 --> 00:07:41.350 to the center of mass, and that takes 00:07:41.350 --> 00:07:45.330 a force, which you sometimes think of as a centrifugal force 00:07:45.330 --> 00:07:45.887 pulling out. 00:07:45.887 --> 00:07:47.720 That's what you're going to have to pull in, 00:07:47.720 --> 00:07:51.800 some m r omega squared. 00:07:51.800 --> 00:07:55.710 So no centripetal term would be expected in this problem. 00:07:55.710 --> 00:07:57.420 Let's go to the next one. 00:07:57.420 --> 00:07:59.500 This one, you had two different conditions. 00:07:59.500 --> 00:08:03.740 Either rolling without slip or rolling with slip. 00:08:03.740 --> 00:08:08.030 And the question is, for which conditions 00:08:08.030 --> 00:08:10.290 are the generalized forces associated 00:08:10.290 --> 00:08:13.730 with a virtual displacement of delta theta, 00:08:13.730 --> 00:08:18.560 the rotation of the wheel, equal to zero? 00:08:18.560 --> 00:08:20.200 So remember, generalized forces are 00:08:20.200 --> 00:08:23.820 the forces that account for the non-conservative forces 00:08:23.820 --> 00:08:25.350 in the problem. 00:08:25.350 --> 00:08:30.070 So if it's rolling down the hill without slip, 00:08:30.070 --> 00:08:35.900 are there any non-conservative forces acting on it? 00:08:35.900 --> 00:08:38.710 Is there a friction force acting on it? 00:08:38.710 --> 00:08:40.360 Yeah, but does it do any work? 00:08:40.360 --> 00:08:44.019 No, because there's no delta r at the point of application 00:08:44.019 --> 00:08:44.560 of the force. 00:08:44.560 --> 00:08:46.300 There's no motion. 00:08:46.300 --> 00:08:50.650 So for which condition does the generalized forces 00:08:50.650 --> 00:08:51.300 equal to zero? 00:08:51.300 --> 00:08:53.960 Certainly for the condition when it does not slip. 00:08:53.960 --> 00:08:57.740 What about if it does slip? 00:08:57.740 --> 00:09:00.480 Would you expect a non-conservative force 00:09:00.480 --> 00:09:01.590 to do work on it? 00:09:01.590 --> 00:09:04.660 What force would that be? 00:09:04.660 --> 00:09:05.800 The friction force. 00:09:05.800 --> 00:09:11.630 So now, as the wheel turns, that point of application where it's 00:09:11.630 --> 00:09:17.020 sliding, you're actually getting a little delta omega, delta 00:09:17.020 --> 00:09:18.260 theta rather. 00:09:18.260 --> 00:09:22.540 You move a little distance r delta theta, 00:09:22.540 --> 00:09:24.140 dotted with the force. 00:09:24.140 --> 00:09:27.620 You get a certain little bit of virtual work 00:09:27.620 --> 00:09:31.560 done, r delta theta times f. 00:09:31.560 --> 00:09:36.370 So the only case a is where you get zero. 00:09:36.370 --> 00:09:37.215 Oh, this one. 00:09:41.070 --> 00:09:44.970 What's the generalized force associated with f? 00:09:44.970 --> 00:09:46.500 That's this applied force. 00:09:46.500 --> 00:09:50.320 It's applied to the rolling thing going down the hill. 00:09:50.320 --> 00:09:53.110 The force is horizontal. 00:09:53.110 --> 00:09:56.730 And you're asked what's the generalized force associated 00:09:56.730 --> 00:10:01.760 with f due to the virtual displacement delta x? 00:10:01.760 --> 00:10:07.980 And x is the motion of the main cart. 00:10:07.980 --> 00:10:11.130 So when you're doing generalized forces, 00:10:11.130 --> 00:10:14.230 you think in terms of virtual work. 00:10:14.230 --> 00:10:17.650 Can you imagine that that cart's moving a little distance 00:10:17.650 --> 00:10:23.100 delta x, the main x in the xyo system? 00:10:23.100 --> 00:10:26.560 It shifts a little distance delta x. 00:10:26.560 --> 00:10:29.360 That force, does it move-- does the point 00:10:29.360 --> 00:10:33.910 of application of that force move with that delta x? 00:10:33.910 --> 00:10:36.092 That's really what the question comes down to. 00:10:44.290 --> 00:10:44.895 This one. 00:11:01.380 --> 00:11:04.310 So this is your inertial system, and it's 00:11:04.310 --> 00:11:06.700 going to account for the motion of this object. 00:11:09.900 --> 00:11:15.300 And then up here, got your wheel. 00:11:15.300 --> 00:11:23.680 And we have some coordinate system attached to this object, 00:11:23.680 --> 00:11:27.470 noting the position of the wheel as it 00:11:27.470 --> 00:11:29.860 goes up and down the hill. 00:11:29.860 --> 00:11:34.380 So this coordinate's relative to the moving cart. 00:11:34.380 --> 00:11:37.240 This coordinate's inertial. 00:11:37.240 --> 00:11:43.580 The question's only asking then, what is the virtual work done 00:11:43.580 --> 00:11:49.440 that's associated width delta x here, some motion of the cart. 00:11:49.440 --> 00:11:55.960 And it's going to be what we're looking for, the Qx delta x. 00:11:55.960 --> 00:11:59.427 And when we say, OK. 00:11:59.427 --> 00:12:01.135 At the point of application of the force, 00:12:01.135 --> 00:12:04.170 there's literally F here. 00:12:04.170 --> 00:12:06.490 And it's in the horizontal direction, which is exactly 00:12:06.490 --> 00:12:15.060 the same direction as capital X. So this is equal to the force. 00:12:15.060 --> 00:12:19.280 And in the problem it's called-- this is just 00:12:19.280 --> 00:12:22.105 called F. It's a vector. 00:12:24.850 --> 00:12:28.310 And we're trying to deduce the deflection 00:12:28.310 --> 00:12:30.140 at the point of application. 00:12:30.140 --> 00:12:34.460 I'm going to call it some-- it can be many forces. 00:12:34.460 --> 00:12:36.350 I'm going to call it F sub i so that we 00:12:36.350 --> 00:12:39.580 think of it as one of many. 00:12:39.580 --> 00:12:57.260 Times the displacement delta ri due to delta x. 00:12:57.260 --> 00:12:58.740 So this is a vector. 00:12:58.740 --> 00:13:00.100 This is a displacement. 00:13:00.100 --> 00:13:01.580 This is a force. 00:13:01.580 --> 00:13:04.390 The amount of work done as this force moves 00:13:04.390 --> 00:13:09.040 through that displacement is some F 00:13:09.040 --> 00:13:14.300 dot delta r, the movement at this point i where it's 00:13:14.300 --> 00:13:19.490 applied, due to this motion. 00:13:19.490 --> 00:13:24.450 So what is the delta r at this point due to that motion? 00:13:29.250 --> 00:13:31.720 So another concept here, when you're doing this, 00:13:31.720 --> 00:13:34.700 you think of these mentally one at a time. 00:13:34.700 --> 00:13:37.120 How many degrees of freedom do we have in this problem? 00:13:40.890 --> 00:13:43.190 Takes two to completely describe the motion. 00:13:43.190 --> 00:13:46.230 x and this one attached to the cart. 00:13:49.050 --> 00:13:51.750 And you think of these one at a time. 00:13:51.750 --> 00:13:53.370 So right now, we're asking what's 00:13:53.370 --> 00:13:59.060 the virtual work done by that force due to movement of this 00:13:59.060 --> 00:14:02.870 coordinate only, assuming this one is frozen? 00:14:02.870 --> 00:14:05.680 So this one is not allowed to move. 00:14:05.680 --> 00:14:07.180 So if this is not allowed to move, 00:14:07.180 --> 00:14:12.020 does the position of this change relative to this cart as you 00:14:12.020 --> 00:14:12.907 move the cart? 00:14:15.530 --> 00:14:16.360 No. 00:14:16.360 --> 00:14:17.910 That one is held fixed. 00:14:17.910 --> 00:14:19.950 It moves with the cart. 00:14:19.950 --> 00:14:23.760 So if you move the cart a little distance here, 00:14:23.760 --> 00:14:29.060 delta x in this frame, then this wheel 00:14:29.060 --> 00:14:31.810 moves with the cart that distance. 00:14:31.810 --> 00:14:35.500 So how much work gets done? 00:14:35.500 --> 00:14:44.430 So this delta ri in this case is equal to delta x. 00:14:44.430 --> 00:14:46.860 And the amount of work that gets done 00:14:46.860 --> 00:15:00.610 is delta w at point i due to delta x here is F dot delta x, 00:15:00.610 --> 00:15:04.170 but F is in the capital I direction. 00:15:04.170 --> 00:15:06.770 This is in the capital I direction. 00:15:06.770 --> 00:15:10.580 So it's just F delta x. 00:15:10.580 --> 00:15:11.930 is the work done. 00:15:11.930 --> 00:15:14.760 And that's Qx delta x. 00:15:14.760 --> 00:15:23.580 So the generalized force is just F. 00:15:23.580 --> 00:15:24.500 All right. 00:15:24.500 --> 00:15:25.310 What else we got? 00:15:28.040 --> 00:15:29.106 OK, wait a minute. 00:15:32.509 --> 00:15:33.800 Oh, this is due to the dashpot. 00:15:36.420 --> 00:15:38.010 What is generalized force associated 00:15:38.010 --> 00:15:42.670 with the dashpot due to a virtual deflection x1? 00:15:42.670 --> 00:15:45.640 Now x1's the coordinate, that thing 00:15:45.640 --> 00:15:46.850 rolling up and down the hill. 00:15:49.490 --> 00:15:52.725 So now the same thing, the dashpot force. 00:16:02.460 --> 00:16:05.142 Here's your cart. 00:16:05.142 --> 00:16:05.975 Here's your dashpot. 00:16:16.960 --> 00:16:20.190 And a free body diagram of the cart 00:16:20.190 --> 00:16:25.740 would show a dashpot force here, bx dot in that direction. 00:16:29.170 --> 00:16:37.890 And if we move the cart, again, a little bit delta 00:16:37.890 --> 00:16:40.655 x, how much work gets done? 00:16:40.655 --> 00:16:42.930 AUDIENCE: [INAUDIBLE]. 00:16:42.930 --> 00:16:45.790 PROFESSOR: This is not the question asked up there. 00:16:45.790 --> 00:16:47.530 Just first this. 00:16:47.530 --> 00:16:52.370 How much virtual work gets done by that in a deflection delta 00:16:52.370 --> 00:16:54.180 x? 00:16:54.180 --> 00:17:01.800 Well, it's some Fi dot delta r due to delta x, which 00:17:01.800 --> 00:17:03.760 is just delta x. 00:17:03.760 --> 00:17:09.349 So this is some minus bx dot in the I direction times delta 00:17:09.349 --> 00:17:10.900 x in the I again. 00:17:10.900 --> 00:17:13.609 It's just in this case it's minus. 00:17:13.609 --> 00:17:28.669 So this virtual work done this time due to just the dashpot. 00:17:28.669 --> 00:17:29.960 I don't know how to write that. 00:17:29.960 --> 00:17:31.250 I don't have a subscript. 00:17:31.250 --> 00:17:35.590 So this dashpot only here, bx dot, 00:17:35.590 --> 00:17:40.020 is bx dot minus in the I direction-- 00:17:40.020 --> 00:17:46.270 that's the force-- dotted with the dr that it moves. 00:17:48.910 --> 00:17:51.870 And the dr that it moves is just delta x. 00:17:51.870 --> 00:18:00.120 So this would be minus bx I dot delta x I is 00:18:00.120 --> 00:18:02.280 your virtual work that's done. 00:18:02.280 --> 00:18:08.750 And this is equal to Qx delta x. 00:18:08.750 --> 00:18:14.270 But this is a part of Qx due only to the dashpot. 00:18:14.270 --> 00:18:20.530 And you get minus bx dot delta x. 00:18:20.530 --> 00:18:26.990 So now we've gotten both parts of the total generalized force 00:18:26.990 --> 00:18:32.375 associated with the motion delta x is F minus bx dot. 00:18:39.550 --> 00:18:44.960 In general, this expression Qx delta x 00:18:44.960 --> 00:18:49.740 is the summation over all of the applied forces 00:18:49.740 --> 00:19:02.640 i dot delta r at i due to, in this case, 00:19:02.640 --> 00:19:07.050 the motion in the x direction. 00:19:07.050 --> 00:19:10.150 And we have two contributions here. 00:19:10.150 --> 00:19:14.190 They come from the applied force F and this. 00:19:14.190 --> 00:19:16.990 So Qx in this problem is going to turn out 00:19:16.990 --> 00:19:22.554 to be F minus bx dot. 00:19:22.554 --> 00:19:23.970 Now, what about the other-- what's 00:19:23.970 --> 00:19:25.720 really asked in this problem is how much-- 00:19:25.720 --> 00:19:29.070 what's the generalized force associated with the motion 00:19:29.070 --> 00:19:32.330 of the wheel down the hill? 00:19:32.330 --> 00:19:34.270 This is in the x1 direction. 00:19:34.270 --> 00:19:37.770 So now you have a little virtual deflection delta x1. 00:19:41.830 --> 00:19:46.580 How much virtual work gets done by the dashpot? 00:19:46.580 --> 00:19:47.480 AUDIENCE: Zero. 00:19:47.480 --> 00:19:48.438 PROFESSOR: I hear zero. 00:19:50.780 --> 00:19:54.800 So a little virtual displacement of delta x1, 00:19:54.800 --> 00:19:58.190 does it move the main cart? 00:19:58.190 --> 00:20:00.750 That's really what's going on here. 00:20:00.750 --> 00:20:05.360 Does the main cart move because the wheel and the main cart 00:20:05.360 --> 00:20:07.615 change relative position a little bit? 00:20:11.054 --> 00:20:11.597 AUDIENCE: No. 00:20:11.597 --> 00:20:12.180 PROFESSOR: No. 00:20:12.180 --> 00:20:13.180 I hear no here. 00:20:13.180 --> 00:20:16.000 I mean, that's the-- the coordinate x1 00:20:16.000 --> 00:20:20.620 is the relative motion between the cart and the wheel. 00:20:20.620 --> 00:20:23.840 And it's independent of the motion of the cart with respect 00:20:23.840 --> 00:20:27.130 to the inertial frame. 00:20:27.130 --> 00:20:29.520 So any motion of the little wheel 00:20:29.520 --> 00:20:31.405 does not affect the main cart. 00:20:35.030 --> 00:20:38.200 So there's no virtual work done by that dashpot 00:20:38.200 --> 00:20:40.901 because the wheel moves up and down the hill. 00:20:40.901 --> 00:20:42.650 And it makes sense, physical sense, right? 00:20:42.650 --> 00:20:44.650 The wheel can sit and roll up and down the hill all day long. 00:20:44.650 --> 00:20:46.150 It's not going to move that dashpot. 00:20:49.310 --> 00:20:51.180 Next. 00:20:51.180 --> 00:20:53.110 This problem. 00:20:53.110 --> 00:20:56.471 I just realized this problem's harder than I thought it was. 00:20:56.471 --> 00:20:59.484 It's one of those things that you look at it, 00:20:59.484 --> 00:21:00.775 oh, that looks straightforward. 00:21:00.775 --> 00:21:02.630 Then I looked at Audrey's solution 00:21:02.630 --> 00:21:05.030 and said, oh, she did it right. 00:21:05.030 --> 00:21:08.140 And this is a little trickier than you might think. 00:21:08.140 --> 00:21:12.330 So are the Axyz axes, which rotate with the hub-- 00:21:12.330 --> 00:21:15.500 so there's a rotating there-- definitely 00:21:15.500 --> 00:21:19.010 principal coordinates of that rod? 00:21:19.010 --> 00:21:19.970 That's not a problem. 00:21:19.970 --> 00:21:21.730 But are they principal coordinates 00:21:21.730 --> 00:21:24.040 for that disk out on the end? 00:21:26.986 --> 00:21:28.361 AUDIENCE: [INAUDIBLE]. 00:21:28.361 --> 00:21:29.110 PROFESSOR: Pardon? 00:21:29.110 --> 00:21:31.090 AUDIENCE: Depends if it's slipping or not. 00:21:31.090 --> 00:21:32.256 PROFESSOR: Depends if it's-- 00:21:32.256 --> 00:21:34.260 AUDIENCE: Slipping or not. 00:21:34.260 --> 00:21:38.275 PROFESSOR: Actually, I don't think so. 00:21:42.370 --> 00:21:44.370 Let's just talk about-- principal coordinates 00:21:44.370 --> 00:21:47.000 need to be attached to the body, right? 00:21:47.000 --> 00:21:51.890 So problems like this do-- let's say we're using Lagrange, 00:21:51.890 --> 00:21:54.930 and we want to calculate the total kinetic-- you 00:21:54.930 --> 00:21:58.030 need to calculate the total kinetic energy of the system. 00:21:58.030 --> 00:22:02.190 So how would you go about breaking this thing down 00:22:02.190 --> 00:22:04.610 to compute the total kinetic energy? 00:22:04.610 --> 00:22:09.420 Would you break it into more than one part? 00:22:09.420 --> 00:22:12.259 What would be the natural things to break it into? 00:22:12.259 --> 00:22:13.550 AUDIENCE: The disk and the rod. 00:22:13.550 --> 00:22:15.174 PROFESSOR: The disk and the rod, right? 00:22:15.174 --> 00:22:19.090 So the kinetic energy of the rod, that's 00:22:19.090 --> 00:22:20.090 pretty straightforward. 00:22:20.090 --> 00:22:28.040 It's 1/2i with respect to the center, theta dot squared, 00:22:28.040 --> 00:22:29.380 omega squared. 00:22:29.380 --> 00:22:32.830 But the kinetic energy of that disk out 00:22:32.830 --> 00:22:41.920 there, you need to account for its rotation and its movement 00:22:41.920 --> 00:22:45.760 of its center of mass in the circle. 00:22:45.760 --> 00:22:53.160 So let's-- I'm winging it now, so bear with me if I make any 00:22:53.160 --> 00:22:53.720 mistakes. 00:22:53.720 --> 00:22:54.640 You can help me out. 00:22:57.780 --> 00:23:03.740 T rod would be-- and our coordinate system 00:23:03.740 --> 00:23:06.810 is an A at the center. 00:23:09.570 --> 00:23:15.330 There actually isn't a-- oh yeah, there's the A. 00:23:15.330 --> 00:23:17.310 So T of the rod, I would argue, is 00:23:17.310 --> 00:23:36.040 1/2M of the rod omega squared ML squared over 3. 00:23:36.040 --> 00:23:38.160 Because it's rotating about one end. 00:23:38.160 --> 00:23:40.490 So apply the parallel axis theorem. 00:23:40.490 --> 00:23:44.630 The kinetic energy of the rod is 1/2M i 00:23:44.630 --> 00:23:49.220 with respect to that central axis, omega squared. 00:23:49.220 --> 00:23:52.880 And so that gets you from ML squared over 12 00:23:52.880 --> 00:23:56.830 to ML squared over 3 because you're swinging around one end. 00:23:56.830 --> 00:23:59.240 But now we need T of the disk. 00:24:01.950 --> 00:24:04.160 And to do T of the disk, I would do it 00:24:04.160 --> 00:24:12.870 by saying 1/2M of the disk velocity of this center of mass 00:24:12.870 --> 00:24:19.220 of the disk in o dot VGo-- so that 00:24:19.220 --> 00:24:24.830 takes care of its kinetic energy due to motion 00:24:24.830 --> 00:24:43.170 of its center of mass-- plus 1/2 omega dot H with respect to G. 00:24:43.170 --> 00:24:45.990 With respect to G, can you express 00:24:45.990 --> 00:24:50.870 H for the disk in terms of a mass moment of inertia 00:24:50.870 --> 00:24:53.487 and some rotations, rotation rates? 00:24:53.487 --> 00:24:54.195 Is it legitimate? 00:25:04.380 --> 00:25:08.440 Remember, I started-- a few days ago, I started the top 00:25:08.440 --> 00:25:12.530 and said, here's the general expression for kinetic energy. 00:25:12.530 --> 00:25:14.170 Full 3D, right? 00:25:14.170 --> 00:25:20.550 And it basically was that expression. 00:25:20.550 --> 00:25:23.380 That works full 3D. 00:25:23.380 --> 00:25:28.220 And then when you fix a point about which something 00:25:28.220 --> 00:25:33.760 is rotating, a rigid body, then you can simplify it. 00:25:33.760 --> 00:25:38.070 But in this case, H, you could represent the angular momentum 00:25:38.070 --> 00:25:42.600 around G as some I omega. 00:25:42.600 --> 00:25:46.920 And then you have to multiply it by-- so this inside of here 00:25:46.920 --> 00:25:51.830 will be-- you can represent this in here 00:25:51.830 --> 00:25:57.650 as some I omega dotted with omega, 00:25:57.650 --> 00:26:00.030 and you will get the kinetic energy 00:26:00.030 --> 00:26:02.350 due to the rotation of the disk. 00:26:02.350 --> 00:26:05.160 What makes this problem a little trickier than I thought-- 00:26:05.160 --> 00:26:06.740 I wasn't thinking really clearly-- 00:26:06.740 --> 00:26:12.520 is that omega is in-- what frame do you express omega 00:26:12.520 --> 00:26:17.000 in when you're computing H in terms 00:26:17.000 --> 00:26:18.320 of mass moments of inertia? 00:26:20.930 --> 00:26:25.580 In order to compute I, you have to use what coordinate system? 00:26:25.580 --> 00:26:26.550 AUDIENCE: [INAUDIBLE]. 00:26:26.550 --> 00:26:27.980 PROFESSOR: Body fixed. 00:26:27.980 --> 00:26:31.610 And when you compute I omega, what's the omega? 00:26:31.610 --> 00:26:35.394 What unit vectors is the omega expressed in? 00:26:35.394 --> 00:26:37.814 AUDIENCE: [INAUDIBLE]. 00:26:37.814 --> 00:26:40.240 It's rotating, but it's [INAUDIBLE]. 00:26:40.240 --> 00:26:42.560 PROFESSOR: But in what-- so the unit vector's 00:26:42.560 --> 00:26:45.330 associated with what coordinate system are the ones that you 00:26:45.330 --> 00:26:47.458 use to define omega? 00:26:47.458 --> 00:26:48.880 AUDIENCE: [INAUDIBLE]. 00:26:48.880 --> 00:26:52.110 PROFESSOR: The one attached to the body or not? 00:26:52.110 --> 00:26:54.940 Generally in terms of the one attached to the body. 00:26:54.940 --> 00:26:58.035 And the disk gets-- that means they're rotating. 00:26:58.035 --> 00:26:59.400 It gets a little messy. 00:26:59.400 --> 00:27:00.860 So I'm not going to do it because I 00:27:00.860 --> 00:27:02.930 can't do it in my head. 00:27:02.930 --> 00:27:07.780 So this one you need to do the kinetic energy of that. 00:27:07.780 --> 00:27:09.950 You need to use this approach. 00:27:09.950 --> 00:27:15.020 And you have to be careful with those. 00:27:15.020 --> 00:27:17.520 Food for thought at office hours and in recitation. 00:27:20.910 --> 00:27:21.445 All right. 00:27:31.160 --> 00:27:35.480 I had meant to start today-- we're well in, 00:27:35.480 --> 00:27:40.297 and I haven't even started where I was going to really go today. 00:27:40.297 --> 00:27:41.630 So I have kind of a broken play. 00:27:41.630 --> 00:27:43.030 I have to decide what to drop. 00:27:43.030 --> 00:27:43.755 Give me a moment. 00:28:30.520 --> 00:28:33.790 So the principal thing I wanted to talk about today 00:28:33.790 --> 00:28:38.210 was to dig a little deeper into finding generalized forces. 00:28:38.210 --> 00:28:40.830 And the way we've been doing it is 00:28:40.830 --> 00:28:42.840 what I would call kind of an intuitive approach. 00:28:47.910 --> 00:28:52.270 So let's imagine we've got a rigid body. 00:28:59.570 --> 00:29:03.356 And I have some forces acting on it, F1. 00:29:18.180 --> 00:29:21.780 And at the point of application of each of these forces, 00:29:21.780 --> 00:29:24.050 I can have a position vector that goes there. 00:29:24.050 --> 00:29:31.580 So there's R1, R2, R3. 00:29:34.910 --> 00:29:37.990 And they're all with respect to my o frame here, 00:29:37.990 --> 00:29:40.700 but I'm going to not write in all the o's. 00:29:51.960 --> 00:29:56.890 Let's say that this is a body in planar motion. 00:29:56.890 --> 00:29:58.300 So it's confined to a plane. 00:29:58.300 --> 00:29:59.280 It's a rigid body. 00:29:59.280 --> 00:30:01.380 It's like my disk there. 00:30:01.380 --> 00:30:03.940 So how many degrees of freedom at most 00:30:03.940 --> 00:30:07.700 would it have if there's no constraints? 00:30:07.700 --> 00:30:08.570 Three, right? 00:30:08.570 --> 00:30:12.480 It could move in the x, in the y, and it can rotate in the z. 00:30:12.480 --> 00:30:14.570 So potentially three degrees of freedom. 00:30:14.570 --> 00:30:20.770 So it'll take, let's say, G is maybe here. 00:30:20.770 --> 00:30:30.910 So we need three generalized coordinates. 00:30:30.910 --> 00:30:36.260 And they would be, say, x, y, and some rotation of theta. 00:30:40.860 --> 00:30:48.360 And with them, we have virtual displacements delta 00:30:48.360 --> 00:30:53.530 x, delta y, and delta theta. 00:30:53.530 --> 00:30:56.150 And we use those to figure out the amount 00:30:56.150 --> 00:31:00.450 of non-conservative work that happens 00:31:00.450 --> 00:31:03.255 when you make those little small motions occur. 00:31:05.980 --> 00:31:16.230 And we're trying to find-- we need to find Qx, Qy, and Q 00:31:16.230 --> 00:31:19.910 theta, the generalized forces associated 00:31:19.910 --> 00:31:24.631 with these small variations, virtual reflections in R3 00:31:24.631 --> 00:31:25.130 coordinates. 00:31:33.310 --> 00:31:34.810 So the jth one. 00:31:34.810 --> 00:31:36.430 I have a bunch of forces maybe up 00:31:36.430 --> 00:31:40.760 to here, someone here that's some Fj. 00:31:40.760 --> 00:31:47.430 So the generalized force-- I don't mean that. 00:31:47.430 --> 00:31:50.470 These are Fi's. 00:31:50.470 --> 00:31:55.420 The j's referred to are generalized coordinates. 00:31:55.420 --> 00:32:05.505 So generalized force Qj delta qj. 00:32:12.760 --> 00:32:19.170 This is the little bit of work done in the virtual deflection 00:32:19.170 --> 00:32:21.500 delta qj. 00:32:21.500 --> 00:32:27.630 And there's work done by all of these applied forces 00:32:27.630 --> 00:32:29.630 in the system, possibly. 00:32:29.630 --> 00:32:33.960 Every one, if I cause there a little delta x of this body, 00:32:33.960 --> 00:32:39.240 it moves over an amount delta x. 00:32:39.240 --> 00:32:41.840 All of those points of application of those forces 00:32:41.840 --> 00:32:44.320 move a little bit. 00:32:44.320 --> 00:32:49.290 And therefore, at every location a little bit of work gets done. 00:32:49.290 --> 00:32:52.360 So in order to account for all of the work, 00:32:52.360 --> 00:33:05.050 I have to do a summation of the Fi dot delta r at i. 00:33:05.050 --> 00:33:14.020 And now this is associated with the virtual displacement 00:33:14.020 --> 00:33:18.030 of generalized coordinate j. 00:33:18.030 --> 00:33:24.290 So the total virtual work done by a displacement 00:33:24.290 --> 00:33:30.090 of one of these is a summation of all the little bits of work 00:33:30.090 --> 00:33:34.660 done at all the points of application of forces 00:33:34.660 --> 00:33:37.190 dotted with the amount that the point 00:33:37.190 --> 00:33:42.740 of application of that force moves caused by delta qj. 00:33:42.740 --> 00:33:48.320 So in general, that's what the total qj would be. 00:33:48.320 --> 00:33:53.070 Now, we've done problems like that more or less intuitively. 00:33:53.070 --> 00:33:55.356 We figured it out and said, OK, if it moves this much, 00:33:55.356 --> 00:33:56.980 then it's going to move over that much. 00:33:56.980 --> 00:34:00.050 And if there is an angle between them, we take the component, 00:34:00.050 --> 00:34:01.350 and we just figure it out. 00:34:01.350 --> 00:34:06.164 Is there a more mathematical way of doing this? 00:34:06.164 --> 00:34:07.580 And so I'm going to show you that. 00:34:12.070 --> 00:34:19.880 So we've been doing this kind of the intuitive approach here, 00:34:19.880 --> 00:34:23.690 the reasoning out each of the deflections and calculating 00:34:23.690 --> 00:34:26.010 the result. That there's a kinematic-- there's 00:34:26.010 --> 00:34:27.874 an explicit kinematic way of doing this. 00:34:27.874 --> 00:34:29.040 AUDIENCE: I have a question. 00:34:29.040 --> 00:34:29.706 PROFESSOR: Yeah. 00:34:29.706 --> 00:34:32.388 AUDIENCE: What is it at the right corner? [INAUDIBLE]? 00:34:32.388 --> 00:34:33.679 PROFESSOR: Yeah, OK, I'm sorry. 00:34:33.679 --> 00:34:37.730 It says-- this is a delta ri. 00:34:37.730 --> 00:34:39.530 It has double subscripts here. 00:34:39.530 --> 00:34:45.360 This is the displacement at the point of application of force 00:34:45.360 --> 00:34:49.366 i due to the virtual displacement 00:34:49.366 --> 00:34:54.120 of generalized coordinate j. 00:34:54.120 --> 00:35:01.350 So that an example might be, if that's delta x over here, 00:35:01.350 --> 00:35:07.440 the summation is over i, is over F1 F2 to Fi. 00:35:07.440 --> 00:35:13.090 And the deflection is the deflection 00:35:13.090 --> 00:35:17.930 at the point of application i due to-- in this case 00:35:17.930 --> 00:35:22.830 I'm talking about delta x-- due to the virtual displacements 00:35:22.830 --> 00:35:25.010 delta x. 00:35:25.010 --> 00:35:27.930 And so that's-- we'd work each one of these out and add them 00:35:27.930 --> 00:35:29.210 up, and we'd have the answer. 00:35:29.210 --> 00:35:33.370 But there is a more explicit mathematical way 00:35:33.370 --> 00:35:34.350 of saying this. 00:35:34.350 --> 00:35:47.720 And that is to say that Qj delta qj is the summation over i 00:35:47.720 --> 00:35:52.760 equals 1 to N, however many there are, 00:35:52.760 --> 00:36:10.450 of Fi dot the derivative of ri with respect to qj delta qj. 00:36:13.240 --> 00:36:14.880 What's that mean? 00:36:14.880 --> 00:36:17.500 So let's look at one of these. 00:36:17.500 --> 00:36:21.100 So force one, there's a position vector R1. 00:36:24.760 --> 00:36:32.210 If I move in the x direction a little delta x, 00:36:32.210 --> 00:36:36.910 this point moves over delta x in that direction, 00:36:36.910 --> 00:36:40.810 in the x direction horizontally. 00:36:40.810 --> 00:36:50.190 Our position vector R1 has potentially components in the y 00:36:50.190 --> 00:36:51.860 as well as components in the x. 00:36:51.860 --> 00:36:55.520 But I'm only changing it in the x direction. 00:36:55.520 --> 00:36:58.590 So that portion of the possible movement 00:36:58.590 --> 00:37:06.110 of R1 due to changes in just one of the coordinates-- 00:37:06.110 --> 00:37:10.010 in this case, I was doing qx-- the derivative of R1 00:37:10.010 --> 00:37:12.850 with respect to qx, so only a part 00:37:12.850 --> 00:37:16.780 of its total possible movement is due to x. 00:37:16.780 --> 00:37:19.480 This gives us that portion. 00:37:19.480 --> 00:37:24.510 Times delta qx is the total movement 00:37:24.510 --> 00:37:29.890 in the direction of qx dotted with the force. 00:37:29.890 --> 00:37:31.610 You get the work done. 00:37:31.610 --> 00:37:37.600 So I find this-- if I were you, this is highly abstract. 00:37:37.600 --> 00:37:40.180 I think we need-- let's do an example of this 00:37:40.180 --> 00:37:43.610 and see how it works out. 00:37:43.610 --> 00:37:48.680 And since we were talking about that problem, 00:37:48.680 --> 00:37:51.840 I'll do this cart by this method. 00:39:13.610 --> 00:39:16.990 So I need a position vector to the point 00:39:16.990 --> 00:39:22.190 of application of this external non-conservative force. 00:39:22.190 --> 00:39:24.540 Because I'm calling this force one, 00:39:24.540 --> 00:39:26.880 I'll call that position vector R1. 00:39:26.880 --> 00:39:31.040 And it goes from here to there. 00:39:31.040 --> 00:39:34.530 But we know that the total motion of this point 00:39:34.530 --> 00:39:36.970 is made up of the motion of the main cart 00:39:36.970 --> 00:39:41.240 plus the motion of the wheel relative to the main cart. 00:39:41.240 --> 00:39:45.100 And so we fall back on our notation. 00:39:45.100 --> 00:39:48.080 So I'll say this is R1 and zero. 00:39:48.080 --> 00:39:51.970 Here's my point A. It's R is the vector. 00:39:51.970 --> 00:39:57.940 This is R of A with respect to o plus-- 00:39:57.940 --> 00:39:59.860 and we'll better give this a name. 00:39:59.860 --> 00:40:10.410 What have I-- so this is my point one here. 00:40:10.410 --> 00:40:15.410 So this is plus R1 with respect to A. 00:40:15.410 --> 00:40:17.730 So we've done that lots of times, this term. 00:40:17.730 --> 00:40:20.970 That's just how-- that point is the sum of this vector 00:40:20.970 --> 00:40:22.735 plus this vector. 00:40:22.735 --> 00:40:29.230 So you have an R-- this is R1 with respect to A from here 00:40:29.230 --> 00:40:30.250 to here. 00:40:30.250 --> 00:40:32.790 And the sum of those two is this one. 00:40:32.790 --> 00:40:36.170 So this is R1 here. 00:40:36.170 --> 00:40:41.070 OK, so let's see if we can come up with an expression for that. 00:40:41.070 --> 00:40:53.800 Well, this point A is just x in the I plus some Y in the J. 00:40:53.800 --> 00:40:56.500 That's this term. 00:40:56.500 --> 00:40:59.700 Then I need this one. 00:41:04.960 --> 00:41:07.500 So I want-- because I'm going to take some derivatives 00:41:07.500 --> 00:41:10.360 and things, I want to get everything in terms 00:41:10.360 --> 00:41:13.380 of unit vectors and one system. 00:41:13.380 --> 00:41:17.010 So I know this one, this is my x1, 00:41:17.010 --> 00:41:21.210 and it will have a unit vector i, lowercase i, 00:41:21.210 --> 00:41:23.390 in this direction. 00:41:23.390 --> 00:41:31.810 So the unit vector i here has components in the capital IJ 00:41:31.810 --> 00:41:33.590 system. 00:41:33.590 --> 00:41:44.790 And this is theta, and that's theta. 00:41:44.790 --> 00:41:51.830 So this has-- i has a component here, 00:41:51.830 --> 00:42:00.360 which is cosine theta cap I, and then this piece is 00:42:00.360 --> 00:42:20.430 minus sine theta J. So this should be x1, 00:42:20.430 --> 00:42:26.990 the distance this thing moves, broken into two pieces, 00:42:26.990 --> 00:42:35.330 cos theta I minus sine theta J. So that's now-- 00:42:35.330 --> 00:42:43.950 the position of this thing is the position of the cart at A 00:42:43.950 --> 00:42:49.940 plus the vector that goes from A to point one, 00:42:49.940 --> 00:42:53.330 which is the distance x1 to here, 00:42:53.330 --> 00:42:55.510 broken into two pieces, an I piece and a J piece. 00:43:01.820 --> 00:43:03.170 So now we're almost done. 00:43:03.170 --> 00:43:06.400 So I would like to find Qx. 00:43:10.900 --> 00:43:18.690 And Qx then should be the summation-- well, let's see. 00:43:37.900 --> 00:43:41.950 So I'm only going to compute the part of the generalized force 00:43:41.950 --> 00:43:46.030 in the Qx direction due to just this one force. 00:43:46.030 --> 00:43:48.260 Now, remember we have other forces, 00:43:48.260 --> 00:43:50.060 non-conservative forces, acting on this. 00:43:50.060 --> 00:43:52.050 We've got a bx dot too. 00:43:52.050 --> 00:43:55.650 But I'm just going to do the contribution to Qx 00:43:55.650 --> 00:43:58.360 that comes from this force F1. 00:44:01.040 --> 00:44:11.440 And so Qx delta x is the virtual work done, 00:44:11.440 --> 00:44:21.180 is F1I dot partial of R1 with respect to x delta x. 00:44:26.740 --> 00:44:35.760 But the derivative of R1 with respect to capital 00:44:35.760 --> 00:44:39.820 X, there's no capital X's over here. 00:44:39.820 --> 00:44:42.080 So nothing comes from that. 00:44:42.080 --> 00:44:43.690 There's one right here. 00:44:43.690 --> 00:44:46.530 So the derivative of x with respect to x gives me 1. 00:44:46.530 --> 00:44:49.610 I just get 1 times I back here. 00:44:55.520 --> 00:45:02.280 F1 I hat dot I hat delta x. 00:45:02.280 --> 00:45:09.800 So it's just F1 delta x, which we knew intuitively 00:45:09.800 --> 00:45:11.599 when we worked this problem earlier, when 00:45:11.599 --> 00:45:12.640 we were talking about it. 00:45:12.640 --> 00:45:16.410 The amount of virtual work that gets 00:45:16.410 --> 00:45:23.100 done by this particular force in a deflection, virtual 00:45:23.100 --> 00:45:26.870 displacement delta x, it's just F1 delta x. 00:45:26.870 --> 00:45:29.320 But we've proven it-- we've done it 00:45:29.320 --> 00:45:34.640 in a very precise, kinematic way where 00:45:34.640 --> 00:45:38.780 we found the position vector, worked the whole thing out. 00:45:38.780 --> 00:45:40.080 So that's the simple one. 00:45:40.080 --> 00:45:49.765 Let's now find the harder one, but not much now. 00:45:49.765 --> 00:46:09.600 So we'd like to find Qx1 due to just this F1 only delta x1. 00:46:09.600 --> 00:46:19.690 Well, that should be F1I dot partial of R1 00:46:19.690 --> 00:46:22.826 with respect to x1 delta x1. 00:46:27.030 --> 00:46:33.120 So the derivative of R1 with respect to x1-- this stuff 00:46:33.120 --> 00:46:35.670 has nothing to do with x1. 00:46:35.670 --> 00:46:37.520 The x1 only appears over here. 00:46:37.520 --> 00:46:40.830 And the derivative of this expression, just cosine I minus 00:46:40.830 --> 00:46:59.170 sine theta J. 00:46:59.170 --> 00:47:02.100 So I dot I is the only part you get back. 00:47:02.100 --> 00:47:06.700 This is-- and I need a delta x1. 00:47:06.700 --> 00:47:13.230 F1I dot I is cosine theta delta x1. 00:47:13.230 --> 00:47:30.170 So Qx1 and F1 only here equals F1 cosine theta. 00:47:30.170 --> 00:47:37.100 So this time the motion delta x1, only part of it 00:47:37.100 --> 00:47:40.530 is in the direction of F1. 00:47:40.530 --> 00:47:49.240 And that portion, by taking this derivative here, 00:47:49.240 --> 00:47:53.640 we get the contribution to this that comes from x1. 00:47:53.640 --> 00:47:57.300 And then dotted with the force, we only 00:47:57.300 --> 00:47:59.350 take that component of that motion 00:47:59.350 --> 00:48:01.650 in the direction of the force. 00:48:01.650 --> 00:48:03.550 And that gives us our total virtual work. 00:48:03.550 --> 00:48:06.650 So here is then the total virtual work 00:48:06.650 --> 00:48:11.860 done by F1 due to the little motion delta x1. 00:48:11.860 --> 00:48:14.120 So now we've got the contribution here 00:48:14.120 --> 00:48:23.300 to the generalized force that is associated with deflections 00:48:23.300 --> 00:48:24.405 of coordinate x1. 00:48:28.180 --> 00:48:31.980 Is that the total generalized force 00:48:31.980 --> 00:48:37.990 associated with generalized coordinate x1 in this problem? 00:48:37.990 --> 00:48:42.140 Are there Any other non-conservative forces 00:48:42.140 --> 00:48:46.516 in the problem that move when delta x1 is moved? 00:48:46.516 --> 00:48:48.310 AUDIENCE: What about friction? 00:48:48.310 --> 00:48:50.136 PROFESSOR: Well, let's see. 00:48:50.136 --> 00:48:50.635 Friction. 00:48:53.890 --> 00:48:56.090 Friction, you're presuming, on the wheel? 00:48:56.090 --> 00:48:56.850 OK. 00:48:56.850 --> 00:48:58.740 So he asked about friction on the wheel. 00:48:58.740 --> 00:49:05.340 Well, let's say that there's no slip in this case. 00:49:05.340 --> 00:49:07.960 Then does the friction at the point of contact 00:49:07.960 --> 00:49:10.021 with the wheel do any work? 00:49:10.021 --> 00:49:10.520 No. 00:49:10.520 --> 00:49:14.230 So do we have to account for it as a generalized force? 00:49:14.230 --> 00:49:15.930 No. 00:49:15.930 --> 00:49:18.115 So how about the dashpot? 00:49:20.740 --> 00:49:23.860 So a little virtual deflection, delta x1, 00:49:23.860 --> 00:49:27.090 does it make the big cart move? 00:49:27.090 --> 00:49:27.590 No. 00:49:27.590 --> 00:49:31.580 So are there any other forces in the problem 00:49:31.580 --> 00:49:37.293 that move when you cause a small movement in x1? 00:49:37.293 --> 00:49:38.160 No. 00:49:38.160 --> 00:49:45.040 So in this case, this is the total Qx1. 00:49:45.040 --> 00:49:48.580 Up here, we found Qx, the generalized force 00:49:48.580 --> 00:49:53.820 due to the motion of a cart, the contribution by F1. 00:49:53.820 --> 00:49:55.580 But is there another contribution? 00:49:55.580 --> 00:49:56.376 AUDIENCE: Yes. 00:49:56.376 --> 00:49:57.250 PROFESSOR: And it is? 00:49:57.250 --> 00:49:58.320 AUDIENCE: The dashpot. 00:49:58.320 --> 00:49:59.278 PROFESSOR: The dashpot. 00:49:59.278 --> 00:50:02.440 So we get additionally the total Qx 00:50:02.440 --> 00:50:07.650 here total would be the summation 00:50:07.650 --> 00:50:12.510 of two pieces, an F1 and an F2, which I'd call minus bx. 00:50:12.510 --> 00:50:15.000 It's in the same direction as delta x. 00:50:15.000 --> 00:50:20.700 So you're going to get a minus bx dot plus F1 would 00:50:20.700 --> 00:50:28.100 be the total generalized force in the capital X direction, 00:50:28.100 --> 00:50:29.630 the movement of the main cart. 00:50:37.010 --> 00:50:43.830 So any time you can actually specify a position vector 00:50:43.830 --> 00:50:47.380 to the point of application of an external non-conservative 00:50:47.380 --> 00:50:55.910 force, then you can just plug it into this. 00:50:55.910 --> 00:50:59.280 You do it at each force that's applied. 00:50:59.280 --> 00:51:00.780 You take the derivative with respect 00:51:00.780 --> 00:51:04.810 to that to coordinate qj, delta qj. 00:51:04.810 --> 00:51:09.570 That is the virtual work done by each of these forces. 00:51:09.570 --> 00:51:13.970 And you add up to get the total virtual work done 00:51:13.970 --> 00:51:19.120 due to a deflection at that particular coordinate, j. 00:51:19.120 --> 00:51:24.530 So in the case of capital of Qx, we had two contributions 00:51:24.530 --> 00:51:33.250 because we had two forces on the main cart, F1 and minus bx dot. 00:51:33.250 --> 00:51:36.120 And so the summation in that problem, 00:51:36.120 --> 00:51:40.020 when this is capital X, delta X, you 00:51:40.020 --> 00:51:41.770 have two contributions, F1 and F2. 00:51:45.100 --> 00:51:53.950 Now, what else can I do in the length of time? 00:51:53.950 --> 00:51:57.910 Actually, let me stop for a moment there, think about this. 00:51:57.910 --> 00:52:01.160 Would you have any questions about this? 00:52:01.160 --> 00:52:04.479 So we've described two ways of getting generalized forces. 00:52:04.479 --> 00:52:06.270 One's kind of the intuitive one, figure out 00:52:06.270 --> 00:52:07.686 how much it moves in the direction 00:52:07.686 --> 00:52:09.270 and do the dot product. 00:52:09.270 --> 00:52:11.990 The other one is straight mathematical way. 00:52:11.990 --> 00:52:12.620 Kinematics. 00:52:12.620 --> 00:52:17.520 Plug it in, take the derivative, same thing will come out. 00:52:17.520 --> 00:52:19.340 So while you're thinking about a question, 00:52:19.340 --> 00:52:21.560 I'll look and think what I was going to do next. 00:52:27.930 --> 00:52:32.160 I know what I'll do next, but do you have any questions on this? 00:52:32.160 --> 00:52:34.428 I'm going to do another example of this. 00:52:34.428 --> 00:52:36.404 AUDIENCE: I have a question. 00:52:36.404 --> 00:52:37.890 AUDIENCE: I have a question. 00:52:37.890 --> 00:52:39.236 PROFESSOR: Ah, Christina, yeah. 00:52:39.236 --> 00:52:42.722 AUDIENCE: So I still don't understand how 00:52:42.722 --> 00:52:46.706 if you're going to grab the wheel and move it, 00:52:46.706 --> 00:52:48.698 how that doesn't move the disk. 00:52:48.698 --> 00:52:52.190 Because they're attached, so I don't get it. 00:52:52.190 --> 00:52:57.200 PROFESSOR: Is the wheel-- it's all in how you specify 00:52:57.200 --> 00:52:59.830 your generalized coordinates. 00:52:59.830 --> 00:53:03.900 So in this problem, the two generalized coordinates 00:53:03.900 --> 00:53:07.990 are this capital X in the inertial system which 00:53:07.990 --> 00:53:11.630 describes the motion of the cart, 00:53:11.630 --> 00:53:15.470 and little x1 describes the motion 00:53:15.470 --> 00:53:18.990 of the wheel relative to the cart. 00:53:18.990 --> 00:53:22.960 And actually that allows you to write this statement. 00:53:22.960 --> 00:53:27.760 This is only relative to the cart. 00:53:27.760 --> 00:53:31.320 So the motion of the cart plus the motion of the point 00:53:31.320 --> 00:53:34.510 relative to the cart gives you the total motion. 00:53:34.510 --> 00:53:37.020 And you've picked two coordinates 00:53:37.020 --> 00:53:40.020 that allow you to describe those two things. 00:53:40.020 --> 00:53:45.070 So if you can-- in this problem, if this is the cart, 00:53:45.070 --> 00:53:51.050 this is the wheel, I even have a spring hooked to it here, 00:53:51.050 --> 00:53:56.130 if I move this a little bit, the cart doesn't have to move. 00:53:56.130 --> 00:53:59.870 This going back and forth accounts for x1 relative 00:53:59.870 --> 00:54:00.780 to this table. 00:54:00.780 --> 00:54:02.279 And the table's the cart. 00:54:02.279 --> 00:54:03.195 AUDIENCE: [INAUDIBLE]. 00:54:09.500 --> 00:54:12.140 PROFESSOR: You mean dynamically because you're putting forces 00:54:12.140 --> 00:54:12.990 on it? 00:54:12.990 --> 00:54:13.900 Yeah, well it might. 00:54:13.900 --> 00:54:20.370 But that's not-- in a way, you're asking the question, 00:54:20.370 --> 00:54:24.470 is the cart capable of moving because you 00:54:24.470 --> 00:54:25.990 put a force in the wheel? 00:54:25.990 --> 00:54:29.250 You move the wheel, which puts more spring force, which maybe 00:54:29.250 --> 00:54:30.835 that causes the cart to move. 00:54:30.835 --> 00:54:32.320 Yeah, that could happen. 00:54:32.320 --> 00:54:35.580 But that's not the problem you're 00:54:35.580 --> 00:54:39.100 solving when you're trying to find the generalized forces. 00:54:39.100 --> 00:54:44.830 You're, in fact, allowing a single motion at a time. 00:54:44.830 --> 00:54:47.760 So if you're talking about this motion, 00:54:47.760 --> 00:54:52.080 you have frozen the motion of the main cart. 00:54:52.080 --> 00:54:54.530 And you figure out what the consequence of that is. 00:54:54.530 --> 00:54:56.960 It does a little virtual work because there's a force. 00:54:56.960 --> 00:55:00.600 And you get one of the generalized forces. 00:55:00.600 --> 00:55:03.760 Then if you move the cart, you fix the wheel, 00:55:03.760 --> 00:55:05.036 and the whole cart moves. 00:55:08.180 --> 00:55:11.150 But the amount that this wheel moves 00:55:11.150 --> 00:55:14.140 is exactly equal to the amount that the cart 00:55:14.140 --> 00:55:19.070 moves because you've now fixed this relative position. 00:55:19.070 --> 00:55:22.040 And that's where you get the first-- that's where 00:55:22.040 --> 00:55:25.350 you get the capital Qx term. 00:55:25.350 --> 00:55:30.850 Even know this force F1 is applied here on the wheel, 00:55:30.850 --> 00:55:34.620 this wheel moves when the table moves. 00:55:34.620 --> 00:55:39.460 But the table doesn't move when this relative coordinate 00:55:39.460 --> 00:55:41.650 between the table and here changes. 00:55:41.650 --> 00:55:42.690 It doesn't have to. 00:55:42.690 --> 00:55:45.370 This is free to move when the table is frozen. 00:55:45.370 --> 00:55:48.370 Remember we talked about complete and independent 00:55:48.370 --> 00:55:50.500 coordinates? 00:55:50.500 --> 00:55:57.180 x1 is independent of capital X. If I freeze x1, 00:55:57.180 --> 00:55:59.360 and I make capital X change, just the whole thing 00:55:59.360 --> 00:56:00.980 moves like that. 00:56:00.980 --> 00:56:04.320 If I freeze capital X, x1 can still move. 00:56:08.220 --> 00:56:12.520 So you have to pick independent coordinates. 00:56:12.520 --> 00:56:13.337 Yeah. 00:56:13.337 --> 00:56:15.348 AUDIENCE: [INAUDIBLE] mass of the whole thing 00:56:15.348 --> 00:56:17.097 is much larger than the mass of the wheel? 00:56:17.097 --> 00:56:18.098 PROFESSOR: Not at all. 00:56:18.098 --> 00:56:21.270 AUDIENCE: Because [INAUDIBLE] if you have two massed connected 00:56:21.270 --> 00:56:22.978 by a spring, you pull the first one, 00:56:22.978 --> 00:56:24.442 they kind of pull each other along. 00:56:24.442 --> 00:56:27.074 So why don't you get the pull-along effect over here? 00:56:27.074 --> 00:56:28.490 PROFESSOR: That's a good question. 00:56:28.490 --> 00:56:30.890 It's similar-- it's essentially the same question 00:56:30.890 --> 00:56:32.710 that Christina asked. 00:56:32.710 --> 00:56:35.310 He asked basically-- let's think about it. 00:56:35.310 --> 00:56:37.030 Let's do a problem like that. 00:56:37.030 --> 00:56:41.440 Let's have two carts and a spring in between them, 00:56:41.440 --> 00:56:42.565 and they're both on wheels. 00:57:01.090 --> 00:57:03.230 This is a planar motion problem. 00:57:03.230 --> 00:57:05.780 Each of these bodies is capable in planar motion can 00:57:05.780 --> 00:57:08.460 have x and y and a rotation. 00:57:08.460 --> 00:57:11.990 But because of constraints, how many degrees of freedom 00:57:11.990 --> 00:57:13.030 does body one have? 00:57:17.140 --> 00:57:18.880 I hear one, right? 00:57:18.880 --> 00:57:20.020 I don't allow it to rotate. 00:57:20.020 --> 00:57:21.680 It's got two wheels. 00:57:21.680 --> 00:57:24.705 I don't allow it to go up because it's on the ground. 00:57:24.705 --> 00:57:26.919 I only allow it to move in this direction. 00:57:26.919 --> 00:57:28.210 Same thing to be said for this. 00:57:28.210 --> 00:57:29.310 Only one there. 00:57:29.310 --> 00:57:31.560 How many degrees of freedom do I have in this problem? 00:57:34.660 --> 00:57:37.555 One for each mass, right? 00:57:37.555 --> 00:57:38.930 So I have two degrees of freedom. 00:57:38.930 --> 00:57:42.530 How many generalized coordinates do I need? 00:57:42.530 --> 00:57:47.990 So my generalized coordinate for this one will be x1, 00:57:47.990 --> 00:57:51.595 and for this one will be some x2. 00:57:55.900 --> 00:58:14.130 If you're going to do this problem by Lagrange, and let's 00:58:14.130 --> 00:58:14.740 see. 00:58:14.740 --> 00:58:18.940 Let's put a force now to get back to your question. 00:58:18.940 --> 00:58:22.220 Let's put a force here on one. 00:58:22.220 --> 00:58:23.120 And we'll call it F1. 00:58:27.490 --> 00:58:45.375 And the potential energy-- actually, no, I 00:58:45.375 --> 00:58:46.558 don't want to do that. 00:59:04.860 --> 00:59:09.630 So the potential energy for this is some 1/2k times 00:59:09.630 --> 00:59:12.060 the amount that you stretch the springs, right? 00:59:12.060 --> 00:59:18.030 So you're going to-- the difference between x1 and x2 00:59:18.030 --> 00:59:20.210 minus the unstretched length. 00:59:20.210 --> 00:59:26.300 So x1 minus x2 minus the unstretched length. 00:59:26.300 --> 00:59:28.810 We'll call it l0. 00:59:28.810 --> 00:59:33.450 So this would be the stretch of the springs. 00:59:33.450 --> 00:59:37.120 If the spring had no length, then it 00:59:37.120 --> 00:59:41.640 would just be the difference in these two positions squared. 00:59:41.640 --> 00:59:44.781 So my potential energy looks something like that. 00:59:44.781 --> 00:59:46.280 Now we want to compute the general-- 00:59:46.280 --> 00:59:48.124 and you could use Lagrange, and you 00:59:48.124 --> 00:59:49.790 could figure out two equations of motion 00:59:49.790 --> 00:59:52.626 for this taking your derivatives. 00:59:52.626 --> 00:59:54.000 But the point of the question was 00:59:54.000 --> 00:59:57.840 about getting to the generalized forces, right? 00:59:57.840 --> 01:00:08.100 So now the generalized force Qx1 delta x1 01:00:08.100 --> 01:00:18.610 is F1 times the derivative of R1 with respect to x1 delta x1. 01:00:18.610 --> 01:00:21.480 So how much does the position vector 01:00:21.480 --> 01:00:27.120 from marking the position of this cart, which would be R1-- 01:00:27.120 --> 01:00:31.400 so R1 is in effect x1, right? 01:00:31.400 --> 01:00:32.930 It's a pretty trivial problem. 01:00:32.930 --> 01:00:36.300 So the derivative of R1 with respect to x1 is just 1, 01:00:36.300 --> 01:00:39.750 and the amount that it then moves is delta x1. 01:00:39.750 --> 01:00:44.910 So the virtual work done by this force on that first cart 01:00:44.910 --> 01:00:52.070 is just Qx1 equals F1. 01:00:52.070 --> 01:00:55.190 That's the generalized force caused 01:00:55.190 --> 01:00:57.480 by this first mass on the cart. 01:00:57.480 --> 01:00:59.885 What's the generalize force Qx2? 01:01:04.310 --> 01:01:09.830 Well, it's some F1. 01:01:09.830 --> 01:01:13.040 Actually, there's a dot here. 01:01:13.040 --> 01:01:26.870 This would be some x2 with respect to x1 delta x1. 01:01:26.870 --> 01:01:31.220 But how much does x2 move when you 01:01:31.220 --> 01:01:34.510 cause a little virtual deflection 01:01:34.510 --> 01:01:47.460 of-- the virtual work done if I cause 01:01:47.460 --> 01:01:50.310 a little deflection of this one is 01:01:50.310 --> 01:01:53.800 equal to the summation of the forces that 01:01:53.800 --> 01:01:56.040 act through delta x2. 01:01:56.040 --> 01:02:02.820 Now, if you move this a little delta x1 here, 01:02:02.820 --> 01:02:08.120 we figured out that that Qx1, the generalized force 01:02:08.120 --> 01:02:10.910 due to that, is indeed this. 01:02:10.910 --> 01:02:14.380 But what's the generalized force associated 01:02:14.380 --> 01:02:16.520 with motion delta x2? 01:02:16.520 --> 01:02:20.060 Let's move this one now a little bit. 01:02:20.060 --> 01:02:23.588 When you move that a little bit, how much work does F1 do? 01:02:29.940 --> 01:02:32.430 So we need to get two equations of motion, right? 01:02:32.430 --> 01:02:36.070 And you're going to get 1 by taking derivatives 01:02:36.070 --> 01:02:40.530 with respect to coordinate x1. 01:02:40.530 --> 01:02:45.630 You're going to get an equation of motion, which-- 01:02:45.630 --> 01:02:49.920 so EOM x1 associated with x1 double dot 01:02:49.920 --> 01:02:56.440 here is going to have on the right hand side some Qx1. 01:02:56.440 --> 01:02:57.990 And we figured out what that is. 01:02:57.990 --> 01:02:59.830 It's just F1. 01:02:59.830 --> 01:03:02.800 And we're going to get a second equation of motion associated 01:03:02.800 --> 01:03:05.630 with x2 double dot, the mass acceleration 01:03:05.630 --> 01:03:07.410 of the mass of the second one. 01:03:07.410 --> 01:03:11.585 And it's going to be equal to some external forces. 01:03:11.585 --> 01:03:13.210 And there's other terms in here, right? 01:03:13.210 --> 01:03:15.450 We have kx. 01:03:15.450 --> 01:03:19.191 You have your k terms and so forth in here. 01:03:19.191 --> 01:03:21.690 But on the right hand side are the external non-conservative 01:03:21.690 --> 01:03:22.190 forces. 01:03:22.190 --> 01:03:25.110 So are there any non-conservative forces 01:03:25.110 --> 01:03:26.760 on the second mass? 01:03:26.760 --> 01:03:27.260 None. 01:03:27.260 --> 01:03:32.670 So what do you expect Qx2 to be? 01:03:32.670 --> 01:03:43.610 So to get back to your point, when you're 01:03:43.610 --> 01:03:47.320 computing the generalized forces, 01:03:47.320 --> 01:03:50.650 you freeze all of the movements except one 01:03:50.650 --> 01:03:52.380 and figure out the work done. 01:03:52.380 --> 01:03:56.370 Even though in the real system, force 01:03:56.370 --> 01:03:59.782 will result in this whole system-- 01:03:59.782 --> 01:04:01.490 that whole system will move to the right. 01:04:01.490 --> 01:04:04.580 If I put a steady force F1 on there, 01:04:04.580 --> 01:04:06.810 the whole system will go off to the right hand side. 01:04:06.810 --> 01:04:10.500 That would be the solution to the equations of motion 01:04:10.500 --> 01:04:12.000 that you end up with. 01:04:12.000 --> 01:04:16.790 But for the purpose of computing the generalized force 01:04:16.790 --> 01:04:22.470 on each mass, you only fix where the masses 01:04:22.470 --> 01:04:24.810 are at some instant in time. 01:04:24.810 --> 01:04:27.380 And then one coordinate at a time 01:04:27.380 --> 01:04:28.930 cause a little virtual deflection 01:04:28.930 --> 01:04:31.650 and figure out how much work gets done. 01:04:31.650 --> 01:04:35.480 So see the distinction between the solution 01:04:35.480 --> 01:04:37.060 to the full equation to motion? 01:04:37.060 --> 01:04:37.600 Yes, indeed. 01:04:37.600 --> 01:04:39.940 Everything's going to move because of that force. 01:04:39.940 --> 01:04:41.550 And a little bit of work that gets 01:04:41.550 --> 01:04:45.690 done due to the motion of just one coordinate and then 01:04:45.690 --> 01:04:52.370 the other coordinate through all of the non-conservative forces 01:04:52.370 --> 01:04:55.150 that are applied. 01:04:55.150 --> 01:04:57.110 Did that get to your question? 01:04:57.110 --> 01:04:59.398 All right. 01:04:59.398 --> 01:05:07.410 Now, I'll set up-- I don't think I'll have time to finish this. 01:05:17.460 --> 01:05:21.410 So last time we had this problem, this is this rod. 01:05:21.410 --> 01:05:24.410 It's got a sleeve, and it's got a spring. 01:05:24.410 --> 01:05:27.760 And we figured out the potential and kinetic energy equation 01:05:27.760 --> 01:05:28.720 to motion. 01:05:28.720 --> 01:05:30.191 This is a planar motion problem. 01:05:30.191 --> 01:05:30.690 It pivots. 01:05:33.580 --> 01:05:35.220 Thing can slide up and down. 01:05:35.220 --> 01:05:39.420 Requires an angle theta and a deflection 01:05:39.420 --> 01:05:43.360 x with respect to the rod. 01:05:43.360 --> 01:05:46.610 And we figured out t and v, and we found 01:05:46.610 --> 01:05:49.810 the equations of motion for it. 01:05:49.810 --> 01:05:53.230 So here's my system. 01:05:53.230 --> 01:06:15.130 Point A here, spring, sleeve, theta, x1, y1. 01:06:22.630 --> 01:06:27.450 And the distance x1 was measured from here to G, 01:06:27.450 --> 01:06:28.510 to the center of mass. 01:06:31.820 --> 01:06:32.425 That's x1. 01:06:35.730 --> 01:06:38.030 This is in the direction here. 01:06:38.030 --> 01:06:42.730 This is the i1 unit vector direction. 01:06:42.730 --> 01:06:45.940 And the coordinate system is my x1, y1. 01:06:45.940 --> 01:06:50.180 So x1, it's down the axis like that. 01:06:50.180 --> 01:06:56.270 And this problem is a planar motion problem. 01:06:56.270 --> 01:07:00.180 There's two rigid bodies, the rod and the sleeve. 01:07:00.180 --> 01:07:02.850 And we can completely describe the motion of the system 01:07:02.850 --> 01:07:05.890 with it has two degrees of freedom. 01:07:05.890 --> 01:07:08.480 Theta defines the position of the rod, 01:07:08.480 --> 01:07:11.170 and x1 defines the position of the sleeve on the rod. 01:07:11.170 --> 01:07:13.100 So we have two degrees of freedom. 01:07:13.100 --> 01:07:16.610 And when we work this out, we end up 01:07:16.610 --> 01:07:20.050 with two equations of motion. 01:07:20.050 --> 01:07:24.690 And this was called M2. 01:07:24.690 --> 01:07:25.380 This was M1. 01:07:30.390 --> 01:07:34.140 So one equation of motion was M2 x1 double 01:07:34.140 --> 01:08:15.780 dot That was one equation of motion. 01:08:15.780 --> 01:08:19.300 And I'm just leaving the generalized force 01:08:19.300 --> 01:08:21.529 out of this for a minute. 01:08:21.529 --> 01:08:22.384 We'll figure it out. 01:08:22.384 --> 01:08:23.800 And the second equation of motion. 01:08:34.258 --> 01:08:35.752 This is the rod. 01:09:16.670 --> 01:09:19.410 So you get two equations of motion. 01:09:19.410 --> 01:09:21.630 We worked it out in the last lecture. 01:09:21.630 --> 01:09:27.010 And we can see this accounts for the mass moment 01:09:27.010 --> 01:09:29.250 inertia of the rod, mass moment of inertia 01:09:29.250 --> 01:09:33.910 of the sleeve with respect to G. So parallel axis theorem 01:09:33.910 --> 01:09:36.479 adds another piece to it. 01:09:36.479 --> 01:09:38.680 The whole thing, theta double dot. 01:09:38.680 --> 01:09:41.090 And then this is just due to gravity, 01:09:41.090 --> 01:09:44.560 gravity acting on the rod, gravity acting on the sleeve. 01:09:44.560 --> 01:09:46.109 And on the right hand side, you need 01:09:46.109 --> 01:09:50.439 to get these Qx, your generalized forces 01:09:50.439 --> 01:09:54.130 for generalized coordinate x1 and generalized 01:09:54.130 --> 01:09:55.850 coordinate theta. 01:09:55.850 --> 01:10:00.540 So we did it, in fact worked it out last time 01:10:00.540 --> 01:10:02.970 doing the intuitive approach. 01:10:02.970 --> 01:10:08.670 And what if we were to try to do this then 01:10:08.670 --> 01:10:11.890 by the kinematic approach that I described here? 01:10:21.880 --> 01:10:25.330 The force was applied here. 01:10:25.330 --> 01:10:26.450 It was horizontal. 01:10:26.450 --> 01:10:29.626 We called it F2 because it was applied to mass two. 01:10:37.930 --> 01:10:40.810 And in order to use this technique, 01:10:40.810 --> 01:10:42.940 we want now to compute Qx1. 01:10:50.880 --> 01:11:02.950 We have a force F2, and it is in the-- which 01:11:02.950 --> 01:11:05.890 way do I want to do it? 01:11:05.890 --> 01:11:07.203 We'll have an inertial system. 01:11:17.840 --> 01:11:28.170 So we need to describe a vector in-- I better not 01:11:28.170 --> 01:11:30.650 draw it out here. 01:11:30.650 --> 01:11:32.660 So I'll just set this problem up, 01:11:32.660 --> 01:11:34.670 and we'll finish it next time. 01:11:34.670 --> 01:11:41.210 I have an inertial system y and x. 01:11:44.480 --> 01:11:46.520 So that in this system, this force 01:11:46.520 --> 01:11:52.550 would be F2 capital J dotted with the derivative 01:11:52.550 --> 01:11:58.970 of some vector that runs here to this point. 01:11:58.970 --> 01:12:04.470 And I'll call that R2 in o. 01:12:04.470 --> 01:12:13.350 So the derivative of R2 with respect to x1 delta x1. 01:12:16.731 --> 01:12:21.840 And if you can work this out, then you're done. 01:12:21.840 --> 01:12:25.830 The key to this is figuring out what is R2, 01:12:25.830 --> 01:12:28.990 and doing it in unit vectors such 01:12:28.990 --> 01:12:30.930 that you can complete this dot product. 01:12:33.950 --> 01:12:42.330 So how would you describe what is this position to here, 01:12:42.330 --> 01:12:46.770 and what are its unit vectors that break it down? 01:12:46.770 --> 01:12:50.560 Take this R2, and you express it in terms 01:12:50.560 --> 01:12:57.300 of unit vectors in the inertial capital I capital J system. 01:12:57.300 --> 01:13:01.590 And once you've done that, you can take the derivative, 01:13:01.590 --> 01:13:05.600 dot it with that, and you're done. 01:13:05.600 --> 01:13:07.019 So we'll finish that next time. 01:13:07.019 --> 01:13:09.060 You might go off and think about it in your notes 01:13:09.060 --> 01:13:09.640 from last time. 01:13:09.640 --> 01:13:10.806 We already figured this out. 01:13:10.806 --> 01:13:12.910 We just did it the intuitive way. 01:13:12.910 --> 01:13:15.420 Drew the F and figured out which part's in the direction 01:13:15.420 --> 01:13:18.980 of x1, which part's in the direction of delta theta. 01:13:18.980 --> 01:13:19.900 And we figured it out. 01:13:19.900 --> 01:13:21.770 So you actually already have the answer. 01:13:21.770 --> 01:13:24.390 So go see if you can do that.