WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.810 Commons license. 00:00:03.810 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.305 at ocw.mit.edu. 00:00:21.800 --> 00:00:26.650 J. KIM VANDIVER: If I wanted you to remember three equations, 00:00:26.650 --> 00:00:29.530 really commit them to memory, that'd allow you to essentially 00:00:29.530 --> 00:00:35.012 do everything that we've done, they'd be the following three. 00:00:35.012 --> 00:00:41.450 The first would essentially be Newton's second law. 00:00:49.640 --> 00:00:53.500 We use that all the time, and in vector form, 00:00:53.500 --> 00:00:58.850 summation of the external forces on a rigid body 00:00:58.850 --> 00:01:02.710 must be equal to the time rate of change 00:01:02.710 --> 00:01:07.630 of the linear momentum of the rigid body. 00:01:07.630 --> 00:01:13.100 And that we know is equal to its mass times the acceleration 00:01:13.100 --> 00:01:19.060 of its center of mass with respect to an inertial frame, 00:01:19.060 --> 00:01:22.770 and this slash o means with respect to an inertial frame. 00:01:22.770 --> 00:01:25.070 That's Newton's second law, and of course we 00:01:25.070 --> 00:01:27.160 use it all the time. 00:01:27.160 --> 00:01:35.712 Now Euler added to what Newton worked out for us. 00:01:35.712 --> 00:01:38.020 And let me actually just point out over here. 00:01:38.020 --> 00:01:44.780 So P with respect to o then is M velocity 00:01:44.780 --> 00:01:49.955 of G with respect to the inertial frame-- vector, 00:01:49.955 --> 00:01:52.260 vector, vector. 00:01:52.260 --> 00:01:57.840 OK, what Euler taught us was that the summation 00:01:57.840 --> 00:02:07.530 of the torques about a point A is the time rate of change 00:02:07.530 --> 00:02:16.050 of the angular momentum with respect to that point 00:02:16.050 --> 00:02:24.900 plus the velocity of the point cross P with respect to o. 00:02:24.900 --> 00:02:32.150 So that's our general equation for angular momentum 00:02:32.150 --> 00:02:33.460 and [? R ?] for torques. 00:02:33.460 --> 00:02:46.330 And remember, H with respect to A is the cross product of R-- 00:02:46.330 --> 00:02:47.741 how do I want to write this guy? 00:02:57.595 --> 00:02:58.678 I don't want to save that. 00:03:15.130 --> 00:03:17.030 So we define angular momentum. 00:03:17.030 --> 00:03:22.090 It's just a cross product of the distance from the point 00:03:22.090 --> 00:03:29.810 that we're at to the-- I didn't write this right. 00:03:29.810 --> 00:03:40.590 This is A with respect to G cross P o, 00:03:40.590 --> 00:03:44.010 so that's our general definition of angular momentum. 00:03:44.010 --> 00:03:46.130 And finally, the third equation is the one 00:03:46.130 --> 00:03:49.760 that I showed you last time, which does appear in the book 00:03:49.760 --> 00:03:52.190 but not until you get into chapter 21, 00:03:52.190 --> 00:03:55.950 and that is that you can write H with respect 00:03:55.950 --> 00:04:11.365 to A as H with respect to G plus [? RG/A ?] cross P with respect 00:04:11.365 --> 00:04:12.082 to o. 00:04:15.910 --> 00:04:18.420 There's something wrong here. 00:04:18.420 --> 00:04:19.839 Matt, what have I done wrong? 00:04:19.839 --> 00:04:22.840 MATT: Do you want a sum that will give you 00:04:22.840 --> 00:04:24.302 the angle [INAUDIBLE]. 00:04:24.302 --> 00:04:26.260 J. KIM VANDIVER: Yeah, it just slipped my brain 00:04:26.260 --> 00:04:27.176 because I don't do it. 00:04:27.176 --> 00:04:28.670 I don't think about this. 00:04:28.670 --> 00:04:31.390 This is the-- yeah, it's a summation. 00:04:31.390 --> 00:04:34.230 There we go. 00:04:34.230 --> 00:04:37.780 In general, this is a summation, and it 00:04:37.780 --> 00:04:45.040 is the location of all the little particles i with respect 00:04:45.040 --> 00:04:55.220 to A cross the linear momentum of each particle [? mi ?] Vi 00:04:55.220 --> 00:04:57.490 with respect to o. 00:04:57.490 --> 00:04:59.890 And we add all that up, that's the definition 00:04:59.890 --> 00:05:01.300 of angular momentum. 00:05:01.300 --> 00:05:04.330 And we generally for rigid bodies then 00:05:04.330 --> 00:05:10.140 reduce that to saying that it's a mass moment of inertia 00:05:10.140 --> 00:05:17.510 matrix computed with respect to A times omega x, omega y, 00:05:17.510 --> 00:05:19.710 omega z. 00:05:19.710 --> 00:05:23.470 That's how we generally express this. 00:05:23.470 --> 00:05:28.400 But the third equation I want to come to is this one. 00:05:28.400 --> 00:05:30.266 Yeah? 00:05:30.266 --> 00:05:34.129 AUDIENCE: That's not [INAUDIBLE], 00:05:34.129 --> 00:05:36.170 but it's not really with respect to A in general. 00:05:39.606 --> 00:05:40.106 [INAUDIBLE] 00:05:45.050 --> 00:05:48.160 J. KIM VANDIVER: This is i with respect to A. Let's just 00:05:48.160 --> 00:05:52.590 do this because I've just slipped a bit this morning. 00:05:52.590 --> 00:05:56.230 OK, I'm thinking too much about where I'm going here. 00:05:56.230 --> 00:05:58.670 This equation I introduced last time. 00:05:58.670 --> 00:06:01.060 The book doesn't make a big deal about this at all, 00:06:01.060 --> 00:06:03.930 but it's really a fundamental equation 00:06:03.930 --> 00:06:08.870 that we proved from scratch very simply last time. 00:06:08.870 --> 00:06:13.120 And this equation allows you to solve lots of problems 00:06:13.120 --> 00:06:17.050 without even, for example, knowing the parallel axis 00:06:17.050 --> 00:06:17.780 theorem. 00:06:17.780 --> 00:06:20.850 It'll just get you through it even without knowing things 00:06:20.850 --> 00:06:21.670 like that. 00:06:21.670 --> 00:06:24.260 So with these three equations, you can do all the problems 00:06:24.260 --> 00:06:27.680 that we've done, and I'm going to do two or three examples 00:06:27.680 --> 00:06:30.160 this morning, and including coming back 00:06:30.160 --> 00:06:33.050 to a discussion of the parallel axis theorem. 00:06:33.050 --> 00:06:39.849 OK, so these three are the ones I recommend that you remember 00:06:39.849 --> 00:06:41.890 because everything else will flow from-- we know, 00:06:41.890 --> 00:06:43.790 like, we have special cases. 00:06:43.790 --> 00:06:58.380 So when A is at G, then this just 00:06:58.380 --> 00:07:01.490 reduces to H with respect to G. When 00:07:01.490 --> 00:07:04.840 there's no vel-- when A doesn't move, this term goes away. 00:07:04.840 --> 00:07:06.944 Those are all special cases, which you can just 00:07:06.944 --> 00:07:09.360 substitute in the numbers and discover for yourself if you 00:07:09.360 --> 00:07:10.796 don't remember those. 00:07:10.796 --> 00:07:12.170 But these are the three equations 00:07:12.170 --> 00:07:14.600 you want to have in your kit to use. 00:07:36.130 --> 00:07:38.940 So last time, I kind of classified problems 00:07:38.940 --> 00:07:40.520 in four different classes depending 00:07:40.520 --> 00:07:42.060 on the simplifications. 00:07:42.060 --> 00:07:45.720 And the class 4 problem was the more general case, 00:07:45.720 --> 00:07:51.630 and that's when the point A is allowed to move in general, 00:07:51.630 --> 00:07:56.160 and that last equation we had can 00:07:56.160 --> 00:07:58.060 be very handy in those cases. 00:07:58.060 --> 00:08:02.080 And I mentioned a problem last time, but I didn't work it out. 00:08:02.080 --> 00:08:06.450 And that's simply this problem, where 00:08:06.450 --> 00:08:11.330 you've got a block, maybe a box on a cart, 00:08:11.330 --> 00:08:15.300 and you're accelerating the cart. 00:08:15.300 --> 00:08:17.960 At what point will the block tip over? 00:08:17.960 --> 00:08:21.690 So this one I've put a nail right here, 00:08:21.690 --> 00:08:27.560 so the block can't slide, but it will tip about this point. 00:08:27.560 --> 00:08:30.730 So if I accelerate it slowly, nothing happens. 00:08:30.730 --> 00:08:35.179 If I accelerate it fast enough, it tips around that point. 00:08:35.179 --> 00:08:36.299 So that's the key. 00:08:36.299 --> 00:08:39.860 Identifying the point's have the work, so that's the key bit. 00:08:39.860 --> 00:08:42.929 What's the greatest acceleration that I can have 00:08:42.929 --> 00:08:47.460 so that this block will just not quite tip? 00:08:47.460 --> 00:08:50.134 That's the problem. 00:08:50.134 --> 00:08:51.550 If you didn't have the nail there, 00:08:51.550 --> 00:08:53.174 there's also the possibility that it'll 00:08:53.174 --> 00:08:54.562 slip without tipping over. 00:08:54.562 --> 00:08:56.380 OK, so that's another problem that we'll 00:08:56.380 --> 00:08:59.740 address at another moment. 00:08:59.740 --> 00:09:01.620 OK, so let's do this problem quickly. 00:09:05.920 --> 00:09:13.490 So this is our problem, and I've got a handle here, 00:09:13.490 --> 00:09:16.426 and I'm pushing on it with a force. 00:09:16.426 --> 00:09:18.480 I've got a couple of wheels. 00:09:18.480 --> 00:09:20.424 This is M1. 00:09:20.424 --> 00:09:28.770 This is M2, and I'll have an inertial frame here X, Y. 00:09:28.770 --> 00:09:32.640 So the question is what's the maximum force 00:09:32.640 --> 00:09:36.220 that I can apply such that the block won't tip over? 00:09:36.220 --> 00:09:41.910 And I've got a little nub there, so this thing can't slide. 00:09:41.910 --> 00:09:45.880 OK, and this is my point A because that's the point 00:09:45.880 --> 00:09:48.720 I expect it to rotate around. 00:09:48.720 --> 00:09:52.560 The X, Y, and Z is coming out of the board. 00:09:52.560 --> 00:09:58.920 OK, so Newton's second, so we apply it first. 00:09:58.920 --> 00:10:00.630 From Newton's second law, we say the sum 00:10:00.630 --> 00:10:05.582 of the forces, in this case, on the system. 00:10:05.582 --> 00:10:10.270 By the system I mean M1 and M2. 00:10:10.270 --> 00:10:12.900 We want to remember that you can collect things together 00:10:12.900 --> 00:10:13.550 at times. 00:10:13.550 --> 00:10:15.930 So the sum of forces on the system, in this case, 00:10:15.930 --> 00:10:26.190 is then M1 plus M2, and it's the acceleration of that point, 00:10:26.190 --> 00:10:33.590 and in this case we are just going to do in the X-direction 00:10:33.590 --> 00:10:37.720 because we know that's the only one where there's any action. 00:10:37.720 --> 00:10:40.190 So the sum of the forces in the X-direction 00:10:40.190 --> 00:10:43.055 is the total mass of the system times the acceleration 00:10:43.055 --> 00:10:44.930 in the X-direction. 00:10:44.930 --> 00:10:47.390 So the X double dot we're looking 00:10:47.390 --> 00:10:53.370 for is F over M1 plus M2. 00:10:53.370 --> 00:10:56.120 So this we're trying to find out what the maximum value for this 00:10:56.120 --> 00:10:56.620 is. 00:11:05.200 --> 00:11:07.400 Now next we can apply the torque equation 00:11:07.400 --> 00:11:09.010 because that's the key one. 00:11:09.010 --> 00:11:11.550 And that's the sum of the torques-- now it's 00:11:11.550 --> 00:11:18.930 just on the box-- at with respect to point A. 00:11:18.930 --> 00:11:20.900 And we can say, well, looking at that, 00:11:20.900 --> 00:11:23.010 we need a free body diagram now of our box. 00:11:26.780 --> 00:11:29.790 You've got an M1g down. 00:11:36.760 --> 00:11:41.350 And you might have a normal force upwards, generally, 00:11:41.350 --> 00:11:44.810 but now we want to think about where 00:11:44.810 --> 00:11:50.000 are the forces going to be acting on the box 00:11:50.000 --> 00:11:53.770 when it's just at this point that it's just barely about-- 00:11:53.770 --> 00:11:55.270 just barely starts to tip. 00:11:58.640 --> 00:12:03.110 So we're hypothesizing that here at A, there'll 00:12:03.110 --> 00:12:07.300 be some upward force, right? 00:12:07.300 --> 00:12:10.330 And it's going to take care of the static equilibrium, 00:12:10.330 --> 00:12:15.690 but it also could enter into our moment calculation 00:12:15.690 --> 00:12:17.490 if we weren't clever with respect to where 00:12:17.490 --> 00:12:18.720 we calculate this point. 00:12:22.960 --> 00:12:25.780 So the sum of the torques about this point, 00:12:25.780 --> 00:12:29.370 we don't have to bring this one into it. 00:12:29.370 --> 00:12:31.320 We do have to consider this one. 00:12:31.320 --> 00:12:39.760 So the [? R ?] here cross that force, so we get a moment, Mg, 00:12:39.760 --> 00:12:42.836 and I need some dimensions on my box. 00:12:42.836 --> 00:12:53.260 We'll call it b and a height h. 00:12:53.260 --> 00:12:57.080 So the moment arm, this axon is b/2, 00:12:57.080 --> 00:12:59.780 and it's in the minus [? z ?] direction, 00:12:59.780 --> 00:13:08.680 so I get a minus M1g b/2 k hat. 00:13:08.680 --> 00:13:10.980 Those are my external torques on the box 00:13:10.980 --> 00:13:14.650 because the normal force supporting it on the floor 00:13:14.650 --> 00:13:15.710 is acting at A now. 00:13:19.060 --> 00:13:29.950 And this must be from our second formula, 00:13:29.950 --> 00:13:34.050 dH with respect to A/dt plus the velocity of A 00:13:34.050 --> 00:13:39.000 with respect to o, cross P with respect to o. 00:13:39.000 --> 00:13:49.990 And P with respect to o was [? M1VG ?] with respect 00:13:49.990 --> 00:13:55.710 to o, all right? 00:13:55.710 --> 00:13:59.060 So we need to know what is the velocity of A 00:13:59.060 --> 00:14:00.620 with respect to o. 00:14:03.650 --> 00:14:08.220 But we're only allowing motions in this direction. 00:14:08.220 --> 00:14:10.950 This has to be evaluated in the inertial frame, 00:14:10.950 --> 00:14:15.200 so we have a coordinate X here. 00:14:15.200 --> 00:14:17.960 So it's going to be some X dot, and I'll 00:14:17.960 --> 00:14:20.180 give you a capital I hat just remind you 00:14:20.180 --> 00:14:23.230 that's in the inertial frame. 00:14:23.230 --> 00:14:26.760 What is the velocity of G with respect to o? 00:14:34.253 --> 00:14:35.544 AUDIENCE: Would it be the same? 00:14:35.544 --> 00:14:37.008 J. KIM VANDIVER: Yeah. 00:14:37.008 --> 00:14:38.740 You know, in this condition we're 00:14:38.740 --> 00:14:41.130 saying we want the block not to quite tip over, 00:14:41.130 --> 00:14:43.810 and that means it's moving with the cart at exactly 00:14:43.810 --> 00:14:45.800 the same velocity, so it's the same. 00:14:45.800 --> 00:14:49.260 It's also equal to A with respect to o, 00:14:49.260 --> 00:14:51.260 but if that's the case, what's the cross product 00:14:51.260 --> 00:14:52.884 between velocity of A with respect to o 00:14:52.884 --> 00:14:54.811 and G with respect to o? 00:14:54.811 --> 00:14:56.330 Ah, so that one goes to zero. 00:14:59.680 --> 00:15:00.770 That's because of this. 00:15:09.170 --> 00:15:15.630 OK, so that leaves us with-- we don't 00:15:15.630 --> 00:15:17.780 have to deal with these terms, but we 00:15:17.780 --> 00:15:25.210 do need to compute H with respect to A. 00:15:25.210 --> 00:15:28.350 And now I'm going to use this formulation because it makes 00:15:28.350 --> 00:15:30.630 this problem easy to do and not have 00:15:30.630 --> 00:15:33.760 to deal with parallel axes or any of that. 00:15:33.760 --> 00:15:36.020 So I'm going to use my third equation, 00:15:36.020 --> 00:15:38.860 and it says that this is H with respect 00:15:38.860 --> 00:15:47.380 to G plus RG with respect to A cross P with respect to o. 00:15:52.280 --> 00:15:56.590 So you notice I chose this problem kind of on purpose 00:15:56.590 --> 00:15:59.250 today because I started saying, here's three 00:15:59.250 --> 00:16:00.990 really important equations. 00:16:00.990 --> 00:16:04.850 We use all three to do this problem. 00:16:04.850 --> 00:16:06.440 OK, so now we're exercising the third. 00:16:19.940 --> 00:16:26.720 So my block here, since we're dealing 00:16:26.720 --> 00:16:28.510 with angular momentum of rigid bodies, 00:16:28.510 --> 00:16:32.130 I now have to think in terms of a coordinate system attached 00:16:32.130 --> 00:16:39.980 to the block, and I'll formulate it 00:16:39.980 --> 00:16:42.840 so it's lined up the same way. 00:16:42.840 --> 00:16:50.480 So this is a x1 I'll call it, y1, and z1. 00:16:50.480 --> 00:17:03.550 But x1 and big X, these are in the same direction, 00:17:03.550 --> 00:17:06.599 and so is z and z1 and capital [? E. ?] 00:17:06.599 --> 00:17:12.020 And that means that my I hats are the same as the unit 00:17:12.020 --> 00:17:15.890 vectors associated with my coordinate system attached 00:17:15.890 --> 00:17:16.690 to the body. 00:17:16.690 --> 00:17:19.425 Remember, this is this body-fixed coordinate system 00:17:19.425 --> 00:17:22.260 that allows us to compute moments of inertia, 00:17:22.260 --> 00:17:25.470 and I'm going to have this located at G, OK? 00:17:49.660 --> 00:17:53.300 So the only rotation that we're allowing in this problem. 00:17:53.300 --> 00:17:57.550 This is basically a 2D planar motion problem. 00:17:57.550 --> 00:18:00.510 It could conceivably have x and y translations 00:18:00.510 --> 00:18:05.600 and a z rotation constrained in the y, 00:18:05.600 --> 00:18:09.650 so it has possible rotation and translation. 00:18:09.650 --> 00:18:16.680 So the only omega we're considering is this omega z, 00:18:16.680 --> 00:18:22.415 and so we'll proceed to use it to compute H. 00:18:22.415 --> 00:18:23.880 So we use this formula. 00:18:29.510 --> 00:18:33.507 So H with respect to G is-- just I'm 00:18:33.507 --> 00:18:35.715 doing a couple of these things sort of carefully just 00:18:35.715 --> 00:18:38.970 to remind you of where these things all come from. 00:18:38.970 --> 00:18:46.580 This is I with respect to G, and in this case, 0 0 omega z. 00:18:46.580 --> 00:18:51.810 And we come out of this then the only-- 00:18:51.810 --> 00:18:55.540 are these principal axes? 00:18:55.540 --> 00:19:00.410 So if they are, this is diagonal, right, 00:19:00.410 --> 00:19:04.460 so I've located these through G. 00:19:04.460 --> 00:19:06.490 Taken into consideration the symmetries, 00:19:06.490 --> 00:19:09.460 I know because the symmetry of these are principal axes. 00:19:09.460 --> 00:19:14.140 Therefore, this is a diagonal mass moment of inertia matrix, 00:19:14.140 --> 00:19:17.140 so when I calculate to do out the multiplication, 00:19:17.140 --> 00:19:27.420 I get Izz with respect to G omega z k-hat. 00:19:27.420 --> 00:19:31.420 There's only one vector component of H 00:19:31.420 --> 00:19:35.250 that comes out of that, and that's Hz 00:19:35.250 --> 00:19:41.680 is that Izz G omega z k-hat. 00:19:41.680 --> 00:19:46.880 And there are two other possible components of H, 00:19:46.880 --> 00:19:48.390 but they're both 0. 00:19:48.390 --> 00:19:52.720 There's no other rotate-- there is no angular momentum 00:19:52.720 --> 00:19:53.950 in the x- or y-directions. 00:19:56.390 --> 00:19:56.890 OK. 00:20:08.970 --> 00:20:14.050 [? RG/A ?] in this problem is from here-- this 00:20:14.050 --> 00:20:29.960 is point A-- two there, so RG respect to A is b/2 i plus 00:20:29.960 --> 00:20:33.190 h/2 k. 00:20:33.190 --> 00:20:37.490 It's the location of just the position of G with respect 00:20:37.490 --> 00:20:44.000 to A. [? b/2 ?] over [? h/2 up. ?] So in order 00:20:44.000 --> 00:20:46.755 to complete my calculation of H of A, 00:20:46.755 --> 00:20:50.085 I need to get the second piece of the [? RG/A ?] cross P/o. 00:21:07.730 --> 00:21:20.170 So [? RG/A ?] is my b/2 i plus h/2 k cross. 00:21:20.170 --> 00:21:30.560 And then my linear momentum is M1 x1 dot 00:21:30.560 --> 00:21:33.064 in the i hat direction. 00:21:33.064 --> 00:21:35.230 And it'll either be a capitalize I or a lowercase i. 00:21:35.230 --> 00:21:37.110 They're the same, all right? 00:21:37.110 --> 00:21:39.970 So this is my second term the RG cross 00:21:39.970 --> 00:21:46.340 P. i cross i, these two-- this pair gives you nothing, 00:21:46.340 --> 00:21:52.260 so it's k cross i is j, and so you get Izz with respect 00:21:52.260 --> 00:21:55.590 to G omega z. 00:21:55.590 --> 00:22:11.790 And then a k term from here M1 h/2 00:22:11.790 --> 00:22:15.390 x dot, and it's also in the k. 00:22:19.190 --> 00:22:25.248 And that's my total angular momentum now with respect to A. 00:22:25.248 --> 00:22:26.660 AUDIENCE: [INAUDIBLE]? 00:22:26.660 --> 00:22:34.700 J. KIM VANDIVER: And I think-- just a sec-- k-- you're right. 00:22:34.700 --> 00:22:35.820 Wait a second. 00:22:35.820 --> 00:22:38.856 k cross i should give me a j, right? 00:22:38.856 --> 00:22:39.772 AUDIENCE: [INAUDIBLE]. 00:22:46.506 --> 00:22:48.430 Professor? 00:22:48.430 --> 00:22:52.890 J. KIM VANDIVER: Yeah, k cross i gives me a positive j, right? 00:22:52.890 --> 00:22:53.870 [INTERPOSING VOICES] 00:22:53.870 --> 00:22:57.496 [? AUDIENCE: RGA ?] should be a positive j based off 00:22:57.496 --> 00:22:58.620 of your [INAUDIBLE] system. 00:22:58.620 --> 00:23:01.340 J. KIM VANDIVER: Ah, yeah, yeah. 00:23:01.340 --> 00:23:06.510 This is a good catch. 00:23:06.510 --> 00:23:08.120 All right, so what happens now? 00:23:08.120 --> 00:23:13.155 Now we get j cross i, gives you a minus k. 00:23:19.670 --> 00:23:23.290 Now that should be OK, and let's see if that 00:23:23.290 --> 00:23:24.601 agrees with what I wrote here. 00:23:24.601 --> 00:23:26.100 Yes, they did it right in the paper. 00:23:26.100 --> 00:23:27.766 Just couldn't put it right on the board. 00:23:50.060 --> 00:23:54.370 So now the sum of the torques with respect to A, 00:23:54.370 --> 00:24:01.030 we've already figured out that it's minus M1 g b/2 k 00:24:01.030 --> 00:24:09.255 hat must be equal to the time rate of change 00:24:09.255 --> 00:24:11.810 from our second equation dH dt. 00:24:11.810 --> 00:24:14.910 And we've already figured out that the second term goes away. 00:24:14.910 --> 00:24:17.775 So it's just then the time rate of change of that expression. 00:24:28.157 --> 00:24:30.830 Oh, I think we could probably do it straight away, 00:24:30.830 --> 00:24:33.810 so the time rate of change, the derivative of this, 00:24:33.810 --> 00:24:35.570 you get omega z dot. 00:24:35.570 --> 00:24:37.080 You get an x double dot. 00:24:37.080 --> 00:24:40.590 The k doesn't rotate, so there's nothing that comes from that. 00:24:40.590 --> 00:24:41.990 If you take that time derivative, 00:24:41.990 --> 00:24:56.260 you get Izz with respect to G omega z dot minus M1 h/2 00:24:56.260 --> 00:25:02.090 x double dot, and these are still all in the k direction. 00:25:02.090 --> 00:25:05.300 And that's an equation that only has 00:25:05.300 --> 00:25:09.020 k hat terms on the left hand side, k on the right, 00:25:09.020 --> 00:25:13.500 so this is just one potential component of the torque. 00:25:13.500 --> 00:25:16.860 And so we can drop the k's at this point. 00:25:16.860 --> 00:25:20.470 And we now have an expression for the external torques 00:25:20.470 --> 00:25:24.850 in terms of the accelerations of the system. 00:25:24.850 --> 00:25:27.480 This is essentially theta double dot [? if it ?] tips, 00:25:27.480 --> 00:25:30.120 and that's the linear acceleration. 00:25:30.120 --> 00:25:33.150 But now then the key to the problem 00:25:33.150 --> 00:25:36.590 is we're trying to find out when just at the point 00:25:36.590 --> 00:25:42.242 it might tip but doesn't, so what's omega dot-- omega z dot? 00:25:42.242 --> 00:25:46.640 OK, so what we're looking for we let omega z 00:25:46.640 --> 00:25:48.985 dot or require that to be 0. 00:25:48.985 --> 00:25:51.150 I mean this makes this even simpler, 00:25:51.150 --> 00:25:57.260 and you find out then that M1 g [? b/2 ?] 00:25:57.260 --> 00:26:01.707 equals M1 h/2 x double dot. 00:26:01.707 --> 00:26:03.540 And I have to add minus signs on both sides, 00:26:03.540 --> 00:26:04.820 so I got rid of them. 00:26:04.820 --> 00:26:10.415 And the M1s go away, and I can solve for h for x double dot, 00:26:10.415 --> 00:26:13.870 and that's going to be g b/h. 00:26:19.140 --> 00:26:23.290 OK, and we started off saying what's the maximum force? 00:26:23.290 --> 00:26:28.700 So this is now x max, the acceleration maximum 00:26:28.700 --> 00:26:38.480 and the force maximum is just equal to M1 plus M2 00:26:38.480 --> 00:26:51.350 x dot max for-- and let's see if that makes sense. 00:26:51.350 --> 00:26:59.950 If b and h were equal-- that's sort of a common sense 00:26:59.950 --> 00:27:05.270 argument here-- b and h were equal, 00:27:05.270 --> 00:27:10.290 that tells you that you could accelerate 1g at 1g-- just 1g 00:27:10.290 --> 00:27:11.320 and that makes sense. 00:27:11.320 --> 00:27:16.182 The Mg here is [? putting a ?] restoring moment down of the MG 00:27:16.182 --> 00:27:21.760 [? b/2. ?] And the acceleration of this is putting 00:27:21.760 --> 00:27:25.870 an overturning moment on it in the other direction that would 00:27:25.870 --> 00:27:30.750 be Mg [? h/2, ?] and if h and b are the same, 00:27:30.750 --> 00:27:33.250 then it would be exactly 1g. 00:27:33.250 --> 00:27:36.240 Now so that's the answer the problem. 00:27:36.240 --> 00:27:41.630 If you want to do live dangerously from the beginning, 00:27:41.630 --> 00:27:45.410 you could have done this with a fictitious force. 00:27:45.410 --> 00:27:47.730 I just mention in passing I'm not recommending 00:27:47.730 --> 00:27:48.910 that you generally go there. 00:27:48.910 --> 00:27:51.260 This was a very straightforward way of doing it. 00:27:51.260 --> 00:27:54.060 We just started with these three laws, 00:27:54.060 --> 00:27:58.890 and we just worked our way through the problem and all 00:27:58.890 --> 00:28:00.980 the way to the end, and it fell out. 00:28:00.980 --> 00:28:03.260 If you didn't want to live dangerously though, 00:28:03.260 --> 00:28:15.930 you could say that when you accelerate this, 00:28:15.930 --> 00:28:19.450 you can think of there being a fictitious force that 00:28:19.450 --> 00:28:25.340 is equal to minus the mass of this object 00:28:25.340 --> 00:28:28.460 times the acceleration on the center of gravity of force 00:28:28.460 --> 00:28:32.400 that is pulling it in the other direction. 00:28:32.400 --> 00:28:40.584 So you could think of there being a force M1X double dot 00:28:40.584 --> 00:28:42.500 opposite to the direction of the acceleration. 00:28:42.500 --> 00:28:45.640 This is minus the mass times the acceleration. 00:28:45.640 --> 00:28:47.710 It's the fictitious force. 00:28:47.710 --> 00:28:54.860 Here's your point A. Here's gravity M1 g. 00:28:54.860 --> 00:28:57.040 This is the distance b/2. 00:28:59.570 --> 00:29:03.960 This is the height h/2, and you could say just 00:29:03.960 --> 00:29:09.140 at that moment of balance, the sum of those two moment 00:29:09.140 --> 00:29:11.445 should be 0-- the sub of the external torques. 00:29:15.030 --> 00:29:21.950 And you would end up with a M1 x double dot 00:29:21.950 --> 00:29:29.100 h/2 minus M1 g [? b/2, ?] and you come up 00:29:29.100 --> 00:29:30.700 with the same answer. 00:29:30.700 --> 00:29:35.790 But for most of this that takes a real leap of faith 00:29:35.790 --> 00:29:39.340 to do that to convince yourself that you're right, right? 00:29:39.340 --> 00:29:40.735 You got to know a lot of dynamics 00:29:40.735 --> 00:29:43.170 and remember all the parts and pieces 00:29:43.170 --> 00:29:47.270 to be able to just go there. 00:29:47.270 --> 00:29:50.746 But if you do it just carefully, those three equations, 00:29:50.746 --> 00:29:51.620 it'll all worked out. 00:29:51.620 --> 00:29:52.280 Yeah? 00:29:52.280 --> 00:29:54.840 AUDIENCE: Why is omega z [INAUDIBLE]? 00:29:54.840 --> 00:29:56.440 J. KIM VANDIVER: Because as long as 00:29:56.440 --> 00:30:00.040 the condition satisfied that it doesn't tip over, 00:30:00.040 --> 00:30:03.336 what's the rotation rate of this block? 00:30:03.336 --> 00:30:04.260 AUDIENCE: 0. 00:30:04.260 --> 00:30:05.830 J. KIM VANDIVER: 0. 00:30:05.830 --> 00:30:09.820 So we're not solving for dynamics in this problem 00:30:09.820 --> 00:30:11.040 of the thing tipping. 00:30:11.040 --> 00:30:12.570 It isn't rolling on us. 00:30:12.570 --> 00:30:15.330 We're coming just up to the point 00:30:15.330 --> 00:30:17.970 that it tips, and say we're not going any farther. 00:30:17.970 --> 00:30:22.820 And so we just require the two moments-- the one essentially 00:30:22.820 --> 00:30:26.990 caused by the acceleration, this fictitious force, that's 00:30:26.990 --> 00:30:30.400 got to be just in equilibrium with the restoring moment 00:30:30.400 --> 00:30:32.230 provided by Mg. 00:30:32.230 --> 00:30:34.400 That's essentially the problem we've worked, 00:30:34.400 --> 00:30:36.880 but we've done it really carefully just using 00:30:36.880 --> 00:30:40.090 these three laws. 00:30:40.090 --> 00:30:43.700 And the real problem though if you didn't have the nail here 00:30:43.700 --> 00:30:48.210 is if now if you need to test your solution, 00:30:48.210 --> 00:30:53.330 could you ever actually reach that rate of acceleration 00:30:53.330 --> 00:30:55.870 before the thing starts sliding on you? 00:30:55.870 --> 00:30:57.070 And, you know, maybe not. 00:30:57.070 --> 00:31:01.305 This thing slides before it tips, OK? 00:31:01.305 --> 00:31:06.247 AUDIENCE: Would the fictitious force take into account M2? 00:31:06.247 --> 00:31:07.830 J. KIM VANDIVER: Take into account M2? 00:31:07.830 --> 00:31:09.934 AUDIENCE: Yes. 00:31:09.934 --> 00:31:11.600 J. KIM VANDIVER: So the angular momentum 00:31:11.600 --> 00:31:15.970 was all calculated with respect to the rigid body M1, right? 00:31:15.970 --> 00:31:19.050 The only reason M2 entered into it 00:31:19.050 --> 00:31:20.959 this is the way things happen in real life. 00:31:20.959 --> 00:31:22.000 It's just a real problem. 00:31:22.000 --> 00:31:24.519 It's a box on a cart, and is it going to tip over, 00:31:24.519 --> 00:31:25.560 and you're pushing on it. 00:31:25.560 --> 00:31:27.770 How hard can I push? 00:31:27.770 --> 00:31:29.790 To do that problem you just have to consider 00:31:29.790 --> 00:31:32.270 these to begin with so that you realize that, 00:31:32.270 --> 00:31:36.970 oh, the acceleration involves both of these masses, 00:31:36.970 --> 00:31:39.230 but the angular momentum part of the problem 00:31:39.230 --> 00:31:43.670 involves only the boxes tipping. 00:31:43.670 --> 00:31:47.440 So problems even as simple as this one looks has 00:31:47.440 --> 00:31:49.370 their little complications, and you've 00:31:49.370 --> 00:31:52.020 got to work through them. 00:31:52.020 --> 00:31:54.610 So just so we had to treat the body as a two-body system 00:31:54.610 --> 00:31:58.460 to start with, and then look at angular momentum for just 00:31:58.460 --> 00:31:59.050 the box. 00:31:59.050 --> 00:32:02.598 AUDIENCE: Does the pin or screw exert 00:32:02.598 --> 00:32:03.990 a force on the [INAUDIBLE]? 00:32:07.880 --> 00:32:10.060 J. KIM VANDIVER: Yes, absolutely, mm-hmm. 00:32:10.060 --> 00:32:12.430 Does it create a moment? 00:32:12.430 --> 00:32:16.380 No, and that's why we don't have to consider it. 00:32:16.380 --> 00:32:19.305 I could have put it in there and probably should have. 00:32:19.305 --> 00:32:27.260 If I had done a free body diagram here, 00:32:27.260 --> 00:32:31.490 you could still say that there's a reaction force upwards here. 00:32:31.490 --> 00:32:35.610 And I'll call it Ry and another one this direction 00:32:35.610 --> 00:32:37.570 I'll call Rx. 00:32:37.570 --> 00:32:41.600 And when computing moments about this point, neither of them 00:32:41.600 --> 00:32:42.580 enter into the problem. 00:32:42.580 --> 00:32:46.880 And that's why we like to do our calculations for angular 00:32:46.880 --> 00:32:53.010 momentum with respect to points around which the body rotates 00:32:53.010 --> 00:32:54.590 because that allows us to not have 00:32:54.590 --> 00:32:59.040 to solve for those unknown forces that pop up there, 00:32:59.040 --> 00:33:01.530 so yeah, indeed there are forces there. 00:33:01.530 --> 00:33:06.520 OK, got it figured out. 00:33:06.520 --> 00:33:09.660 So we've talked a little tiny bit about parallel axis 00:33:09.660 --> 00:33:13.900 theorem, and so I'm going to consider this a slender rod. 00:33:17.640 --> 00:33:22.700 I can spin it about one of its printable axes and calculate 00:33:22.700 --> 00:33:24.790 its angular momentum of whatever I need, 00:33:24.790 --> 00:33:29.190 and if I move it over to an axis that's-- I just move it over 00:33:29.190 --> 00:33:30.230 by a little bit. 00:33:30.230 --> 00:33:37.350 Let's say this is the x-axis here in my body, 00:33:37.350 --> 00:33:40.990 and this is the z, so it's at omega z. 00:33:40.990 --> 00:33:45.835 And I move my point and around which I am rotating over 00:33:45.835 --> 00:33:48.460 by a small amount A, and that's this distance between these two 00:33:48.460 --> 00:33:49.850 holes. 00:33:49.850 --> 00:33:52.200 And by parallel axis theorem, we could say, well, 00:33:52.200 --> 00:33:55.520 now there's a mass moment of inertia with respect to A, 00:33:55.520 --> 00:33:58.700 which is the mass moment of inertia with respect to g, 00:33:58.700 --> 00:34:03.530 which is [? Ml ?] squared over 12, if this is the length, 00:34:03.530 --> 00:34:06.040 plus the mass times this distance squared. 00:34:06.040 --> 00:34:08.760 We know how to do that. 00:34:08.760 --> 00:34:17.080 But now what happens if I take this stick, my slender rod, 00:34:17.080 --> 00:34:24.920 and here's my original x, y, z. 00:34:24.920 --> 00:34:29.270 And I want to move it, my point A, my stick now 00:34:29.270 --> 00:34:38.520 I'm going to move it over by an amount a and up by an amount c 00:34:38.520 --> 00:34:43.480 so that it's now up here with respect to this point. 00:34:43.480 --> 00:34:45.889 So this stick has been like this, 00:34:45.889 --> 00:34:47.960 and now I'm going to then rotate it about that. 00:34:47.960 --> 00:34:49.350 That's the simple case. 00:34:49.350 --> 00:34:51.250 Now I want to rotate it about this axis 00:34:51.250 --> 00:34:54.409 when it's moved over to like that. 00:34:57.780 --> 00:35:04.280 So not only have I moved it-- so here's the original problem-- 00:35:04.280 --> 00:35:15.675 like this, but now I move it up and over-- maybe 00:35:15.675 --> 00:35:17.990 I'll do it there-- and now I spin it. 00:35:17.990 --> 00:35:21.270 And I'm interested in angular momentum about this point, 00:35:21.270 --> 00:35:24.990 so now it's dynamically unbalanced for sure. 00:35:24.990 --> 00:35:27.900 This thing is trying to wobble, and it's 00:35:27.900 --> 00:35:31.550 been pushed up and over, so is there 00:35:31.550 --> 00:35:34.160 a way to get-- how do we solve that problem? 00:35:34.160 --> 00:35:38.140 How do we compute the torques for this problem? 00:35:38.140 --> 00:35:41.010 Well, we could try to go at it with parallel access 00:35:41.010 --> 00:35:44.110 to start with, but you don't know how to do that in general 00:35:44.110 --> 00:35:45.130 for this problem. 00:35:45.130 --> 00:35:48.140 So what do we go to? 00:35:48.140 --> 00:35:49.970 We go back to those three equations, 00:35:49.970 --> 00:35:52.140 and just start there, and it'll all fall out. 00:35:52.140 --> 00:35:56.800 I'm going to do this kind of briefly for this problem, 00:35:56.800 --> 00:36:04.420 so I want to compute H with respect to A H with respect 00:36:04.420 --> 00:36:12.520 to G plus RG with respect to A cross P. 00:36:12.520 --> 00:36:19.150 And now this is [? RG/A ?] is this vector here, 00:36:19.150 --> 00:36:21.232 that position vector, OK? 00:36:24.430 --> 00:36:28.600 And the only rotation is omega z. 00:36:28.600 --> 00:36:30.710 This is our z-axis. 00:36:30.710 --> 00:36:39.030 So I can write this as Izz with respect to G, 0, 0, 00:36:39.030 --> 00:36:48.800 omega z plus my [? RG/A ?] cross P/o, and that in this problem-- 00:36:48.800 --> 00:36:50.990 let's get the vector right this time-- it's 00:36:50.990 --> 00:36:59.780 a in the i direction plus c in the k direction cross. 00:36:59.780 --> 00:37:06.360 Now we need to know what's P with respect to o, 00:37:06.360 --> 00:37:12.910 but that's just the velocity of this so the center of mass, 00:37:12.910 --> 00:37:16.270 which is now here. 00:37:16.270 --> 00:37:20.520 The velocity is omega cross r, so when 00:37:20.520 --> 00:37:22.810 you do that-- you've done that problem lots of times-- 00:37:22.810 --> 00:37:28.140 it's this distance, so perpendicular distance 00:37:28.140 --> 00:37:30.110 [? across ?] k. 00:37:30.110 --> 00:37:31.740 It's got to be going into the board 00:37:31.740 --> 00:37:32.864 if it's spinning like this. 00:37:32.864 --> 00:37:35.090 We know the answer's got to come out plus j, 00:37:35.090 --> 00:37:47.810 and it's r-- it's A omega z in the j-direction cross Ma omega 00:37:47.810 --> 00:37:50.610 z, but it is in the j hat direction. 00:37:50.610 --> 00:37:54.240 That's your linear momentum, and this is [? RG/A ?] 00:37:54.240 --> 00:37:55.535 cross with a linear momentum. 00:37:59.440 --> 00:38:04.150 This here is M velocity of G here with respect to o. 00:38:04.150 --> 00:38:11.890 That's this term, so we carry out this calculation 00:38:11.890 --> 00:38:19.000 and we get-- and I need a j, OK. 00:38:19.000 --> 00:38:25.280 So i cross j gives me a k, so I'll get 2, 00:38:25.280 --> 00:38:30.030 and this is a diagonal matrix because we 00:38:30.030 --> 00:38:34.710 began with a set of principal axes for our stick, 00:38:34.710 --> 00:38:40.180 so this term is Izz with respect to G. 00:38:40.180 --> 00:38:43.390 I shouldn't have written zz here, just i with respect to z. 00:38:43.390 --> 00:38:44.560 We multiply this out. 00:38:44.560 --> 00:38:54.190 We pick up just this term omega z k hat plus i cross j is k, 00:38:54.190 --> 00:39:03.370 so that's Ma squared omega [? zk, ?] and k cross j 00:39:03.370 --> 00:39:12.645 is minus i minus M ac omega z in the i hat direction. 00:39:15.330 --> 00:39:20.140 So this is my angular momentum, and reminding, it is a vector. 00:39:20.140 --> 00:39:22.715 This has got components in the k and the i directions. 00:39:52.570 --> 00:39:54.090 I've rewritten it here. 00:39:54.090 --> 00:39:58.970 This also could be written-- this H with respect to A 00:39:58.970 --> 00:40:11.192 is I with respect to A times the rotation vector. 00:40:11.192 --> 00:40:12.900 If we had just set out with this problem, 00:40:12.900 --> 00:40:15.420 and said we're going-- if we know the mass moment of inertia 00:40:15.420 --> 00:40:18.790 matrix computed with respect to this point 00:40:18.790 --> 00:40:23.130 and multiplied it by our rotation rate vector, 00:40:23.130 --> 00:40:26.570 we should have gotten the same answer. 00:40:26.570 --> 00:40:27.780 So we didn't do that. 00:40:27.780 --> 00:40:30.820 We just worked it out using just this formula, 00:40:30.820 --> 00:40:33.430 but we've essentially discovered the answer. 00:40:33.430 --> 00:40:35.250 This is the rotation around k. 00:40:37.760 --> 00:40:42.470 This is the Izz term with respect to A, 00:40:42.470 --> 00:40:45.850 and it really looks like the familiar parallel axis piece. 00:40:45.850 --> 00:40:49.640 It's the amount that you moved this axis over, 00:40:49.640 --> 00:40:53.509 and the axis you're spinning around, you moved it over A. 00:40:53.509 --> 00:40:55.300 And you know the parallel axis theorem just 00:40:55.300 --> 00:40:57.980 says add Ma squared, and you've got the answer. 00:40:57.980 --> 00:41:01.350 But we also moved it up, and it gave us another term. 00:41:01.350 --> 00:41:04.250 What do you suppose in that term is? 00:41:04.250 --> 00:41:11.207 So this is Izz with respect to A, this term. 00:41:11.207 --> 00:41:13.290 By parallel axis theorem, you're familiar with it. 00:41:13.290 --> 00:41:22.410 This one is I, and this is xz with respect 00:41:22.410 --> 00:41:26.040 in the A coordinate system. 00:41:26.040 --> 00:41:32.240 OK, we've created-- we've made this thing unbalanced. 00:41:32.240 --> 00:41:34.010 The angular momentum vector no longer 00:41:34.010 --> 00:41:36.640 points in the same direction as the rotation. 00:41:36.640 --> 00:41:38.755 It has a k component and an I component, 00:41:38.755 --> 00:41:40.880 and sure enough, when I spend that thing like that, 00:41:40.880 --> 00:41:47.610 I feel that additional torque out around my hand 00:41:47.610 --> 00:41:49.230 down here where I'm holding it. 00:41:49.230 --> 00:41:52.620 This thing is going-- trying to pull it back and forth. 00:41:52.620 --> 00:41:55.090 So we have essentially just shown-- 00:41:55.090 --> 00:41:58.790 we have just arrived at a more complicated parallel axis 00:41:58.790 --> 00:42:02.860 theorem by just using this basic formula. 00:42:02.860 --> 00:42:07.830 So Williams in the dynamics-- that second handout 00:42:07.830 --> 00:42:12.660 from Williams has actually given us a general formulation 00:42:12.660 --> 00:42:14.810 for parallel axis theorem. 00:42:18.720 --> 00:42:20.070 So this is just handy to know. 00:42:20.070 --> 00:42:28.200 I'll just give it to you, and if you have a problem some day, 00:42:28.200 --> 00:42:30.770 where you just like to go directly to this statement, 00:42:30.770 --> 00:42:32.650 so now I'm just going to say, in general, we 00:42:32.650 --> 00:42:37.410 could have moved over by A, up by c and actually 00:42:37.410 --> 00:42:42.980 into the board by b, so x, I'm moving A. y, I'm moving b. 00:42:42.980 --> 00:42:44.240 z, I'm moving c. 00:42:44.240 --> 00:42:48.310 OK, so you could have all those possible moves of the new point 00:42:48.310 --> 00:42:49.540 about which you're rotating. 00:42:49.540 --> 00:42:50.948 Yes? 00:42:50.948 --> 00:42:55.214 AUDIENCE: For the [INAUDIBLE] matrix i with respect to A, 00:42:55.214 --> 00:42:57.435 does that mean that you're rotating about point A? 00:42:57.435 --> 00:42:58.060 Or what is it-- 00:42:58.060 --> 00:43:00.796 J. KIM VANDIVER: You are rotating about point A. 00:43:00.796 --> 00:43:02.245 AUDIENCE: OK. 00:43:02.245 --> 00:43:04.660 But isn't point A just the [? immersion? ?] 00:43:11.520 --> 00:43:12.805 J. KIM VANDIVER: Let me think. 00:43:12.805 --> 00:43:14.180 I'm trying to think of a good way 00:43:14.180 --> 00:43:18.570 to-- when you started this problem, 00:43:18.570 --> 00:43:22.680 we're spending about the middle-- about G, OK? 00:43:22.680 --> 00:43:24.840 And we can calculate angular momentum 00:43:24.840 --> 00:43:29.210 of this object with respect to G and learn some things, right? 00:43:29.210 --> 00:43:30.950 But now we've built a different device. 00:43:30.950 --> 00:43:36.520 This is an object now that is, in fact, spinning. 00:43:42.380 --> 00:43:44.140 Let's imagine it's a fan blade, and you've 00:43:44.140 --> 00:43:46.440 broken one blade off. 00:43:46.440 --> 00:43:49.230 The motor is here. 00:43:49.230 --> 00:43:52.590 A proper fan blade would have blades on both sides, 00:43:52.590 --> 00:43:53.760 so it's nice and balanced. 00:43:53.760 --> 00:43:56.460 The shaft has some length and it sticks out, 00:43:56.460 --> 00:43:58.430 and now you break off one of the blades. 00:43:58.430 --> 00:44:01.800 And so you have a system that's doing this, 00:44:01.800 --> 00:44:04.730 but this is going to put a lot of unusual loads on this point 00:44:04.730 --> 00:44:05.710 down here. 00:44:05.710 --> 00:44:11.790 So it is rotating at the same rotation rate about this point, 00:44:11.790 --> 00:44:15.820 but it is now a system whose center of mass is over here, 00:44:15.820 --> 00:44:16.690 and it's up. 00:44:16.690 --> 00:44:20.140 It's above this point, and it's out 00:44:20.140 --> 00:44:21.940 from this point, the center of mass. 00:44:21.940 --> 00:44:24.720 And now you compute the angular momentum with this point, 00:44:24.720 --> 00:44:26.460 and take its time derivative, you 00:44:26.460 --> 00:44:31.390 will find all the torques required to make this happen. 00:44:31.390 --> 00:44:32.306 AUDIENCE: [INAUDIBLE]? 00:44:34.710 --> 00:44:36.085 J. KIM VANDIVER: A little louder. 00:44:36.085 --> 00:44:39.550 AUDIENCE: The movement A, [? i-hat ?] Mc [? k-hat ?] 00:44:39.550 --> 00:44:40.050 [INAUDIBLE] 00:44:40.050 --> 00:44:42.770 J. KIM VANDIVER: Movement that you've moved the center of mass 00:44:42.770 --> 00:44:46.560 away from this axis of rotation. 00:44:46.560 --> 00:44:51.660 And you've moved it up and over. 00:44:51.660 --> 00:44:54.342 So actually, let's think about that for a minute. 00:44:54.342 --> 00:44:55.800 This is our beginning point, right? 00:44:55.800 --> 00:45:03.090 We're at G. If I just move it up, does it cause any problems? 00:45:03.090 --> 00:45:05.100 It's still balanced, and you'll find out 00:45:05.100 --> 00:45:07.760 that nothing's happened in this problem. 00:45:07.760 --> 00:45:10.700 H with respect to A is the same as it was before. 00:45:14.190 --> 00:45:15.875 But now you move it up and over, it 00:45:15.875 --> 00:45:17.230 starts causing complications. 00:45:17.230 --> 00:45:17.750 Yeah? 00:45:17.750 --> 00:45:24.185 AUDIENCE: [INAUDIBLE] H of A equals [? A ?] times omega z. 00:45:24.185 --> 00:45:28.145 So i doesn't have like direction, 00:45:28.145 --> 00:45:31.610 but you have an [? ixz ?] terms and [? xc ?] term, 00:45:31.610 --> 00:45:34.580 and you're multiplying it by a vector with an omega z term, 00:45:34.580 --> 00:45:37.550 so how are you getting an i hat from that? 00:45:37.550 --> 00:45:41.890 J. KIM VANDIVER: OK, so I haven't gone into this, 00:45:41.890 --> 00:45:44.090 and I'm not going to. 00:45:44.090 --> 00:45:48.530 The [? i ?] matrix is a thing called a tensor. 00:45:48.530 --> 00:45:53.590 And so this is a convention that you 00:45:53.590 --> 00:45:57.710 can make it a very mathematical and assigned unit vectors 00:45:57.710 --> 00:46:01.150 to each of these terms inside, but I haven't done that. 00:46:01.150 --> 00:46:07.810 But the convention here is that this, the H, angular momentum 00:46:07.810 --> 00:46:09.650 is a vector. 00:46:09.650 --> 00:46:12.526 It consists of three components. 00:46:17.930 --> 00:46:23.590 It has three potential vector components, Hx in the I, 00:46:23.590 --> 00:46:30.950 Hy in the j, and Hz in the k directions. 00:46:30.950 --> 00:46:34.630 And we, in a compact form, say we can express that 00:46:34.630 --> 00:46:37.490 by multiplying its mass moment of inertia matrix 00:46:37.490 --> 00:46:43.570 with respect to A times the vector that is its rotation. 00:46:43.570 --> 00:46:45.850 And I only have one component of this omega, 00:46:45.850 --> 00:46:49.140 but it is in the k direction. 00:46:49.140 --> 00:46:55.050 And when you take this vector multiplied by the top row, that 00:46:55.050 --> 00:46:58.900 gives you everything that's [? I. ?] You multiply 00:46:58.900 --> 00:47:00.710 this vector times the second row, 00:47:00.710 --> 00:47:02.950 it gives you everything is in the j direction, 00:47:02.950 --> 00:47:06.160 and the result is part of Hy in the j direction. 00:47:06.160 --> 00:47:09.400 You take this vector, and you multiply it by the third row, 00:47:09.400 --> 00:47:13.500 it gives you the z-directed components, OK? 00:47:13.500 --> 00:47:22.110 So this one, because when we multiply this out 00:47:22.110 --> 00:47:24.790 what's in this matrix-- is this matrix diagonal? 00:47:27.680 --> 00:47:28.870 No way. 00:47:28.870 --> 00:47:32.201 And we move it, and we've gone and done something strange 00:47:32.201 --> 00:47:32.700 here. 00:47:32.700 --> 00:47:35.110 It's not diagonal, and when you multiply 00:47:35.110 --> 00:47:38.730 this times this top row, there's a term here, 00:47:38.730 --> 00:47:41.010 which is not 0-- shouldn't write 0. 00:47:41.010 --> 00:47:42.060 Make an x here. 00:47:42.060 --> 00:47:46.715 This times this gives you an omega z times that in the i hat 00:47:46.715 --> 00:47:47.215 direction. 00:47:50.430 --> 00:47:55.020 OK this term comes from there being something there. 00:47:55.020 --> 00:47:57.970 OK, and I'm just going to give you a generalization that you 00:47:57.970 --> 00:48:01.120 can go back and read the Williams handout, which 00:48:01.120 --> 00:48:10.620 he says, if you want to rotate about some fixed point A. 00:48:10.620 --> 00:48:15.120 Then you can say that this matrix 00:48:15.120 --> 00:48:19.170 is equal to the one with respect to G, which in our case 00:48:19.170 --> 00:48:21.910 is diagonal. 00:48:21.910 --> 00:48:24.060 The original one that we had doesn't have to be, 00:48:24.060 --> 00:48:27.590 but if you pick principal axes, this will be diagonal. 00:48:27.590 --> 00:48:40.580 Plus M b squared plus c squared minus ab minus ac. 00:48:40.580 --> 00:48:41.900 It's symmetric. 00:48:41.900 --> 00:48:45.555 This is a squared plus c squared. 00:49:02.070 --> 00:49:04.220 So if we're rotating a system if you 00:49:04.220 --> 00:49:06.280 know its mass moment of inertia matrix 00:49:06.280 --> 00:49:10.270 with respect to G [INAUDIBLE] perhaps principal axes, 00:49:10.270 --> 00:49:13.650 and you want to rotate it about any other point 00:49:13.650 --> 00:49:17.510 where you've moved it by an a, by a b, by a c, 00:49:17.510 --> 00:49:20.560 this is essentially the general parallel axis theorem. 00:49:20.560 --> 00:49:23.450 And you can just plug it in and use that. 00:49:23.450 --> 00:49:26.600 And you can see what would happen in this case 00:49:26.600 --> 00:49:29.120 if you multiplied-- if you add these two together, 00:49:29.120 --> 00:49:30.330 they add term by term. 00:49:30.330 --> 00:49:33.050 This goes into this point on. 00:49:33.050 --> 00:49:39.820 This one adds to this, so in this problem, 00:49:39.820 --> 00:49:46.150 in our particular problem, this becomes 00:49:46.150 --> 00:49:55.490 you have an Ixx, Iyy, Izz with respect to G terms. 00:49:55.490 --> 00:50:01.945 It's diagonal to start with, and then in our problem 00:50:01.945 --> 00:50:05.220 that we did there, we didn't have a b. 00:50:05.220 --> 00:50:08.270 No b terms, so the b terms are all 0, 00:50:08.270 --> 00:50:12.570 but you do end up with a minus ac term over here. 00:50:18.160 --> 00:50:24.280 And if you multiply this by 0, 0, omega z, see what happens. 00:50:24.280 --> 00:50:29.590 You get minus ac in the I direction, 00:50:29.590 --> 00:50:32.695 and you need an M here. 00:50:38.110 --> 00:50:41.270 And you come down here, you get the Izz G omega 00:50:41.270 --> 00:50:43.870 z in the k direction, and that's those two terms. 00:50:43.870 --> 00:50:44.510 Here they are. 00:50:49.120 --> 00:50:51.280 Yeah, oops, I need the plus [? A ?] squared. 00:50:55.950 --> 00:50:56.930 Yeah? 00:50:56.930 --> 00:51:00.360 AUDIENCE: When you move it over to the end of the stick, 00:51:00.360 --> 00:51:05.260 normally like you're just shifting it parallely, 00:51:05.260 --> 00:51:06.730 but now there's gravity. 00:51:06.730 --> 00:51:08.531 Is that why you have those little things? 00:51:08.531 --> 00:51:09.655 J. KIM VANDIVER: Say again? 00:51:09.655 --> 00:51:11.613 AUDIENCE: So normally, if you just take a stick 00:51:11.613 --> 00:51:15.328 and you have a principal axis and you 00:51:15.328 --> 00:51:17.496 can't move that parallel to the end, 00:51:17.496 --> 00:51:18.966 you wouldn't get all of this but-- 00:51:18.966 --> 00:51:21.340 J. KIM VANDIVER: Yeah, and if you just move it-- so yeah, 00:51:21.340 --> 00:51:22.714 I'm glad you asked that question. 00:51:22.714 --> 00:51:28.870 So normally the types of problems in which we usually 00:51:28.870 --> 00:51:31.470 use the parallel axis theorem is you've 00:51:31.470 --> 00:51:34.150 been spinning it around one axis. 00:51:34.150 --> 00:51:36.600 It's a principal axis, everything's fine, 00:51:36.600 --> 00:51:40.060 and you want to just move it over, 00:51:40.060 --> 00:51:43.570 so we do that when we do pendulum problems. 00:51:43.570 --> 00:51:45.220 Here's the thing around the center, 00:51:45.220 --> 00:51:48.170 and we just want to know, how does this thing behave? 00:51:48.170 --> 00:51:50.870 When we move it up here, it becomes a pendulum, 00:51:50.870 --> 00:51:53.270 and we would like to compute the angular 00:51:53.270 --> 00:51:54.780 momentum around this point. 00:51:54.780 --> 00:51:57.260 We say, oh, we've moved it a distance d. 00:51:57.260 --> 00:52:02.870 The new mass moment of inertia in the z direction 00:52:02.870 --> 00:52:07.820 is the original plus Md squared, right? 00:52:07.820 --> 00:52:11.520 So when you only make one move, you only 00:52:11.520 --> 00:52:14.190 have an a, a b, or a c. 00:52:14.190 --> 00:52:16.290 You never get any of the off-diagonal terms 00:52:16.290 --> 00:52:18.690 because their products, see? 00:52:18.690 --> 00:52:21.950 But if you start doing more than one, 00:52:21.950 --> 00:52:24.130 you've introduced complications, and that's 00:52:24.130 --> 00:52:26.550 part of the reason I put this up is 00:52:26.550 --> 00:52:32.570 there's limits to the simple parallel axis theorem, 00:52:32.570 --> 00:52:34.800 but there is a general way of doing it. 00:52:34.800 --> 00:52:35.300 Yes? 00:52:35.300 --> 00:52:39.478 AUDIENCE: I believe her actual question involved 00:52:39.478 --> 00:52:40.891 what causes the torques? 00:52:40.891 --> 00:52:41.763 Is it gravity? 00:52:41.763 --> 00:52:42.304 For example-- 00:52:42.304 --> 00:52:43.720 AUDIENCE: Yeah, I was asking why-- 00:52:43.720 --> 00:52:44.910 AUDIENCE: [INAUDIBLE]. 00:52:44.910 --> 00:52:47.900 J. KIM VANDIVER: Ah, so what causes these torques when 00:52:47.900 --> 00:52:49.067 you have two? 00:52:49.067 --> 00:52:49.650 AUDIENCE: Yes. 00:52:49.650 --> 00:52:52.920 J. KIM VANDIVER: OK, we've talked about this a little bit 00:52:52.920 --> 00:52:53.420 before. 00:52:56.340 --> 00:52:59.370 When we calculate the angular momentum, 00:52:59.370 --> 00:53:01.540 it never involves gravity. 00:53:04.470 --> 00:53:08.410 Never gets in there, not in the angular momentum expression. 00:53:08.410 --> 00:53:11.810 So when you go to compute d, the time derivative of the angular 00:53:11.810 --> 00:53:13.643 momentum, you will not find torques 00:53:13.643 --> 00:53:15.532 that are caused by gravity. 00:53:15.532 --> 00:53:16.490 You just can't do them. 00:53:16.490 --> 00:53:19.830 It's not part-- but are the torques in the System I mean, 00:53:19.830 --> 00:53:20.830 why just hold this here. 00:53:20.830 --> 00:53:22.246 There's [? Mg ?] down, and there's 00:53:22.246 --> 00:53:25.870 a static moment caused by this thing trying 00:53:25.870 --> 00:53:27.420 to twist this down. 00:53:27.420 --> 00:53:29.300 When you sum the torques-- when you 00:53:29.300 --> 00:53:32.060 do the sum of the torques in the equation, 00:53:32.060 --> 00:53:36.390 you do your free body diagram, that term will appear. 00:53:36.390 --> 00:53:40.450 But it is balanced by some physical torque over here 00:53:40.450 --> 00:53:41.920 that balances it. 00:53:41.920 --> 00:53:46.480 You just won't find it from doing [? dh ?] dt. 00:53:46.480 --> 00:53:49.501 It's just part of the static equilibrium of the system. 00:53:49.501 --> 00:53:52.868 AUDIENCE: But the reason when you move the pen over, 00:53:52.868 --> 00:53:56.235 the reason you even have the shift is because of 00:53:56.235 --> 00:53:57.170 [? gravity? ?] 00:53:57.170 --> 00:53:58.170 J. KIM VANDIVER: No, no. 00:53:58.170 --> 00:54:04.670 So she's asking if is gravity the reason you-- 00:54:04.670 --> 00:54:07.670 I'm not even quite sure. 00:54:07.670 --> 00:54:12.725 AUDIENCE: How did you shift the z component of your [? plan? ?] 00:54:12.725 --> 00:54:16.200 J. KIM VANDIVER: OK, so let's just make 00:54:16.200 --> 00:54:18.100 this a real physical problem. 00:54:18.100 --> 00:54:21.470 I'm making an airplane engine, and I'm making an airplane 00:54:21.470 --> 00:54:23.750 with a propeller on it. 00:54:23.750 --> 00:54:28.320 And the bearing that supports the propeller shaft, 00:54:28.320 --> 00:54:31.200 if I put it really close to the propeller, 00:54:31.200 --> 00:54:35.290 OK, then it's doing this, and everything's fine. 00:54:35.290 --> 00:54:39.080 What if I extended the propeller shaft? 00:54:39.080 --> 00:54:42.800 OK, so now the bearing is back here, 00:54:42.800 --> 00:54:45.130 and now the propeller spins. 00:54:45.130 --> 00:54:50.260 Does that cause any torques, any loads back here on the bearing 00:54:50.260 --> 00:54:52.350 because I've extended it? 00:54:52.350 --> 00:54:55.170 No, and you could prove that just by going through this. 00:54:55.170 --> 00:54:57.740 You do [? dh ?] dt, and you find out the only torques 00:54:57.740 --> 00:55:00.330 when it's nice and balanced are those required 00:55:00.330 --> 00:55:04.700 to drive this propeller in the direction 00:55:04.700 --> 00:55:07.460 of the axis of rotation. 00:55:07.460 --> 00:55:14.020 But as soon as you do something to that propeller, like you 00:55:14.020 --> 00:55:18.510 mount it off-center, which rarely happens, 00:55:18.510 --> 00:55:21.350 but if you dinged-- if you broke off 00:55:21.350 --> 00:55:23.970 the tip of one of the blades of this propeller, 00:55:23.970 --> 00:55:27.650 you'd now have a propeller that looks like that, right? 00:55:27.650 --> 00:55:29.800 And now, even in the original system 00:55:29.800 --> 00:55:34.530 if your bearing is right close, this is spinning around, 00:55:34.530 --> 00:55:35.930 but is it putting a load? 00:55:35.930 --> 00:55:38.240 Sure there's a centrifugal force that 00:55:38.240 --> 00:55:39.760 is going around and around. 00:55:39.760 --> 00:55:41.960 It has nothing to do with gravity, 00:55:41.960 --> 00:55:45.264 nothing at all to do with gravity. 00:55:45.264 --> 00:55:48.960 OK, just because you now have the mass centers out here, 00:55:48.960 --> 00:55:51.310 it has momentum, the time rate of change 00:55:51.310 --> 00:55:56.502 of that linear momentum is a force making 00:55:56.502 --> 00:55:57.710 this thing going in a circle. 00:55:57.710 --> 00:56:01.770 OK, now if I extend the propeller shaft 00:56:01.770 --> 00:56:05.920 so it's like that, and I foolishly designed my airplane 00:56:05.920 --> 00:56:08.700 so it had a long propeller shaft sticking out there, 00:56:08.700 --> 00:56:11.860 if there's a little bit of unbalanced in this blade 00:56:11.860 --> 00:56:14.970 so that you're not spinning about the mass center. 00:56:14.970 --> 00:56:17.650 You're spinning about some other point. 00:56:17.650 --> 00:56:20.620 Now that centrifugal force going around is pulling 00:56:20.620 --> 00:56:24.360 is trying to bend this back and forth around this bearing. 00:56:24.360 --> 00:56:28.930 And because of the way in which we formulate angular momentum, 00:56:28.930 --> 00:56:31.340 if you formulate it about that point 00:56:31.340 --> 00:56:36.330 and take its time derivative, it will reveal those moments. 00:56:36.330 --> 00:56:38.880 It's really amazing that it can do that for you, 00:56:38.880 --> 00:56:42.433 but it has [? zero, ?] nothing to do with gravity, OK? 00:56:45.340 --> 00:56:51.400 So handy, hard to remember this from just a blackboard 00:56:51.400 --> 00:56:55.090 presentation, but it's in that second reading by Williams. 00:56:55.090 --> 00:56:57.645 He does it in a very-- he proves it 00:56:57.645 --> 00:57:01.849 in a very simple way using the summations of the MI [? RI's ?] 00:57:01.849 --> 00:57:02.390 and so forth. 00:57:02.390 --> 00:57:04.970 Proves it in a very simple way, but a very general 00:57:04.970 --> 00:57:08.880 handy formula that you can use. 00:57:08.880 --> 00:57:14.600 OK, so let's have another topic, which I'm not going to do. 00:57:14.600 --> 00:57:15.790 We've got a few minutes. 00:57:15.790 --> 00:57:17.059 You have [? money ?] cards. 00:57:17.059 --> 00:57:18.350 We've been thinking about that. 00:57:18.350 --> 00:57:19.741 And let's ask questions, yeah? 00:57:19.741 --> 00:57:20.616 AUDIENCE: [INAUDIBLE] 00:57:23.637 --> 00:57:25.053 J. KIM VANDIVER: Where did I the-- 00:57:25.053 --> 00:57:27.886 AUDIENCE: The plus [? MA ?] squared and the plus A squared. 00:57:27.886 --> 00:57:30.010 J. KIM VANDIVER: Ah, I was doing that kind of fast, 00:57:30.010 --> 00:57:31.410 and I probably even messed it up, 00:57:31.410 --> 00:57:44.550 so let me check the-- so I was doing for the example 00:57:44.550 --> 00:57:55.100 that [? RG/A ?] equals-- we moved it over by ai and up 00:57:55.100 --> 00:58:00.840 by ck, so and plus 0j. 00:58:00.840 --> 00:58:03.830 We didn't move it in the j direction, OK? 00:58:03.830 --> 00:58:06.450 So this is my amount that I've moved it. 00:58:06.450 --> 00:58:09.520 I've moved it an amount a, and an amount c. 00:58:09.520 --> 00:58:11.770 So the Williams formula would say, ah, 00:58:11.770 --> 00:58:15.760 that new mass moment of inertia matrix with respect to a 00:58:15.760 --> 00:58:20.950 becomes the original plus-- and now every place there's 00:58:20.950 --> 00:58:23.010 a b here, it goes to 0. 00:58:31.810 --> 00:58:41.660 And the remaining bits I add with those, 00:58:41.660 --> 00:58:51.940 so I should get a M-- whoops, this is a c squared, 00:58:51.940 --> 00:58:53.420 no second term. 00:58:53.420 --> 00:58:57.190 The third term is minus Mac. 00:58:57.190 --> 00:59:00.640 Over here, that one is 0. 00:59:00.640 --> 00:59:03.520 This one becomes a squared plus c squared-- 00:59:03.520 --> 00:59:05.670 AUDIENCE: --minus [INAUDIBLE]? 00:59:05.670 --> 00:59:08.120 J. KIM VANDIVER: Minus, yep, M. 00:59:08.120 --> 00:59:13.960 And this one is minus Mac 0 Izz and a squared plus b squared, 00:59:13.960 --> 00:59:15.090 so it's just a squared. 00:59:15.090 --> 00:59:17.658 So this is now correct. 00:59:17.658 --> 00:59:18.610 AUDIENCE: [INAUDIBLE]? 00:59:18.610 --> 00:59:19.610 J. KIM VANDIVER: Pardon? 00:59:19.610 --> 00:59:20.840 AUDIENCE: Times M in the bottom of it? 00:59:20.840 --> 00:59:22.298 J. KIM VANDIVER: Yeah, all of these 00:59:22.298 --> 00:59:29.750 needs M. Ma squared's M, M, M, M, 00:59:29.750 --> 00:59:33.800 so that's our problem when we did two shifts. 00:59:33.800 --> 00:59:37.350 But a second we shifted it in two directions, 00:59:37.350 --> 00:59:41.310 we get these off-diagonal terms, which 00:59:41.310 --> 00:59:44.940 means that if you are actually making this device rotate 00:59:44.940 --> 00:59:47.954 about point A, you will get these unbalanced moments-- 00:59:47.954 --> 00:59:49.120 this dynamically unbalanced. 00:59:51.970 --> 00:59:53.414 Yeah? 00:59:53.414 --> 00:59:56.039 AUDIENCE: Will we be trying to find our [? mass ?] moments 00:59:56.039 --> 01:00:01.776 of inertia directly, or should we just do [INAUDIBLE] equation 01:00:01.776 --> 01:00:05.185 and have them fall out like the integral with x times y 01:00:05.185 --> 01:00:07.679 [? vM? ?] [INAUDIBLE] 01:00:07.679 --> 01:00:10.220 J. KIM VANDIVER: He's kind of asking advice in whether or not 01:00:10.220 --> 01:00:15.590 we ought to be trying to find the inertia 01:00:15.590 --> 01:00:18.280 matrix about other points, right, 01:00:18.280 --> 01:00:21.290 like [? the I ?] with respect to A. 01:00:21.290 --> 01:00:24.850 Where I started today, I said you can basically 01:00:24.850 --> 01:00:27.360 do all the problems with these three equations, 01:00:27.360 --> 01:00:32.080 and this doesn't mention parallel axis theorem. 01:00:32.080 --> 01:00:36.937 We use this to find, in fact, to solve this problem. 01:00:36.937 --> 01:00:38.770 We didn't talk parallel axis theorem at all, 01:00:38.770 --> 01:00:42.220 it just-- the answer dropped out. 01:00:42.220 --> 01:00:43.800 And if we looked into that careful, 01:00:43.800 --> 01:00:45.030 we could generalize that. 01:00:45.030 --> 01:00:47.270 We could work now with that a bit and say, ah, 01:00:47.270 --> 01:00:49.030 there's a pattern to this. 01:00:49.030 --> 01:00:55.650 I'll bet this can be recast like this, and it can. 01:00:55.650 --> 01:00:59.200 So if you know this exists, and you 01:00:59.200 --> 01:01:02.500 don't want to have to grind through finding 01:01:02.500 --> 01:01:08.730 these terms that come from here, you 01:01:08.730 --> 01:01:14.850 can do the problems by finding the mass moment of inertia 01:01:14.850 --> 01:01:17.470 matrix with respect to A. 01:01:17.470 --> 01:01:26.460 You can use this form only when you have fixed axis rotation. 01:01:26.460 --> 01:01:30.220 The thing is the point A is [? about ?] which 01:01:30.220 --> 01:01:35.060 this rotation is occurring, then you can write down the formula 01:01:35.060 --> 01:01:37.700 this directly like that. 01:01:37.700 --> 01:01:41.780 So you that got to be careful when you apply 01:01:41.780 --> 01:01:43.750 the parallel axis theorem. 01:01:46.990 --> 01:01:51.530 The nice thing about these three forms 01:01:51.530 --> 01:01:54.120 is they're generally true. 01:01:54.120 --> 01:01:55.670 Point A can move. 01:01:55.670 --> 01:01:57.940 Point A can be accelerating. 01:01:57.940 --> 01:02:02.820 The problem that we did here, point A is accelerating. 01:02:02.820 --> 01:02:06.890 Not only moving, it's not even an inertial frame. 01:02:06.890 --> 01:02:11.031 We solved this problem-- this is a non-inertial frame problem. 01:02:11.031 --> 01:02:15.520 We went right at it and solved it directly, OK? 01:02:15.520 --> 01:02:18.600 So I would say, if you have any doubt to answer your question, 01:02:18.600 --> 01:02:21.760 when you're in doubt just use the formula. 01:02:21.760 --> 01:02:25.020 And then you will need to find moments of inertia 01:02:25.020 --> 01:02:26.940 with respect to the center of mass, 01:02:26.940 --> 01:02:29.830 and it's in your interest to use principal axes because they're 01:02:29.830 --> 01:02:30.330 easier. 01:02:30.330 --> 01:02:30.830 Yeah? 01:02:30.830 --> 01:02:32.780 AUDIENCE: If you were to say [INAUDIBLE]? 01:02:38.662 --> 01:02:40.120 J. KIM VANDIVER: Ah, so if you took 01:02:40.120 --> 01:02:43.930 the derivative of this, [? dh ?] dt, 01:02:43.930 --> 01:02:48.050 this one just gives you omega z dot. 01:02:48.050 --> 01:02:50.900 That's you're spin in the direction of rotation. 01:02:50.900 --> 01:02:54.530 This term, that unit vector rotates. 01:02:54.530 --> 01:03:01.720 This gives you two terms-- gives you an i term, omega dot i, 01:03:01.720 --> 01:03:07.600 and it gives you an omega j term, OK? 01:03:07.600 --> 01:03:10.730 And what those are, those are two torques. 01:03:10.730 --> 01:03:14.040 This problem, there's a torque caused 01:03:14.040 --> 01:03:16.190 by the centrifugal force trying to bend 01:03:16.190 --> 01:03:20.410 this in the j direction. 01:03:20.410 --> 01:03:24.650 And if this is accelerating, you know, a theta double dot term, 01:03:24.650 --> 01:03:28.415 there is a torque trying to bend this thing backwards. 01:03:28.415 --> 01:03:30.410 OK, get them both. 01:03:30.410 --> 01:03:36.610 All right [? money ?] cards and see you on Thursday. 01:03:36.610 --> 01:03:42.230 Thursday we're going to do a nasty problem, 01:03:42.230 --> 01:03:44.530 and it is it a lead in to a problem. 01:03:44.530 --> 01:03:47.130 You can do the same problem extraordinarily easily 01:03:47.130 --> 01:03:47.940 using energy. 01:03:47.940 --> 01:03:49.731 And that's kind of the purpose of doing it, 01:03:49.731 --> 01:03:52.060 so you see both ways of doing it.