WEBVTT 00:00:00.070 --> 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 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.258 at ocw.mit.edu. 00:00:27.130 --> 00:00:28.630 PROFESSOR: So today I'm mostly going 00:00:28.630 --> 00:00:31.780 to talk about two formulas that we are 00:00:31.780 --> 00:00:34.870 going to make a lot of use of. 00:00:34.870 --> 00:00:37.740 One is the time derivative of linear momentum, 00:00:37.740 --> 00:00:40.890 which will be used a lot, has got to be equal to, 00:00:40.890 --> 00:00:41.716 for a particle. 00:00:45.840 --> 00:00:49.160 We'll just do particles today and then rigid bodies soon. 00:00:49.160 --> 00:00:52.546 So for a particle, this is just the mass times 00:00:52.546 --> 00:00:53.920 the acceleration of the particle. 00:00:56.600 --> 00:00:59.830 The rigid body, it's the total mass times acceleration 00:00:59.830 --> 00:01:00.930 in the center of gravity. 00:01:00.930 --> 00:01:05.650 And the other formula that we've learned, that we've come up to, 00:01:05.650 --> 00:01:14.770 is the sum of the external torques with respect to a point 00:01:14.770 --> 00:01:22.770 A on a particle is the time derivative of the angular 00:01:22.770 --> 00:01:29.670 momentum of the particle at B with respect to A dt. 00:01:29.670 --> 00:01:34.160 And then it's got this nuisance term, this velocity of A 00:01:34.160 --> 00:01:40.820 with respect to O cross the linear momentum at the point B 00:01:40.820 --> 00:01:44.950 with respect to O. So we talked about this last time. 00:01:44.950 --> 00:01:51.880 So this is a particle that's located out here 00:01:51.880 --> 00:01:59.410 at B. There's some intermediate point A. And we're 00:01:59.410 --> 00:02:03.630 computing the angular momentum of this particle, m, 00:02:03.630 --> 00:02:06.880 with respect to A. And A can be moving. 00:02:06.880 --> 00:02:09.280 And if it is, when you go to compute torques, 00:02:09.280 --> 00:02:11.820 you have to use this kind of messy formula. 00:02:11.820 --> 00:02:15.220 We often try to simplify the problems 00:02:15.220 --> 00:02:19.450 that we do so that we can either make A a fixed 00:02:19.450 --> 00:02:21.860 axis of rotation-- make it rotate 00:02:21.860 --> 00:02:25.480 about this point, in which case this goes to zero-- 00:02:25.480 --> 00:02:29.100 or there's another time when you can make 00:02:29.100 --> 00:02:32.940 the velocity of this point parallel 00:02:32.940 --> 00:02:34.420 to the velocity of that point. 00:02:34.420 --> 00:02:37.560 And the most common one of all is 00:02:37.560 --> 00:02:43.020 when you use this formula about the center of mass. 00:02:43.020 --> 00:02:46.240 Then the velocity of the center of mass 00:02:46.240 --> 00:02:49.037 is the same direction as the momentum. 00:02:49.037 --> 00:02:50.620 And therefore, this term goes to zero. 00:02:50.620 --> 00:02:54.890 So we have these two cases where this formula simplifies. 00:02:54.890 --> 00:02:57.410 So the way we most often use it is 00:02:57.410 --> 00:03:00.920 the summation of the torques with respect to A 00:03:00.920 --> 00:03:07.370 is just the derivative of the angular momentum dt. 00:03:07.370 --> 00:03:08.250 And that's it. 00:03:08.250 --> 00:03:26.900 This is when vA is zero, or vA is parallel to P, such 00:03:26.900 --> 00:03:36.560 as A is at the center of mass. 00:03:36.560 --> 00:03:39.500 So these are the two formulas that we want to use. 00:03:56.580 --> 00:03:58.480 Let's just do a real simple case first. 00:04:31.130 --> 00:04:38.070 And this is just my mass on a string spinning 00:04:38.070 --> 00:04:39.640 around, constant speed. 00:04:44.080 --> 00:04:47.199 I'm going to pretend I'm on a frictionless table 00:04:47.199 --> 00:04:48.740 so I don't have to deal with gravity. 00:04:48.740 --> 00:04:51.420 This thing is horizontal, just spinning around. 00:04:51.420 --> 00:04:56.636 And I'll call this point A. This is my point O. This is x, z. 00:04:59.920 --> 00:05:01.900 No, y. 00:05:01.900 --> 00:05:04.710 It's looking down on it. 00:05:04.710 --> 00:05:09.310 So it's going to have a velocity in this theta hat direction. 00:05:09.310 --> 00:05:13.977 And you've got your R hat. 00:05:13.977 --> 00:05:15.810 We'll just use polar coordinates to do this. 00:05:15.810 --> 00:05:19.940 So the velocity of A with respect to O we 00:05:19.940 --> 00:05:25.310 know-- and we'll give this just a length magnitude r1. 00:05:25.310 --> 00:05:30.923 So this is going to be theta dot equals a constant. 00:05:34.880 --> 00:05:36.910 r dot equals 0. 00:05:36.910 --> 00:05:38.690 So it's just fixed length. 00:05:38.690 --> 00:05:40.710 Fixed length is r1. 00:05:40.710 --> 00:05:43.090 So we know the simple formula for the velocity 00:05:43.090 --> 00:05:50.350 of that is r1 theta dot in the theta hat direction. 00:05:50.350 --> 00:05:53.910 [INAUDIBLE] hat, excuse me. 00:05:53.910 --> 00:06:00.180 And the momentum then is just the mass times that. 00:06:04.300 --> 00:06:13.365 And the time derivative of the linear momentum 00:06:13.365 --> 00:06:16.420 with respect to this fixed reference frame 00:06:16.420 --> 00:06:18.385 is the time derivative of this. 00:06:18.385 --> 00:06:19.430 This is constant. 00:06:19.430 --> 00:06:20.380 This is constant. 00:06:20.380 --> 00:06:21.790 This is constant. 00:06:21.790 --> 00:06:25.050 The only thing you have to take is the derivative of theta hat. 00:06:25.050 --> 00:06:34.989 So this is mr1 theta dot theta hat time derivative. 00:06:34.989 --> 00:06:37.030 But we've worked that one out two or three times. 00:06:37.030 --> 00:06:45.600 It's theta dot-- minus theta dot r hat. 00:06:56.030 --> 00:06:57.170 We'll put that in. 00:06:57.170 --> 00:07:06.290 You get minus mr1 theta dot squared r hat. 00:07:06.290 --> 00:07:11.240 And this must be the sum of the external forces. 00:07:11.240 --> 00:07:13.140 And if we're, looking down on this, 00:07:13.140 --> 00:07:20.690 draw a free body diagram of our mass, looking down on it, 00:07:20.690 --> 00:07:25.110 here's r hat. 00:07:25.110 --> 00:07:28.780 The theta direction like this, z coming out of the board. 00:07:28.780 --> 00:07:33.950 So this is x, y, and o. 00:07:33.950 --> 00:07:37.540 There is a inward force on it that we 00:07:37.540 --> 00:07:39.830 call the tension in the string. 00:07:39.830 --> 00:07:45.830 And some of external force is just minus Tr hat. 00:07:45.830 --> 00:07:50.540 So T is mr1 theta dot squared, which 00:07:50.540 --> 00:07:53.460 is what we talked about just a second ago. 00:07:53.460 --> 00:07:56.320 So this is just a demonstration of what 00:07:56.320 --> 00:07:59.050 we were talking about with the survey questions, 00:07:59.050 --> 00:08:02.030 that when you spin something, you're 00:08:02.030 --> 00:08:05.760 changing its linear momentum. 00:08:05.760 --> 00:08:07.570 Not in magnitude. 00:08:07.570 --> 00:08:08.900 It stays constant. 00:08:08.900 --> 00:08:12.080 But it's constantly changing in direction. 00:08:12.080 --> 00:08:15.850 And the direction change-- you have 00:08:15.850 --> 00:08:20.910 to take this derivative-- gives you the [INAUDIBLE]. 00:08:20.910 --> 00:08:26.090 Direction changes causes the centripetal acceleration. 00:08:26.090 --> 00:08:29.710 To cause that acceleration, you have to put a force on it. 00:08:29.710 --> 00:08:33.059 And the force is that inward tension in the string. 00:08:33.059 --> 00:08:37.610 OK, so that's the first really simple one. 00:08:37.610 --> 00:08:41.479 Now, let's take a [INAUDIBLE]. 00:09:01.580 --> 00:09:03.710 Now let's do the same problem, but let it 00:09:03.710 --> 00:09:05.440 be a little bit more general. 00:09:05.440 --> 00:09:07.900 I've got a tabletop. 00:09:07.900 --> 00:09:10.720 This problem's described in the book as an example. 00:09:10.720 --> 00:09:12.175 You've got a hole in the table. 00:09:15.176 --> 00:09:17.380 You've got your mass out here. 00:09:17.380 --> 00:09:19.010 So this is looking down on the table 00:09:19.010 --> 00:09:21.070 and down through the hole, coming out 00:09:21.070 --> 00:09:23.610 the bottom of the hole, you have this string. 00:09:23.610 --> 00:09:25.070 And you pull on it. 00:09:31.000 --> 00:09:37.280 So r dot-- so we're looking down on it. 00:09:37.280 --> 00:09:40.565 This thing, again, has some theta. 00:09:44.060 --> 00:09:48.755 r hat directions like this, theta hat directions like that. 00:09:48.755 --> 00:09:55.855 This thing's spinning, going around and around my table top. 00:09:55.855 --> 00:09:57.820 But there's a hole in the tabletop, 00:09:57.820 --> 00:09:59.880 and I can pull the string and shorten the string. 00:10:02.820 --> 00:10:04.040 So r dot's a constant. 00:10:08.700 --> 00:10:11.548 Theta dot will not be a constant. 00:10:15.030 --> 00:10:16.160 it will change. 00:10:22.743 --> 00:10:24.390 This is my vector r. 00:10:27.610 --> 00:10:30.745 So the velocity, we'll call this A. This is my origin, 00:10:30.745 --> 00:10:35.790 O. The velocity in this case of the particle with respect 00:10:35.790 --> 00:10:47.940 to the fixed inertial frame is r dot r hat plus r theta 00:10:47.940 --> 00:10:50.005 dot theta hat. 00:10:50.005 --> 00:10:53.620 So now it has two possible velocities, 00:10:53.620 --> 00:10:55.319 and I'm going to be pulling it in, 00:10:55.319 --> 00:10:57.110 and it's going to be going round and round. 00:11:10.110 --> 00:11:41.040 So the linear momentum is just mvA with respect to O. 00:11:41.040 --> 00:11:43.930 So with this problem, if I take the time 00:11:43.930 --> 00:11:48.920 derivative of the linear momentum, you recognize this. 00:11:48.920 --> 00:11:52.020 This is just velocity in polar coordinates. 00:11:52.020 --> 00:11:55.930 We've done this derivative before to get the acceleration. 00:11:55.930 --> 00:11:58.650 So the acceleration, it's the time derivative of A 00:11:58.650 --> 00:12:06.160 with respect to O is acceleration 00:12:06.160 --> 00:12:11.620 of A with respect to O. And that's a messy formula, 00:12:11.620 --> 00:12:14.270 and we've derived it before. 00:12:14.270 --> 00:12:22.870 So that looks like-- and it's got four terms. 00:12:22.870 --> 00:12:29.520 r double dot minus r theta dot squared 00:12:29.520 --> 00:12:39.400 and the r hat plus r theta double dot plus 2r dot theta 00:12:39.400 --> 00:12:43.730 dot in the theta hat direction. 00:12:43.730 --> 00:12:47.740 And if multiplied by the mass, that's 00:12:47.740 --> 00:12:49.830 the mass times the acceleration. 00:12:49.830 --> 00:12:54.830 And this would be equal to the sum of all the external forces 00:12:54.830 --> 00:12:56.470 acting on that particle. 00:12:56.470 --> 00:12:57.720 That's Newton's second law. 00:13:02.740 --> 00:13:07.671 But in our problem here, I've said this is constant. 00:13:07.671 --> 00:13:10.095 So I can throw out this term that's going to be zero. 00:13:13.360 --> 00:13:14.415 So this term goes away. 00:13:16.935 --> 00:13:20.760 But this term's certainly not zero. 00:13:20.760 --> 00:13:24.070 This term is not necessarily zero. 00:13:24.070 --> 00:13:27.820 This term is certainly not zero. 00:13:27.820 --> 00:13:31.300 So what can we do with that? 00:13:31.300 --> 00:13:34.450 So now's the time you draw some free body diagrams. 00:13:34.450 --> 00:13:37.500 And let's look at the side view. 00:13:41.200 --> 00:13:44.930 The side view, here's your particle, 00:13:44.930 --> 00:13:52.460 mg down, some normal force up, and a tension pulling in. 00:13:55.280 --> 00:13:58.490 So the table's supporting it, gravity down, and just 00:13:58.490 --> 00:14:01.390 the string force pulling it. 00:14:01.390 --> 00:14:08.192 In the top view, you're looking down on it. 00:14:08.192 --> 00:14:09.650 I'm not allowing them to-- assuming 00:14:09.650 --> 00:14:12.500 it's a frictionless table, so there's no friction. 00:14:12.500 --> 00:14:16.090 So the only force seen from the top is the tension. 00:14:16.090 --> 00:14:24.220 There are no forces in the plane in the direction theta hat. 00:14:24.220 --> 00:14:27.200 So the sum of the external forces in the theta hat 00:14:27.200 --> 00:14:30.810 direction are in this direction. 00:14:30.810 --> 00:14:39.770 This is-- so the total forces on this thing from your free body 00:14:39.770 --> 00:14:41.755 diagram in the theta hat direction are? 00:14:41.755 --> 00:14:42.570 AUDIENCE: Zero. 00:14:42.570 --> 00:14:43.410 PROFESSOR: Zero. 00:14:43.410 --> 00:14:47.180 And that allows us to take this term and set it equal to 0. 00:14:57.860 --> 00:14:59.165 This is the x. 00:15:03.040 --> 00:15:07.370 So the theta hat piece is equal to 0 00:15:07.370 --> 00:15:17.610 is equal to m r theta double dot plus 2r dot theta dot. 00:15:17.610 --> 00:15:24.180 [INAUDIBLE] solve this for-- that's equal to 0. 00:15:24.180 --> 00:15:31.410 So r theta double dot equals minus 2r dot theta dot. 00:15:34.910 --> 00:15:41.470 So in order for this thing to satisfy Newton's law, 00:15:41.470 --> 00:15:47.150 it happens to be that as you pull it in, 00:15:47.150 --> 00:15:50.480 the product of r dot and theta dot 00:15:50.480 --> 00:15:53.230 gives you the angular acceleration. 00:15:58.340 --> 00:16:00.830 And the other term, we just went through that. 00:16:00.830 --> 00:16:03.260 So what we did a couple minutes ago. 00:16:03.260 --> 00:16:08.790 The force in the r hat direction, 00:16:08.790 --> 00:16:12.020 sum of the forces-- this is a derivative 00:16:12.020 --> 00:16:18.810 of the linear momentum-- is minus mr theta dot 00:16:18.810 --> 00:16:21.450 squared r hat. 00:16:21.450 --> 00:16:24.040 And we know from the free body diagram 00:16:24.040 --> 00:16:29.680 that that better be equal to minus T in the r hat direction. 00:16:29.680 --> 00:16:31.140 So just like before, it tells you 00:16:31.140 --> 00:16:34.900 that T equals mr theta dot squared. 00:16:34.900 --> 00:16:38.210 So the mass times the centripetal acceleration, 00:16:38.210 --> 00:16:41.340 in order to make that centripetal motion happen, 00:16:41.340 --> 00:16:48.190 you have to pull on it with a force mr theta dot squared. 00:16:48.190 --> 00:16:56.933 So what can you say about the angular momentum 00:16:56.933 --> 00:17:01.230 of this particle with respect to O? 00:17:08.960 --> 00:17:11.700 So let's write it out. 00:17:11.700 --> 00:17:17.520 So H of the particle with respect to O 00:17:17.520 --> 00:17:45.810 is rAO cross P. This is r r hat cross P. 00:17:45.810 --> 00:17:49.760 So r cross r, you get nothing from that term. 00:17:49.760 --> 00:17:53.770 r hat cross theta hat gives you a k, a positive k. 00:17:53.770 --> 00:18:15.096 So you get r r squared theta dot in the k direction. 00:18:30.950 --> 00:18:32.180 And I'm missing something. 00:18:36.120 --> 00:18:37.510 There we go. 00:18:37.510 --> 00:18:41.000 m r squared theta dot. 00:18:41.000 --> 00:18:44.290 mr theta dot is the linear momentum times 00:18:44.290 --> 00:18:46.200 another r gives you the angular momentum, 00:18:46.200 --> 00:18:47.950 m r squared theta dot. 00:18:47.950 --> 00:18:51.830 And because r hat cross theta dot is k, 00:18:51.830 --> 00:18:56.450 this angular momentum is directed upward 00:18:56.450 --> 00:18:59.840 about the center of rotation. 00:18:59.840 --> 00:19:02.730 So that's our expression for our angular momentum. 00:19:02.730 --> 00:19:11.780 But if you take the time derivative of it, 00:19:11.780 --> 00:19:12.745 what should it tell us? 00:19:12.745 --> 00:19:17.450 Let's go back to our formulas we started with. 00:19:17.450 --> 00:19:21.340 When the center of rotation's not moving, 00:19:21.340 --> 00:19:24.580 when the point with respect to which you're taking the angular 00:19:24.580 --> 00:19:27.740 momentum doesn't move, then that's 00:19:27.740 --> 00:19:31.520 one of the cases where you can get rid of those extra terms. 00:19:31.520 --> 00:19:35.120 So in this case, we're computing the angular momentum 00:19:35.120 --> 00:19:37.330 with respect to O, which doesn't move. 00:19:37.330 --> 00:19:41.560 The velocity of O here is zero. 00:19:41.560 --> 00:19:46.560 So this allows us just to say that the sum of the torques 00:19:46.560 --> 00:19:50.090 with respect about O is just equal to the time 00:19:50.090 --> 00:19:53.820 rate of change of H with respect to O. 00:19:53.820 --> 00:20:03.960 And to do that, this term can change, 00:20:03.960 --> 00:20:05.310 and this term can change. 00:20:05.310 --> 00:20:07.790 But how about the derivative of k? 00:20:07.790 --> 00:20:08.950 It's just constant, right? 00:20:08.950 --> 00:20:11.270 So we're going to get two terms out of this. 00:20:11.270 --> 00:20:32.170 We're going to get m2r r dot theta dot k plus r 00:20:32.170 --> 00:20:36.115 squared theta double dot k. 00:20:44.210 --> 00:20:49.134 What are the external torques in this problem? 00:20:49.134 --> 00:20:50.800 You're going to have to pretend that I'm 00:20:50.800 --> 00:20:53.710 on a frictionless table top here. 00:20:53.710 --> 00:20:57.440 When I'm going around like this, what 00:20:57.440 --> 00:21:02.877 are the torques about the center point? 00:21:02.877 --> 00:21:04.710 So what are the-- first, what are the forces 00:21:04.710 --> 00:21:07.590 acting on the mass? 00:21:07.590 --> 00:21:09.040 Just the tension. 00:21:09.040 --> 00:21:12.190 And the tension cross the moment arm, 00:21:12.190 --> 00:21:13.940 the tension's in the r hat direction. 00:21:13.940 --> 00:21:16.020 The string is in the r direction. 00:21:16.020 --> 00:21:17.720 r hat cross r hat is-- 00:21:17.720 --> 00:21:18.370 AUDIENCE: Zero. 00:21:18.370 --> 00:21:19.036 PROFESSOR: Zero. 00:21:19.036 --> 00:21:20.870 So there's no torques. 00:21:20.870 --> 00:21:25.770 So for this problem, this is equal to 0. 00:21:28.500 --> 00:22:06.270 And that then allows us to write-- the m cancels out, 00:22:06.270 --> 00:22:08.560 obviously, and I can get rid of-- one 00:22:08.560 --> 00:22:10.350 of these r's goes away. 00:22:10.350 --> 00:22:17.070 And I'm left with an r dot theta dot. 00:22:17.070 --> 00:22:18.740 And I move this to the other side 00:22:18.740 --> 00:22:21.300 equal to minus r theta double dot. 00:22:21.300 --> 00:22:23.700 And that's what we came up with a minute ago 00:22:23.700 --> 00:22:28.624 when when we did the time derivative 00:22:28.624 --> 00:22:30.290 of the linear momentum, we learned this. 00:22:30.290 --> 00:22:31.920 So we haven't learned much more. 00:22:31.920 --> 00:22:34.200 It's just telling us that, well, this thing's 00:22:34.200 --> 00:22:35.630 going to accelerate. 00:22:35.630 --> 00:22:37.650 It should pull it in. 00:22:37.650 --> 00:22:40.690 And it'll accelerate at any instant 00:22:40.690 --> 00:22:43.250 in time, whatever r dot is, theta dot is, this'll 00:22:43.250 --> 00:22:46.465 be the angular acceleration. 00:22:51.030 --> 00:23:02.310 So what will happen then? 00:23:02.310 --> 00:23:05.275 And let's do it at two points in time. 00:23:21.670 --> 00:23:26.704 And we'll let r2 equals r1 divided by 2. 00:23:26.704 --> 00:23:28.370 So I'm just going to pull this thing in. 00:23:34.580 --> 00:23:35.705 So let's do the experiment. 00:23:38.800 --> 00:23:41.770 I'm going to pull it in about half its length. 00:23:41.770 --> 00:23:44.765 It can speed up, slow down, stay the same speed. 00:23:49.460 --> 00:23:56.402 I'll get it going and then-- and I'll try not to hit you. 00:23:56.402 --> 00:23:59.900 I'll move over here so I hit him instead, OK? 00:23:59.900 --> 00:24:03.571 Let's try it again. 00:24:03.571 --> 00:24:04.070 All right. 00:24:04.070 --> 00:24:06.500 But there's no torques on it. 00:24:06.500 --> 00:24:09.130 There's no torque being applied. 00:24:09.130 --> 00:24:12.820 The angular momentum is constant, 00:24:12.820 --> 00:24:14.530 and yet the thing speeds up. 00:24:14.530 --> 00:24:18.290 So I want to ask you a question. 00:24:18.290 --> 00:24:21.760 Do you think the kinetic energy is staying the same 00:24:21.760 --> 00:24:22.840 or changing? 00:24:22.840 --> 00:24:28.740 So how many think the kinetic energy as I go from out there 00:24:28.740 --> 00:24:32.500 to in here, the kinetic energy stays the same? 00:24:32.500 --> 00:24:33.120 OK. 00:24:33.120 --> 00:24:35.820 How many think it's different? 00:24:35.820 --> 00:24:36.320 All right. 00:24:36.320 --> 00:24:38.655 How many are not so sure? 00:24:38.655 --> 00:24:39.280 Let's find out. 00:24:42.950 --> 00:24:47.030 So we've determined that if this is zero, 00:24:47.030 --> 00:24:52.943 then that means that h with respect to O is a constant. 00:24:56.950 --> 00:25:06.314 So h-- this is at T1-- is r1. 00:25:10.100 --> 00:25:12.970 Be faster if I look at my notes. 00:25:12.970 --> 00:25:23.240 So the angular momentum at r1 is mr1 squared 00:25:23.240 --> 00:25:27.420 theta 1 dot in the k direction. 00:25:27.420 --> 00:25:33.580 And that had better be equal to h at r2. 00:25:36.990 --> 00:25:46.010 And that'll be mr2 squared theta 2 dot. 00:25:46.010 --> 00:25:48.760 And that's also in the k direction. 00:25:48.760 --> 00:25:54.350 But we know that r2 is r1 divided by 2. 00:25:54.350 --> 00:25:55.735 So we can plug that in. 00:26:07.686 --> 00:26:09.890 I'll bring this over here. 00:26:09.890 --> 00:26:20.720 So mr1 squared theta 1 dot is mr1/4 theta 2 dot. 00:26:20.720 --> 00:26:22.230 So I can solve for theta 2. 00:26:31.780 --> 00:26:35.265 So if I shorten the length by a factor of 2, 00:26:35.265 --> 00:26:37.770 the angular velocity goes up by a factor of 4. 00:26:52.590 --> 00:26:54.230 And let's check the kinetic energy. 00:26:54.230 --> 00:27:00.915 Kinetic energy state 1, 1/2mv1 squared. 00:27:04.170 --> 00:27:12.786 That's 1/2mv1 is r theta 1 dot quantity squared. 00:27:32.204 --> 00:27:34.080 So what do I have? 00:27:34.080 --> 00:27:38.000 I have some expressions for this. 00:27:38.000 --> 00:27:51.150 r2 is r1/2, and theta 2 is 4 theta 1 dot quantity squared. 00:27:51.150 --> 00:28:27.440 And if you multiply that out-- yeah. 00:28:27.440 --> 00:28:33.740 So where's the kinetic energy come from? 00:28:33.740 --> 00:28:35.700 AUDIENCE: You're adding energy into the system 00:28:35.700 --> 00:28:37.087 by pulling down. 00:28:37.087 --> 00:28:39.170 PROFESSOR: So she says we add energy to the system 00:28:39.170 --> 00:28:39.850 by pulling down. 00:28:39.850 --> 00:28:41.530 So we're doing some work, right? 00:28:41.530 --> 00:28:46.692 There's tension in that string equal to mr omega squared. 00:28:46.692 --> 00:28:49.430 If you pull it down a certain distance-- in fact, 00:28:49.430 --> 00:28:52.560 r1/2-- you're going to do work that's 00:28:52.560 --> 00:28:54.960 the integral of the tension times 00:28:54.960 --> 00:28:57.350 dr. You integrate it, right? 00:28:57.350 --> 00:29:01.280 And that work goes into-- there's conservation 00:29:01.280 --> 00:29:02.280 of energy in the system. 00:29:02.280 --> 00:29:04.970 That goes into speeding up the rotation, 00:29:04.970 --> 00:29:07.730 and yet the angular momentum has stayed constant 00:29:07.730 --> 00:29:10.740 throughout the action. 00:29:10.740 --> 00:29:12.290 So when I first saw this years ago, 00:29:12.290 --> 00:29:13.854 I thought, that's really cool. 00:29:13.854 --> 00:29:15.020 That's really quite amazing. 00:29:19.050 --> 00:29:21.190 So a nice application of conservation 00:29:21.190 --> 00:29:23.580 of angular momentum. 00:29:23.580 --> 00:29:27.080 An application of using this formula, the time 00:29:27.080 --> 00:29:29.140 rate of change of angular momentum 00:29:29.140 --> 00:29:32.080 with respect to a point, it tells you 00:29:32.080 --> 00:29:34.980 about the torques applied to the system. 00:29:34.980 --> 00:29:37.770 And this is in fact a pretty simple case. 00:29:37.770 --> 00:29:43.040 So the last-- let's move on though to doing 00:29:43.040 --> 00:29:45.900 a little more complicated case. 00:29:45.900 --> 00:29:51.700 And this is similar to the last problem in the homework. 00:30:01.547 --> 00:30:02.755 So this is like the homework. 00:30:14.380 --> 00:30:17.410 The homework, you got this monkey running up the shaft, 00:30:17.410 --> 00:30:17.910 right? 00:30:17.910 --> 00:30:20.870 So I don't have a monkey, and it's not running up the shaft. 00:30:20.870 --> 00:30:31.810 But I do have just this particle on a shaft 00:30:31.810 --> 00:30:35.896 rotating about a central axis. 00:30:35.896 --> 00:30:37.940 Now, this is a mechanical necessity 00:30:37.940 --> 00:30:38.940 to hold it all together. 00:30:38.940 --> 00:30:40.865 But let's just ignore the mass of this center 00:30:40.865 --> 00:30:41.740 piece for the moment. 00:30:41.740 --> 00:30:47.340 Just think of this as a massless arm with a particle on it. 00:30:47.340 --> 00:30:50.680 And I want to calculate angular momentum. 00:30:50.680 --> 00:30:52.500 I want to calculate forces. 00:30:52.500 --> 00:30:55.245 I want to calculate torques and see what happens. 00:31:14.590 --> 00:31:25.340 So I'm going to start by putting my O, x, z frame right 00:31:25.340 --> 00:31:27.910 on the level with this mass. 00:31:31.165 --> 00:31:32.540 What I'm going to show you now is 00:31:32.540 --> 00:31:38.200 that where you put the reference point about which you compute 00:31:38.200 --> 00:31:41.510 the angular momentum matters. 00:31:41.510 --> 00:31:44.530 You get different answers depending on where you put it. 00:31:44.530 --> 00:31:48.810 So I'm going to start by putting it here. 00:31:48.810 --> 00:31:50.760 And this we'll call [INAUDIBLE] out here's 00:31:50.760 --> 00:31:59.275 my point A. Here's O. This is some angle phi here. 00:32:02.390 --> 00:32:10.420 And if this has some length l, then this up here is my r 00:32:10.420 --> 00:32:16.270 equals l cosine phi. 00:32:16.270 --> 00:32:21.186 And this side would be l sine phi. 00:32:21.186 --> 00:32:22.810 These are just the two lengths, but I'm 00:32:22.810 --> 00:32:24.910 going to use polar coordinates. 00:32:24.910 --> 00:32:28.240 So this is going to be my r hat direction. 00:32:28.240 --> 00:32:29.580 Theta hat's into the board. 00:32:40.070 --> 00:32:43.120 So let's compute-- and it's a particle, 00:32:43.120 --> 00:32:49.050 so I'll continue to use lowercase h of A 00:32:49.050 --> 00:32:57.840 with respect to O-- it's a vector-- is r of A with respect 00:32:57.840 --> 00:33:02.450 to O cross P linear momentum with respect 00:33:02.450 --> 00:33:08.840 to O. This is r r hat. 00:33:08.840 --> 00:33:11.230 We just call this l cosine theta, just calling it r. 00:33:11.230 --> 00:33:13.920 It's in the r hat direction. 00:33:13.920 --> 00:33:18.930 Cross with P, and P, we'd done this two or three times now 00:33:18.930 --> 00:33:19.670 today. 00:33:19.670 --> 00:33:26.290 It's the mass times r times theta dot. 00:33:26.290 --> 00:33:28.040 That's its speed. 00:33:28.040 --> 00:33:31.325 And it's in the theta hat direction. 00:33:38.190 --> 00:33:42.640 So taking the r hat cross theta hat gives me k. 00:33:45.220 --> 00:33:53.970 So I get m r squared theta dot k. 00:34:00.347 --> 00:34:01.305 Very simple expression. 00:34:05.980 --> 00:34:08.560 Now, this is a fixed axis rotation. 00:34:08.560 --> 00:34:12.440 So I want to compute the torques. 00:34:12.440 --> 00:34:14.260 Look at my formula. 00:34:14.260 --> 00:34:19.860 The vA in this case is the velocity of O. 00:34:19.860 --> 00:34:23.870 The point of the axis of rotation doesn't move. 00:34:23.870 --> 00:34:26.590 So the second terms go away, and I 00:34:26.590 --> 00:34:34.790 can say that the torque of my particle at A with respect to O 00:34:34.790 --> 00:34:38.730 is just dhA dt. 00:34:43.210 --> 00:34:45.530 So m's a constant. 00:34:45.530 --> 00:34:46.830 r's a constant. 00:34:46.830 --> 00:34:48.719 k's a constant. 00:34:48.719 --> 00:34:51.534 The only thing that has a time derivative is theta dot, 00:34:51.534 --> 00:34:54.136 and it becomes theta double dot. 00:34:54.136 --> 00:35:01.690 This is m r squared theta double dot k. 00:35:01.690 --> 00:35:07.630 And that is equal-- well, that's equal to the sum 00:35:07.630 --> 00:35:10.790 of the external torques. 00:35:10.790 --> 00:35:14.140 So what physically does that mean? 00:35:14.140 --> 00:35:17.010 What physically is that telling us? 00:35:17.010 --> 00:35:20.110 It's telling us theta double dot is the angular 00:35:20.110 --> 00:35:22.850 acceleration of this thing speeding up, 00:35:22.850 --> 00:35:25.270 going faster and faster. 00:35:25.270 --> 00:35:29.590 It takes torque to make that happen. 00:35:29.590 --> 00:35:34.310 If it's going at constant rate, what's theta double dot? 00:35:34.310 --> 00:35:35.650 Zero. 00:35:35.650 --> 00:35:38.390 So at constant rate, the torque required 00:35:38.390 --> 00:35:42.070 to make this thing go constant rate is zero. 00:35:42.070 --> 00:35:43.080 Makes sense. 00:35:43.080 --> 00:35:45.200 But if it were speeding up, if you're 00:35:45.200 --> 00:35:47.330 making it go faster and faster and faster, 00:35:47.330 --> 00:35:48.890 it requires torque to drive it. 00:35:48.890 --> 00:35:50.260 And that's the amount of torque. 00:35:50.260 --> 00:35:55.512 And the torque is around the axis of spin. 00:35:55.512 --> 00:35:56.470 Pretty straightforward. 00:36:02.670 --> 00:36:07.670 So if I did dP [? d, ?] if I took the time 00:36:07.670 --> 00:36:11.760 derivative of just the linear momentum for this particle, 00:36:11.760 --> 00:36:13.180 we've done it here today. 00:36:13.180 --> 00:36:17.810 If I took the time derivative of it, what would I get? 00:36:17.810 --> 00:36:18.440 It's a force. 00:36:18.440 --> 00:36:19.314 And what's the force? 00:36:25.730 --> 00:36:27.650 It's a constant rotation rate. 00:36:27.650 --> 00:36:32.930 Take the time derivative of P. dP dt gives me 00:36:32.930 --> 00:36:34.340 AUDIENCE: [INAUDIBLE]. 00:36:34.340 --> 00:36:35.781 PROFESSOR: Mass times? 00:36:35.781 --> 00:36:37.890 AUDIENCE: [INAUDIBLE]. 00:36:37.890 --> 00:36:40.720 PROFESSOR: Mv squared over r is mass times acceleration. 00:36:40.720 --> 00:36:42.574 The acceleration is which kind? 00:36:42.574 --> 00:36:43.490 AUDIENCE: [INAUDIBLE]. 00:36:43.490 --> 00:36:44.060 PROFESSOR: Centripetal. 00:36:44.060 --> 00:36:46.030 So you just get the same thing back again. 00:36:46.030 --> 00:36:51.660 So there's a force acting inwards words this thing 00:36:51.660 --> 00:36:56.630 to make it go in a circle that is the mr theta dot squared 00:36:56.630 --> 00:37:00.270 term that we've seen so many times. 00:37:00.270 --> 00:37:09.420 But now what I want to do is move the point about which I 00:37:09.420 --> 00:37:10.940 compute this angular momentum. 00:37:16.740 --> 00:37:19.070 And now I'm going to put it here, 00:37:19.070 --> 00:37:22.220 the point of attachment of the arm. 00:37:22.220 --> 00:37:27.280 So here's O, x, z. 00:37:27.280 --> 00:37:28.670 Everything else stays the same. 00:37:28.670 --> 00:37:31.460 All I've done is move the point. 00:37:31.460 --> 00:37:35.960 And I want to compute the angular momentum 00:37:35.960 --> 00:37:43.570 of this A with respect to O. 00:37:43.570 --> 00:37:44.650 Well, that's r. 00:37:44.650 --> 00:37:54.380 This is now r of A with respect to O, this vector. 00:37:54.380 --> 00:37:59.140 This distance here, in polar cylindrical coordinates, is z. 00:38:04.210 --> 00:38:05.430 And this is r. 00:38:05.430 --> 00:38:06.290 Just as before. 00:38:06.290 --> 00:38:08.870 The r hasn't changed. 00:38:08.870 --> 00:38:11.960 And there is an r hat in this direction, 00:38:11.960 --> 00:38:17.660 theta hat into the board, and a k hat in the z direction. 00:38:17.660 --> 00:38:19.750 Those are our unit vectors. 00:38:19.750 --> 00:38:34.240 So this is-- rAO is r r hat plus z k hat. 00:38:34.240 --> 00:38:38.660 That's the position vector. 00:38:38.660 --> 00:38:43.610 And I'm going to cross that with the momentum of A 00:38:43.610 --> 00:38:55.030 with respect to O. And the momentum 00:38:55.030 --> 00:39:04.180 is mr. Theta dot is the velocity. 00:39:04.180 --> 00:39:05.450 And what's its direction? 00:39:08.000 --> 00:39:10.410 Theta hat, right? 00:39:10.410 --> 00:39:12.820 Mass times velocity, momentum. 00:39:12.820 --> 00:39:17.540 And now we need to carry this out. 00:39:17.540 --> 00:39:22.230 The r hat term times theta hat gives you a k. 00:39:27.195 --> 00:39:35.760 m r squared theta dot k. 00:39:35.760 --> 00:39:44.000 And this term, k cross theta hat, gives me a minus r. 00:39:44.000 --> 00:39:57.915 Minus mrz theta dot k. 00:40:03.616 --> 00:40:05.820 AUDIENCE: [INAUDIBLE]. 00:40:05.820 --> 00:40:06.820 PROFESSOR: You're right. 00:40:06.820 --> 00:40:08.510 Thank you. 00:40:08.510 --> 00:40:13.240 Because I would have a disaster if I let that progress. 00:40:13.240 --> 00:40:16.745 This looks like that, right? k cross theta is minus r hat. 00:40:19.260 --> 00:40:21.346 r cross theta is a k. 00:40:21.346 --> 00:40:23.650 k cross theta is a minus r. 00:40:23.650 --> 00:40:26.850 I get two terms. 00:40:26.850 --> 00:40:30.505 And this is now an expression for the angular momentum of A 00:40:30.505 --> 00:40:34.960 with respect to O. And let's see if I have enough room 00:40:34.960 --> 00:40:36.360 to draw it here. 00:40:36.360 --> 00:40:41.120 There is a piece of it here. 00:40:41.120 --> 00:40:49.180 This is h in the z direction, is this arrow. 00:40:49.180 --> 00:40:55.245 And then there's a piece in the r hat direction, like this. 00:40:55.245 --> 00:40:59.780 This is h in the r hat direction. 00:40:59.780 --> 00:41:05.420 And the sum of those two is that. 00:41:05.420 --> 00:41:10.930 So this is hA with respect to O, this guy. 00:41:10.930 --> 00:41:15.829 Perpendicular to the shaft. 00:41:15.829 --> 00:41:17.620 It will turn out it really is perpendicular 00:41:17.620 --> 00:41:20.530 if you work out the numbers. 00:41:20.530 --> 00:41:27.860 Totally different result than when I did it here. 00:41:27.860 --> 00:41:32.550 So this one, m r squared theta dot k, m r squared theta dot k. 00:41:32.550 --> 00:41:34.450 Hey, that term's the same. 00:41:34.450 --> 00:41:38.360 So when I did this, I got just the k term. 00:41:38.360 --> 00:41:40.900 And now I've moved this thing down, and I get a second term. 00:41:44.550 --> 00:41:47.712 So now what we want to know is, what about the torques 00:41:47.712 --> 00:41:48.295 in the system? 00:41:58.110 --> 00:42:01.000 So I want to take the time derivative 00:42:01.000 --> 00:42:04.980 of this guy with respect to-- the time 00:42:04.980 --> 00:42:08.290 derivative of the angular momentum, which 00:42:08.290 --> 00:42:11.160 is going to be equal to the summation 00:42:11.160 --> 00:42:14.870 of the external torques with respect to O, 00:42:14.870 --> 00:42:18.110 but O is now in a different place. 00:42:18.110 --> 00:42:21.100 And so I have to carry out these derivatives. 00:42:21.100 --> 00:42:23.130 And this one I did before. 00:42:23.130 --> 00:42:32.045 This one just gives me my m r squared theta double dot k hat. 00:42:32.045 --> 00:42:36.530 Now, this term, m is a constant. 00:42:36.530 --> 00:42:38.800 r is a constant. 00:42:38.800 --> 00:42:42.260 z is a constant. 00:42:42.260 --> 00:42:49.036 Theta dot is not necessarily a constant. 00:42:49.036 --> 00:42:51.600 We're going to let that be a variable. 00:42:51.600 --> 00:42:55.540 And r hat is certainly changing direction. 00:42:55.540 --> 00:42:57.320 So when I take the derivative of this, 00:42:57.320 --> 00:42:59.700 I'm going to get two terms. 00:42:59.700 --> 00:43:10.700 So the first one is minus mrz theta double dot r hat. 00:43:10.700 --> 00:43:13.505 And that's taking the derivative of this multiplied by that. 00:43:13.505 --> 00:43:15.520 And the second term is the derivative 00:43:15.520 --> 00:43:17.810 of this multiplied by that. 00:43:17.810 --> 00:43:22.200 And so the derivative of r hat is? 00:43:26.480 --> 00:43:27.960 AUDIENCE: [INAUDIBLE]. 00:43:27.960 --> 00:43:29.410 PROFESSOR: Theta dot theta hat. 00:43:29.410 --> 00:43:30.080 Right. 00:43:30.080 --> 00:43:30.870 OK. 00:43:30.870 --> 00:43:45.210 So minus mrz theta dot theta dot theta hat. 00:43:47.720 --> 00:43:50.390 So in other words, this is squared. 00:43:50.390 --> 00:43:54.210 Now I have three terms to mess with. 00:43:54.210 --> 00:43:58.100 We know what the first term means. 00:43:58.100 --> 00:44:00.140 We talked about that. 00:44:00.140 --> 00:44:01.220 This is the torque. 00:44:01.220 --> 00:44:02.910 These are all torques. 00:44:02.910 --> 00:44:06.060 So this is the torque required to do what, the lead term? 00:44:09.030 --> 00:44:11.792 AUDIENCE: [INAUDIBLE]. 00:44:11.792 --> 00:44:13.000 AUDIENCE: [INAUDIBLE] circle. 00:44:13.000 --> 00:44:15.064 PROFESSOR: To make it? 00:44:15.064 --> 00:44:16.030 AUDIENCE: [INAUDIBLE]. 00:44:16.030 --> 00:44:16.904 PROFESSOR: Go faster. 00:44:16.904 --> 00:44:18.990 Change its angular speed, right? 00:44:18.990 --> 00:44:23.860 It's just building up the angular momentum in that spin. 00:44:23.860 --> 00:44:29.190 So this is the angular spin up. 00:44:29.190 --> 00:44:33.405 These other two terms, these are strange things. 00:44:33.405 --> 00:44:39.100 Well first, let's take a look at this one. r theta dot squared. 00:44:39.100 --> 00:44:40.528 What's that remind you of? 00:44:43.280 --> 00:44:49.200 What kind of-- torque is usually some force times a moment arm, 00:44:49.200 --> 00:44:51.920 crossed with a moment arm, right? 00:44:51.920 --> 00:44:56.040 So we know that there's some forces acting in this system. 00:44:56.040 --> 00:44:58.080 It's spinning. 00:44:58.080 --> 00:45:03.160 We know that there is a-- in order 00:45:03.160 --> 00:45:07.030 to make this thing go round and around-- it 00:45:07.030 --> 00:45:09.060 has centripetal acceleration. 00:45:09.060 --> 00:45:13.720 Therefore, there must be a force being applied by this shaft 00:45:13.720 --> 00:45:17.840 inward that's equal to the mass times 00:45:17.840 --> 00:45:22.035 the centripetal acceleration, mr theta dot squared. 00:45:26.480 --> 00:45:33.786 So here this guy is mr theta dot squared. 00:45:37.060 --> 00:45:39.650 That's the force. 00:45:39.650 --> 00:45:42.110 And let's do these. 00:45:42.110 --> 00:45:44.770 Let's call this A. Let's Call. 00:45:44.770 --> 00:45:55.140 This term here B, this term C. So the C term is-- torque C, 00:45:55.140 --> 00:46:01.780 I'll call it-- is some r cross some F. 00:46:01.780 --> 00:46:10.110 And the F, I'm telling you, is the centripetal acceleration 00:46:10.110 --> 00:46:11.260 times the mass. 00:46:11.260 --> 00:46:18.110 And that'll probably be like a minus mr 00:46:18.110 --> 00:46:24.200 theta dot squared r hat, right? 00:46:24.200 --> 00:46:30.650 And what about the moment arm that that acts about? 00:46:30.650 --> 00:46:32.740 What moment arm is perpendicular-- so 00:46:32.740 --> 00:46:35.450 that's a force that's acting in. 00:46:35.450 --> 00:46:37.244 What moment arm is perpendicular to that? 00:46:37.244 --> 00:46:39.160 Because the only thing that's perpendicular to 00:46:39.160 --> 00:46:42.187 it lead to torques. 00:46:42.187 --> 00:46:42.770 PROFESSOR: Hm? 00:46:42.770 --> 00:46:43.550 AUDIENCE: [INAUDIBLE]. 00:46:43.550 --> 00:46:44.091 PROFESSOR: z. 00:46:52.670 --> 00:46:58.525 k cross r should give me a theta. 00:47:02.100 --> 00:47:07.280 Sure enough, there's a minus, sure enough. 00:47:07.280 --> 00:47:09.600 And there's the z. 00:47:09.600 --> 00:47:12.690 You multiply this out, you get this term. 00:47:12.690 --> 00:47:14.820 So this is a strange term. 00:47:14.820 --> 00:47:19.880 It's in the theta hat direction. 00:47:19.880 --> 00:47:21.100 What is that? 00:47:21.100 --> 00:47:25.080 So it's spinning, and it's lined up like this. 00:47:25.080 --> 00:47:27.900 Theta hat's in that direction. 00:47:27.900 --> 00:47:33.370 Positive k, positive theta, positive r hat. 00:47:33.370 --> 00:47:34.510 k's in that direction. 00:47:34.510 --> 00:47:36.670 It's telling you there's a torque 00:47:36.670 --> 00:47:42.684 being applied about this point in the minus theta direction. 00:47:42.684 --> 00:47:44.670 Does that make sense? 00:47:44.670 --> 00:47:49.850 You have a-- there's a centripetal acceleration 00:47:49.850 --> 00:47:50.660 times a mass. 00:47:50.660 --> 00:47:53.860 There's a force times a moment arm. 00:47:53.860 --> 00:47:59.317 This force is trying to bend this thing out. 00:47:59.317 --> 00:48:01.900 If this thing had a hinge down here, and I started to spin it, 00:48:01.900 --> 00:48:02.608 what would it do? 00:48:02.608 --> 00:48:04.960 It would just flop out, right? 00:48:04.960 --> 00:48:08.670 There's got to be a torque keeping that from happening. 00:48:08.670 --> 00:48:10.190 And that's a torque. 00:48:10.190 --> 00:48:11.870 If it wants to go that way, there's 00:48:11.870 --> 00:48:15.360 got to be a torque going the other way keeping it in place. 00:48:15.360 --> 00:48:18.330 And that's what that term is. 00:48:18.330 --> 00:48:18.990 So now we know. 00:48:18.990 --> 00:48:20.710 So this is the Euler. 00:48:20.710 --> 00:48:22.630 This is the spin up. 00:48:22.630 --> 00:48:26.410 This is keeping this thing from flopping out. 00:48:26.410 --> 00:48:27.990 What's this guy? 00:48:27.990 --> 00:48:33.640 This is yet another torque, and it's in the r, minus r, 00:48:33.640 --> 00:48:34.140 direction. 00:48:36.960 --> 00:48:38.780 So let's see if we can intuitively 00:48:38.780 --> 00:48:40.600 figure this one out. 00:48:40.600 --> 00:48:44.560 Well, there's an r theta double dot. 00:48:44.560 --> 00:48:46.260 That should look familiar. 00:48:46.260 --> 00:48:49.350 r theta double dot. 00:48:49.350 --> 00:48:53.060 If this thing is accelerating, angular acceleration, 00:48:53.060 --> 00:48:59.650 speeding up, out here that mass says, I'm here right now. 00:48:59.650 --> 00:49:01.765 In order for me to go a little bit faster, 00:49:01.765 --> 00:49:04.320 I'm accelerating in that direction. 00:49:04.320 --> 00:49:06.070 There must be a force being applied 00:49:06.070 --> 00:49:09.260 by this rod in that direction. 00:49:09.260 --> 00:49:15.390 And that force times a moment arm perpendicular to it, z, 00:49:15.390 --> 00:49:20.870 is a moment in the r direction. 00:49:20.870 --> 00:49:23.770 So as this thing spins up, if this thing could, 00:49:23.770 --> 00:49:25.740 it would fall back. 00:49:25.740 --> 00:49:27.820 But this rod is stiff and won't let it do that. 00:49:27.820 --> 00:49:30.980 So this is the torque down here that 00:49:30.980 --> 00:49:33.660 is required to keep this thing moving. 00:49:36.200 --> 00:49:38.690 So this is really quite amazing. 00:49:41.200 --> 00:49:46.550 Do either of these torques, these second ones, this one 00:49:46.550 --> 00:49:50.880 and this one, do they contribute to-- do they do any work? 00:49:53.460 --> 00:49:56.330 Do they add energy to the system? 00:49:56.330 --> 00:49:59.720 See, work means force through a distance. 00:49:59.720 --> 00:50:02.020 They don't do actually any work. 00:50:02.020 --> 00:50:04.260 They are static torques just required 00:50:04.260 --> 00:50:06.370 to hold the system together. 00:50:06.370 --> 00:50:07.590 This one does some work. 00:50:07.590 --> 00:50:10.715 It actually makes-- this one leads to energy accumulating, 00:50:10.715 --> 00:50:13.000 just going faster and faster faster. 00:50:13.000 --> 00:50:15.060 These are just holding the thing together. 00:50:15.060 --> 00:50:18.720 But the amazing thing is that you 00:50:18.720 --> 00:50:21.930 can use angular momentum to calculate things 00:50:21.930 --> 00:50:24.040 like these torques. 00:50:24.040 --> 00:50:30.520 So if you were designing this, these forces 00:50:30.520 --> 00:50:35.490 acting on this lump out here are producing torques 00:50:35.490 --> 00:50:39.950 about this point, which are the same thing as-- in 2.001 you'll 00:50:39.950 --> 00:50:42.200 be doing bending moments. 00:50:42.200 --> 00:50:45.124 It creates a bending moment in the shaft. 00:50:45.124 --> 00:50:47.040 And if you don't make the shaft strong enough, 00:50:47.040 --> 00:50:49.270 it'll break it off. 00:50:49.270 --> 00:50:54.250 So the torque about this point is the bending moment 00:50:54.250 --> 00:50:58.230 in the r direction and in the theta direction. 00:50:58.230 --> 00:51:00.680 It's trying to be bent in two different ways. 00:51:00.680 --> 00:51:03.480 And you can calculate the stresses down here caused 00:51:03.480 --> 00:51:05.020 by those moments. 00:51:05.020 --> 00:51:08.760 And that would help you design the thing. 00:51:08.760 --> 00:51:11.590 So you not only get dynamics information 00:51:11.590 --> 00:51:13.770 out of taking things like the time rate of change 00:51:13.770 --> 00:51:15.250 of angular momentum. 00:51:15.250 --> 00:51:20.050 You get some of the static information as well. 00:51:20.050 --> 00:51:24.200 The other thing to remember, the really important point 00:51:24.200 --> 00:51:29.720 of the lecture, is that angular momentum 00:51:29.720 --> 00:51:34.910 changes depending on where you pick a reference point. 00:51:34.910 --> 00:51:38.450 So when we picked the reference point just opposite it, 00:51:38.450 --> 00:51:41.500 we got none of the information about the torques down here. 00:51:41.500 --> 00:51:43.720 Because with respect to this point, 00:51:43.720 --> 00:51:47.890 there are no torques except the one speeded up. 00:51:47.890 --> 00:51:51.940 The centripetal force here doesn't cause torques. 00:51:51.940 --> 00:51:57.179 The force out here-- there are no other torques, just 00:51:57.179 --> 00:51:58.470 the one to make it spin faster. 00:51:58.470 --> 00:52:00.220 But as soon as I move it down here, 00:52:00.220 --> 00:52:04.350 I learn something of considerable value. 00:52:04.350 --> 00:52:07.404 So the homework problem has some of the same things 00:52:07.404 --> 00:52:08.820 on it, except the monkey's moving. 00:52:08.820 --> 00:52:11.780 So you get even a little bit more 00:52:11.780 --> 00:52:14.760 interesting information out of it. 00:52:14.760 --> 00:52:15.260 All right. 00:52:28.090 --> 00:52:32.050 So that went faster than I thought. 00:52:32.050 --> 00:52:34.040 So that gives some time for some questions. 00:52:34.040 --> 00:52:35.230 I could see several. 00:52:35.230 --> 00:52:38.048 So we'll start there and then go here. 00:52:38.048 --> 00:52:39.920 AUDIENCE: What exactly [INAUDIBLE] 00:52:39.920 --> 00:52:42.730 torque B is balancing out? 00:52:42.730 --> 00:52:44.704 PROFESSOR: So say again? 00:52:44.704 --> 00:52:48.600 AUDIENCE: Torque B [INAUDIBLE], what 00:52:48.600 --> 00:52:51.515 exactly is that balancing out? 00:52:51.515 --> 00:52:52.390 PROFESSOR: This term? 00:52:52.390 --> 00:52:54.017 You're asking about this term? 00:52:54.017 --> 00:52:55.267 AUDIENCE: [INAUDIBLE]. 00:52:55.267 --> 00:52:55.850 PROFESSOR: OK. 00:52:55.850 --> 00:52:57.915 You want me to explain again what this one means? 00:53:02.850 --> 00:53:09.130 This is a term associated with increasing the angular speed. 00:53:09.130 --> 00:53:21.910 So let's see if we can't-- so this B term, 00:53:21.910 --> 00:53:27.510 I'll call it the torque associated with B, 00:53:27.510 --> 00:53:34.120 minus mrz theta double dot r hat. 00:53:34.120 --> 00:53:42.260 Now, that is going to be some r cross some F. 00:53:42.260 --> 00:53:44.310 And if we can get some physical insight, 00:53:44.310 --> 00:53:46.820 if we could figure out what they are. 00:53:46.820 --> 00:53:55.740 So the mass, the force, is the mass times an acceleration. 00:53:55.740 --> 00:53:58.450 There's an acceleration r theta double dot, which 00:53:58.450 --> 00:54:03.690 is the speed of this thing is increasing speed. 00:54:03.690 --> 00:54:07.490 And the r is going to be z. 00:54:07.490 --> 00:54:11.860 z k hat cross-- and I'm guessing that it's a force that 00:54:11.860 --> 00:54:18.820 looks like mr theta double dot. 00:54:18.820 --> 00:54:26.415 And this is in the theta hat direction. 00:54:29.550 --> 00:54:33.440 k cross theta hat gives me minus r hat. 00:54:33.440 --> 00:54:34.964 r hat and the minus. 00:54:38.130 --> 00:54:40.940 So this looks like a plausible explanation for where 00:54:40.940 --> 00:54:42.080 this might have come from. 00:54:42.080 --> 00:54:48.360 So this is the force speeding up, attempting 00:54:48.360 --> 00:54:49.615 to speed up this mass. 00:54:49.615 --> 00:54:53.450 There's a force pushing on it that's given to it by this rod. 00:54:53.450 --> 00:54:56.950 This rod is pushing on it to make it go faster. 00:54:56.950 --> 00:55:01.600 Mass times acceleration would be mr theta double dot. 00:55:01.600 --> 00:55:04.110 And the moment arm is this distance 00:55:04.110 --> 00:55:06.440 from here down to the point at which 00:55:06.440 --> 00:55:12.240 I've been computing my reference point from here to here at z. 00:55:12.240 --> 00:55:16.600 So force times z, r cross F, puts it 00:55:16.600 --> 00:55:19.985 in the-- ends up in the minus r direction. 00:55:19.985 --> 00:55:23.636 It's got to be this direction, k cross theta hat. 00:55:23.636 --> 00:55:25.070 The force is this way. 00:55:31.542 --> 00:55:32.250 Think about this. 00:55:32.250 --> 00:55:34.040 Force is that way. 00:55:34.040 --> 00:55:37.240 The r is this way. 00:55:37.240 --> 00:55:41.840 The r cross F is this way. 00:55:41.840 --> 00:55:43.920 And that's where you get the minus sign. 00:55:43.920 --> 00:55:47.130 It's in the minus r hat direction. 00:55:47.130 --> 00:55:51.890 But it comes from trying to speed up this mass. 00:55:51.890 --> 00:55:58.260 And if this was a floppy, weak link, as it tries to speed up, 00:55:58.260 --> 00:56:00.300 it would try to bend back. 00:56:00.300 --> 00:56:04.070 It would flop back as this thing tries to make it go faster. 00:56:04.070 --> 00:56:05.600 It will say, no, I don't want to go. 00:56:05.600 --> 00:56:06.710 Lay back on me. 00:56:06.710 --> 00:56:11.320 And that would we going in the-- this way, 00:56:11.320 --> 00:56:12.710 to keep from doing that, you have 00:56:12.710 --> 00:56:16.420 to put a torque on it this way. 00:56:16.420 --> 00:56:17.460 Just trying to speed up. 00:56:17.460 --> 00:56:19.220 It's trying to lay back, and you're saying nope, 00:56:19.220 --> 00:56:19.940 can't do that. 00:56:19.940 --> 00:56:21.985 Go like this. 00:56:21.985 --> 00:56:22.860 So that's the B term. 00:56:26.020 --> 00:56:30.194 AUDIENCE: If you do the problem with a situation like that, 00:56:30.194 --> 00:56:32.150 how do you know where to set it? 00:56:32.150 --> 00:56:33.180 PROFESSOR: How do you know where to set it? 00:56:33.180 --> 00:56:34.959 Well, that's a good question. 00:56:34.959 --> 00:56:37.250 He's saying, when you're doing a problem like this, how 00:56:37.250 --> 00:56:39.570 do you know where to pick the reference frame? 00:56:39.570 --> 00:56:42.856 Well, ask yourself what it is you want to know. 00:56:42.856 --> 00:56:46.980 And in fact, now that you know that from angular momentum 00:56:46.980 --> 00:56:50.470 of mechanical things, you can actually 00:56:50.470 --> 00:56:53.940 get static torques on the system, 00:56:53.940 --> 00:56:56.640 ask yourself where you want to know those torques. 00:56:56.640 --> 00:56:58.480 In this case, if you're designing this, 00:56:58.480 --> 00:57:01.200 you want to know whether you're going to break this thing off. 00:57:01.200 --> 00:57:04.760 And it's probably going to break right down at the bottom 00:57:04.760 --> 00:57:07.140 where the moment arms are the greatest. 00:57:07.140 --> 00:57:08.650 So that's why you pick that point. 00:57:11.910 --> 00:57:18.460 It comes a lot with experience will help you choose. 00:57:18.460 --> 00:57:20.990 But the amazing thing is this information's 00:57:20.990 --> 00:57:23.200 all stored in the angular momentum 00:57:23.200 --> 00:57:25.120 if you pick it in the right place. 00:57:25.120 --> 00:57:25.700 Another hand. 00:57:31.100 --> 00:57:32.240 OK. 00:57:32.240 --> 00:57:34.390 Yes, Phillip. 00:57:34.390 --> 00:57:37.785 AUDIENCE: I had a question about the direction for r hat. 00:57:37.785 --> 00:57:40.210 I thought it had to come out of the origin. 00:57:40.210 --> 00:57:42.635 But you have it going in the x direction. 00:57:47.510 --> 00:57:49.350 PROFESSOR: So I've been using polar-- he's 00:57:49.350 --> 00:57:51.230 asking the direction of r hat. 00:57:51.230 --> 00:57:53.590 So I've just been using polar coordinates. 00:57:53.590 --> 00:57:58.710 And polar coordinates is cylindrical, technically. 00:57:58.710 --> 00:58:05.380 This problem has a z direction upwards, r direction 00:58:05.380 --> 00:58:09.960 radially outwards, r hat, and theta as drawn here 00:58:09.960 --> 00:58:14.110 would be into the board, given the position of the mass. 00:58:14.110 --> 00:58:20.530 Looking down on it, here's my O. And looking down, 00:58:20.530 --> 00:58:23.930 this would be my x and my y. 00:58:23.930 --> 00:58:27.810 And this is some random arbitrary position here. 00:58:27.810 --> 00:58:31.260 And this is theta. 00:58:31.260 --> 00:58:33.770 Looking down on it, in this plane 00:58:33.770 --> 00:58:41.690 is r hat theta hat, k hat coming out of the board. 00:58:41.690 --> 00:58:53.130 Side view, x, z, and my system is like this. 00:58:53.130 --> 00:59:02.130 Now the theta is into the board, and the r direction 00:59:02.130 --> 00:59:02.950 is this way. 00:59:02.950 --> 00:59:03.880 That's r hat. 00:59:03.880 --> 00:59:05.850 And this is z. 00:59:05.850 --> 00:59:09.840 This is the z-coordinate upwards. 00:59:09.840 --> 00:59:15.260 So the position vector, the thing we call rA with respect 00:59:15.260 --> 00:59:21.410 to O, is indeed the length of this whole thing. 00:59:21.410 --> 00:59:29.740 But it is made up of a component in the z direction 00:59:29.740 --> 00:59:39.688 plus a component here that we call r in the r hat direction. 00:59:39.688 --> 00:59:42.592 AUDIENCE: [INAUDIBLE] example where we calculated from 00:59:42.592 --> 00:59:45.980 the bottom rather than the top circle, 00:59:45.980 --> 00:59:50.094 then we got a value for the angular momentum that 00:59:50.094 --> 00:59:51.788 doesn't have a theta hat component, 00:59:51.788 --> 00:59:53.040 but as the thing spins-- 00:59:53.040 --> 00:59:54.300 PROFESSOR: Which are you referring to? 00:59:54.300 --> 00:59:54.883 Are you talk-- 00:59:54.883 --> 00:59:58.660 AUDIENCE: h about the point O down-- 00:59:58.660 --> 01:00:00.350 PROFESSOR: Which example, this guy or-- 01:00:00.350 --> 01:00:01.516 AUDIENCE: Yeah, [INAUDIBLE]. 01:00:01.516 --> 01:00:03.000 PROFESSOR: OK, so point. 01:00:03.000 --> 01:00:04.240 Tell me what you mean here. 01:00:04.240 --> 01:00:05.656 AUDIENCE: Right here. [INAUDIBLE]. 01:00:05.656 --> 01:00:08.326 PROFESSOR: When we computed, not here, but down here. 01:00:08.326 --> 01:00:11.302 AUDIENCE: We got that there was no theta component, 01:00:11.302 --> 01:00:14.774 but as this spins around, theta is changing. 01:00:14.774 --> 01:00:16.975 And if it's always opposite, shouldn't there 01:00:16.975 --> 01:00:17.850 be a theta component? 01:00:17.850 --> 01:00:18.860 PROFESSOR: A theta component of what? 01:00:18.860 --> 01:00:20.280 AUDIENCE: Angular momentum. 01:00:20.280 --> 01:00:22.620 PROFESSOR: Angular momentum. 01:00:22.620 --> 01:00:24.710 She's asking, shouldn't there be a theta component 01:00:24.710 --> 01:00:27.870 of angular momentum? 01:00:27.870 --> 01:00:31.080 So we compute our angular momentum with this formula 01:00:31.080 --> 01:00:32.360 at the top. 01:00:32.360 --> 01:00:43.350 It's an r cross P. So the r consists of the z part, 01:00:43.350 --> 01:00:46.680 and the r part is exactly this right here. 01:00:46.680 --> 01:00:49.080 And the P is only into the board. 01:00:49.080 --> 01:00:51.350 It's only in the theta hat direction. 01:00:51.350 --> 01:00:58.000 So you have a term that's r hat cross theta hat gives you a k, 01:00:58.000 --> 01:01:01.915 and you have a term k cross theta 01:01:01.915 --> 01:01:03.350 hat, which gives you an r. 01:01:03.350 --> 01:01:05.300 There just are no cross products that 01:01:05.300 --> 01:01:09.020 come out of this that are in the theta hat direction. 01:01:09.020 --> 01:01:09.994 Yeah. 01:01:09.994 --> 01:01:12.388 AUDIENCE: So if it's just at this one position, 01:01:12.388 --> 01:01:13.792 then you don't have it. 01:01:13.792 --> 01:01:15.196 But as it spins-- 01:01:15.196 --> 01:01:16.520 PROFESSOR: Ah. 01:01:16.520 --> 01:01:20.210 In this case, that's why polar coordinates are nice 01:01:20.210 --> 01:01:23.830 because as it spins, the theta hat's 01:01:23.830 --> 01:01:26.380 just constantly going with it. 01:01:26.380 --> 01:01:29.920 The r hat's constantly going with it. 01:01:29.920 --> 01:01:32.720 And so the beauty of this thing is 01:01:32.720 --> 01:01:36.730 this is an axially symmetric problem. 01:01:36.730 --> 01:01:43.890 It goes round and round, and the torques are given out 01:01:43.890 --> 01:01:47.550 in this rotating frame. 01:01:47.550 --> 01:01:49.250 I think maybe what's confusing you, 01:01:49.250 --> 01:01:52.400 is if you wanted to know the torques in a fixed inertial 01:01:52.400 --> 01:01:56.879 frame, you'd have to break them down into ijk components, 01:01:56.879 --> 01:01:57.670 which you could do. 01:01:57.670 --> 01:01:59.200 A little tedious. 01:01:59.200 --> 01:02:03.730 But the answers in this one came out in r hat, theta hat, k hat 01:02:03.730 --> 01:02:04.230 terms. 01:02:12.100 --> 01:02:14.540 Happy to answer. 01:02:14.540 --> 01:02:17.760 This is good stuff, but thick. 01:02:17.760 --> 01:02:20.155 So keep-- other questions? 01:02:31.790 --> 01:02:35.270 So what do you think will happen in that final homework problem 01:02:35.270 --> 01:02:39.010 with the monkey running-- now he has some velocity. 01:02:39.010 --> 01:02:43.390 How will that problem differ from what we've done? 01:02:43.390 --> 01:02:44.406 Do you have a question? 01:02:44.406 --> 01:02:45.530 Do you want to answer that? 01:02:45.530 --> 01:02:46.372 Yeah. 01:02:46.372 --> 01:02:48.330 AUDIENCE: Well, [INAUDIBLE] monkey [INAUDIBLE]. 01:02:51.790 --> 01:02:53.040 PROFESSOR: It's going to what? 01:02:53.040 --> 01:02:53.910 AUDIENCE: Look like a circle. 01:02:53.910 --> 01:02:55.730 PROFESSOR: It's going to look like a circle, OK. 01:02:55.730 --> 01:02:57.360 He'll be going in a circle, at any-- 01:02:57.360 --> 01:02:59.857 AUDIENCE: [INAUDIBLE] as in a spiral. 01:02:59.857 --> 01:03:00.690 PROFESSOR: A spiral. 01:03:00.690 --> 01:03:02.980 He'll be going like a helix, huh? 01:03:02.980 --> 01:03:04.000 All right. 01:03:04.000 --> 01:03:07.290 Yeah, the monkey will be going in a helix, yeah. 01:03:07.290 --> 01:03:08.230 AUDIENCE: [INAUDIBLE]. 01:03:08.230 --> 01:03:11.862 PROFESSOR: What forces will act on that monkey? 01:03:11.862 --> 01:03:13.502 AUDIENCE: [INAUDIBLE]. 01:03:13.502 --> 01:03:14.960 PROFESSOR: In what direction do you 01:03:14.960 --> 01:03:17.890 think there will be forces acting on that-- he's hanging 01:03:17.890 --> 01:03:19.060 on for dear life, you know. 01:03:19.060 --> 01:03:20.143 This thing's going around. 01:03:20.143 --> 01:03:22.460 They could throw him off, right? 01:03:22.460 --> 01:03:23.930 So what forces act? 01:03:23.930 --> 01:03:25.700 And if you could figure out what forces 01:03:25.700 --> 01:03:29.030 act-- so you draw a free body diagram of the monkey. 01:03:29.030 --> 01:03:31.490 There's going to be possibly forces in the theta hat 01:03:31.490 --> 01:03:36.780 direction, in the z direction, in the r direction. 01:03:36.780 --> 01:03:40.110 But just think physically where they come from. 01:03:40.110 --> 01:03:41.930 So we now know that certainly he's 01:03:41.930 --> 01:03:45.640 hanging on because he is undergoing 01:03:45.640 --> 01:03:48.080 centripetal acceleration. 01:03:48.080 --> 01:03:50.640 And in order to force him to go in that circle, 01:03:50.640 --> 01:03:54.870 there has to be an inward force applied by this shaft to him. 01:03:54.870 --> 01:03:56.400 So there's a force like that because 01:03:56.400 --> 01:04:00.260 of the centripetal acceleration. 01:04:00.260 --> 01:04:02.970 If it's speeding up, there's a force pushing 01:04:02.970 --> 01:04:05.370 him to make it go faster. 01:04:05.370 --> 01:04:06.700 But he's running up the shaft. 01:04:06.700 --> 01:04:07.710 What else is there? 01:04:13.056 --> 01:04:14.526 Are there any other accelerations? 01:04:17.819 --> 01:04:20.110 AUDIENCE: [INAUDIBLE] angular acceleration [INAUDIBLE]? 01:04:20.110 --> 01:04:21.540 PROFESSOR: Yeah, there's angular acceleration. 01:04:21.540 --> 01:04:22.890 That's the thing trying to speed up. 01:04:22.890 --> 01:04:25.306 And so he's hanging on because this thing is accelerating, 01:04:25.306 --> 01:04:27.970 like pushing you back in the car seat, right? 01:04:27.970 --> 01:04:29.270 Linear acceleration. 01:04:29.270 --> 01:04:31.690 Well, this is trying to-- at any instant in time, 01:04:31.690 --> 01:04:33.020 it's trying to go faster. 01:04:33.020 --> 01:04:35.420 So he's having to hang on because of that. 01:04:35.420 --> 01:04:38.400 So that's one of those terms. 01:04:38.400 --> 01:04:40.070 But now he's moving. 01:04:40.070 --> 01:04:42.820 He's also running up the shaft. 01:04:42.820 --> 01:04:45.410 Will that lead to any other accelerations? 01:04:45.410 --> 01:04:47.450 And force equals mass times acceleration. 01:04:47.450 --> 01:04:49.230 So every time you can-- if you can 01:04:49.230 --> 01:04:51.750 account for all the accelerations in the system, 01:04:51.750 --> 01:04:52.790 multiply it by m. 01:04:52.790 --> 01:04:58.000 You've accounted for all of the forces, sum 01:04:58.000 --> 01:05:00.580 of the forces of the mass times acceleration. 01:05:00.580 --> 01:05:03.320 If you add a new acceleration, you better add a new force. 01:05:03.320 --> 01:05:05.990 If you add a new force, you'll probably add a new torque. 01:05:05.990 --> 01:05:07.910 AUDIENCE: So we have to account for gravity? 01:05:07.910 --> 01:05:10.220 PROFESSOR: Well, gravity, yeah. 01:05:10.220 --> 01:05:12.241 What else? 01:05:12.241 --> 01:05:14.730 AUDIENCE: [INAUDIBLE]. 01:05:14.730 --> 01:05:17.250 PROFESSOR: So he's suggesting there might be a Coriolis 01:05:17.250 --> 01:05:19.490 acceleration. 01:05:19.490 --> 01:05:21.720 And the Coriolis acceleration, in order 01:05:21.720 --> 01:05:25.920 to make the monkey accelerate, according to that term, 01:05:25.920 --> 01:05:28.750 there will have to be yet another force. 01:05:28.750 --> 01:05:31.650 And I think-- so we'll see if that turns up 01:05:31.650 --> 01:05:34.584 in the calculation. 01:05:34.584 --> 01:05:35.750 You've got a couple minutes. 01:05:35.750 --> 01:05:38.680 I want you to do the money cards. 01:05:38.680 --> 01:05:45.050 Think about-- and then on your way out, 01:05:45.050 --> 01:05:47.900 I think just pile them up down here on the table, 01:05:47.900 --> 01:05:50.605 or hand them to me or one of the TAs. 01:05:53.570 --> 01:05:57.610 And that'll help me understand what you understood 01:05:57.610 --> 01:05:59.872 or didn't understand today.