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 --> 00:00:17.260 at ocw.mit.edu. 00:00:21.465 --> 00:00:23.840 PROFESSOR: So we were doing velocities and accelerations. 00:00:23.840 --> 00:00:33.070 We came up with-- I guess I ought to continue to remind us. 00:00:33.070 --> 00:00:36.140 We're talking about velocities and accelerations 00:00:36.140 --> 00:00:40.630 of a point with respect to another point, to which we've 00:00:40.630 --> 00:00:44.480 attached-- I'll make this point here-- 00:00:44.480 --> 00:00:49.760 a reference frame, x prime, y prime, z prime. 00:00:49.760 --> 00:00:52.750 And there's a vector that goes between these two 00:00:52.750 --> 00:00:56.420 and a vector that goes there so that we can say r of B 00:00:56.420 --> 00:01:01.660 with respect to O, my inertial frame, is r of A 00:01:01.660 --> 00:01:06.050 with respect to O plus r of B with respect 00:01:06.050 --> 00:01:07.550 to A. These are all vectors. 00:01:07.550 --> 00:01:10.070 And then from these, we derive velocities and acceleration 00:01:10.070 --> 00:01:10.570 formulas. 00:01:10.570 --> 00:01:15.040 And so we've come up with a couple of very handy formulas. 00:01:15.040 --> 00:01:21.920 The velocity formula, velocity of B with respect to O, 00:01:21.920 --> 00:01:23.920 is the velocity of A with respect 00:01:23.920 --> 00:01:31.247 to O plus you have to take a time derivative of this. 00:01:31.247 --> 00:01:33.580 And so I'm going to give you just the general expression 00:01:33.580 --> 00:01:35.870 here just as a brief reminder. 00:01:35.870 --> 00:01:39.760 It's the derivative of B with respect 00:01:39.760 --> 00:01:46.490 to A as seen in the Axyz frame plus omega with respect to O 00:01:46.490 --> 00:01:49.279 cross rBA. 00:01:49.279 --> 00:01:50.195 These are all vectors. 00:01:52.760 --> 00:01:55.030 So that's the velocity formula, remember? 00:01:55.030 --> 00:01:57.640 This is if you're in the frame rotating 00:01:57.640 --> 00:01:59.650 and translating with it, this is the change 00:01:59.650 --> 00:02:02.860 of length of the vector. 00:02:02.860 --> 00:02:05.430 And this, then, is the contribution to the velocity 00:02:05.430 --> 00:02:10.199 that you see in the fixed frame that comes from the rotation. 00:02:10.199 --> 00:02:12.820 And we then got into polar coordinates. 00:02:12.820 --> 00:02:15.750 And we found out that if you use polar coordinates, 00:02:15.750 --> 00:02:20.080 then you can express this as the velocity of A 00:02:20.080 --> 00:02:27.780 with respect to O plus r dot r hat. 00:02:27.780 --> 00:02:30.530 And I should really say cylindrical coordinates, 00:02:30.530 --> 00:02:43.250 z dot k hat plus r theta dot theta hat. 00:02:43.250 --> 00:02:45.070 So that's exactly the same thing. 00:02:45.070 --> 00:02:47.610 This is full 3D vector notation. 00:02:47.610 --> 00:02:50.680 This is a special case of a coordinate system which 00:02:50.680 --> 00:02:53.730 we call polar coordinates. 00:02:53.730 --> 00:02:58.010 And we came up with another formula for accelerations, 00:02:58.010 --> 00:03:04.030 the full 3D vector version of that, A with respect 00:03:04.030 --> 00:03:10.610 to O plus-- this is the acceleration of B 00:03:10.610 --> 00:03:14.620 with respect to A, but as seen in the Axyz frame. 00:03:19.000 --> 00:03:27.320 2 omega cross-- and this is v, the velocity, 00:03:27.320 --> 00:03:30.700 as seen in the xyz frame. 00:03:30.700 --> 00:03:34.100 So these things, this is no contribution from acceleration. 00:03:34.100 --> 00:03:36.000 This is no contribution from acceleration. 00:03:39.020 --> 00:03:47.810 Plus omega naught cross rBA plus omega cross 00:03:47.810 --> 00:03:55.750 omega cross rBA-- all vectors. 00:03:59.190 --> 00:04:03.130 This is a movement of the frame, the acceleration of it 00:04:03.130 --> 00:04:04.010 with respect to this. 00:04:04.010 --> 00:04:06.330 This is pure translation. 00:04:06.330 --> 00:04:10.800 This is the acceleration of the dog on the merry go 00:04:10.800 --> 00:04:14.535 round with respect to the center of the coordinate system there. 00:04:14.535 --> 00:04:17.170 It does not involve rotation. 00:04:17.170 --> 00:04:21.050 This is this term that we call Coriolis. 00:04:21.050 --> 00:04:22.650 This is a term we call Euler. 00:04:22.650 --> 00:04:26.080 This is the angular speed up of the system, the angular 00:04:26.080 --> 00:04:27.280 acceleration. 00:04:27.280 --> 00:04:28.450 And this is centripetal. 00:04:32.100 --> 00:04:33.900 And if you do these, if you want to go 00:04:33.900 --> 00:04:37.240 to cylindrical coordinates-- and what we're going to do next 00:04:37.240 --> 00:04:39.610 is just do some applications here. 00:04:39.610 --> 00:04:42.490 The acceleration of B with respect to O and cylindrical 00:04:42.490 --> 00:04:46.670 coordinates-- it's a translation piece. 00:04:46.670 --> 00:04:49.970 Plus now in cylindrical, it's useful to group these. 00:04:56.550 --> 00:05:02.650 So I'm going to put together this term and this term. 00:05:02.650 --> 00:05:05.024 Because they're both in the r hat direction. 00:05:05.024 --> 00:05:06.940 And I'm going to put together these two terms. 00:05:06.940 --> 00:05:11.090 Because they're both in the theta hat direction. 00:05:11.090 --> 00:05:16.990 And I need a z double dot k, because it's cylindrical. 00:05:16.990 --> 00:05:20.000 And then I have the theta hat piece, 00:05:20.000 --> 00:05:29.180 r theta double dot plus 2r dot theta dot theta hat. 00:05:29.180 --> 00:05:32.880 This is the same thing as this. 00:05:32.880 --> 00:05:37.587 Except this is expressed in cylindrical coordinates. 00:05:37.587 --> 00:05:39.420 And cylindrical coordinates are particularly 00:05:39.420 --> 00:05:42.740 good for doing the kind of problems that are mostly 00:05:42.740 --> 00:05:45.780 done at the level of this course, which we call planar 00:05:45.780 --> 00:05:50.190 motion problems, confined to an x, y plane, 00:05:50.190 --> 00:05:54.650 and confined to single axis rotation in z. 00:05:54.650 --> 00:05:57.620 This is a coordinate system that's ideally suited 00:05:57.620 --> 00:05:58.810 to do problems like that. 00:05:58.810 --> 00:06:00.410 And that's why we use it. 00:06:00.410 --> 00:06:04.550 So now let's do some examples. 00:06:04.550 --> 00:06:11.550 This is really quite a powerful-- 00:06:11.550 --> 00:06:13.260 now that you have these two equations, 00:06:13.260 --> 00:06:15.625 you can do a lot of kinematics and dynamics. 00:06:25.270 --> 00:06:31.010 So we started last time right at the very end. 00:06:31.010 --> 00:06:36.100 I said, OK, let's do this problem really quickly, right? 00:06:36.100 --> 00:06:39.650 Constant rotation rate, constant radius, 00:06:39.650 --> 00:06:42.030 no angular acceleration, no change 00:06:42.030 --> 00:06:44.340 in length of the thing-- pretty simple problem. 00:06:44.340 --> 00:06:46.630 And we zipped it off really fast. 00:06:46.630 --> 00:06:50.870 And I wanted to start there, do that really quickly. 00:06:50.870 --> 00:06:53.760 So this is my case one. 00:06:53.760 --> 00:06:55.495 It's my ball on the string. 00:07:01.260 --> 00:07:08.060 Here's point A. Here's point B. Here's the theta hat 00:07:08.060 --> 00:07:12.930 direction, r hat direction. 00:07:12.930 --> 00:07:17.880 And it has some constant length R here. 00:07:17.880 --> 00:07:24.140 So r dot r double dot are 0. 00:07:24.140 --> 00:07:33.370 There's no also z, z dot, z double dot, no z motion at all. 00:07:33.370 --> 00:07:36.090 Those are all 0. 00:07:36.090 --> 00:07:39.980 Theta dot is a constant. 00:07:39.980 --> 00:07:45.460 I'll call it cap omega in the k hat direction, right hand rule. 00:07:45.460 --> 00:07:50.080 Theta double dot is 0. 00:07:50.080 --> 00:07:53.177 So the easy way to use these formulas 00:07:53.177 --> 00:07:55.010 is you just start knocking out all the terms 00:07:55.010 --> 00:07:56.490 that you don't need. 00:07:56.490 --> 00:08:00.800 But let's, just to give you a little quick review of how 00:08:00.800 --> 00:08:05.560 to use these things, use the vector full 00:08:05.560 --> 00:08:08.770 3D version of the velocity equation for a second. 00:08:08.770 --> 00:08:11.930 The full 3D version says the velocity of A 00:08:11.930 --> 00:08:16.250 with respect to O, what's that in this problem? 00:08:16.250 --> 00:08:19.630 That's the translating term. 00:08:19.630 --> 00:08:22.060 So I was standing still, not going anywhere. 00:08:22.060 --> 00:08:25.070 But if I were walking along, I'd still spin that thing. 00:08:25.070 --> 00:08:28.070 OK, so this problem, this term is 0. 00:08:28.070 --> 00:08:33.409 This problem, r dot-- let's go here. 00:08:33.409 --> 00:08:35.580 This is the rate of change of the length 00:08:35.580 --> 00:08:40.321 of the string in the coordinate system of the string walking 00:08:40.321 --> 00:08:40.820 along. 00:08:40.820 --> 00:08:43.110 So what's r dot? 00:08:43.110 --> 00:08:45.180 OK, so that term is 0. 00:08:45.180 --> 00:08:51.170 And omega cross rBA-- well, r is in the R hat, capital R R hat, 00:08:51.170 --> 00:08:56.030 cross with omega k. 00:08:56.030 --> 00:09:01.883 k cross R is-- k cross of R hat? 00:09:01.883 --> 00:09:02.630 STUDENT: Theta. 00:09:02.630 --> 00:09:03.505 PROFESSOR: Theta hat. 00:09:03.505 --> 00:09:07.230 So we get our familiar-- this first term is 0. 00:09:07.230 --> 00:09:08.860 The second term is 0. 00:09:08.860 --> 00:09:12.830 The third term, we just get our R omega 00:09:12.830 --> 00:09:14.920 in the theta hat direction. 00:09:14.920 --> 00:09:16.050 We know that to be true. 00:09:16.050 --> 00:09:17.633 The point I'm just trying to make here 00:09:17.633 --> 00:09:21.530 is you can always just fall back and use the vector formula, 00:09:21.530 --> 00:09:23.230 full 3D formulas of both. 00:09:23.230 --> 00:09:26.370 Just plug it in, and everything will just drop out. 00:09:26.370 --> 00:09:30.270 You can also then go to the cylindrical coordinate terms 00:09:30.270 --> 00:09:32.750 when you happen to have it all expressed 00:09:32.750 --> 00:09:37.710 in coordinates of that kind to make it simpler for you. 00:09:37.710 --> 00:09:43.400 OK, so now let's quickly then do the acceleration of B 00:09:43.400 --> 00:09:49.300 with respect to O. And let's use the formulation already 00:09:49.300 --> 00:09:50.890 in cylindrical coordinates. 00:09:50.890 --> 00:09:53.760 What's A with respect to O? 00:09:53.760 --> 00:09:54.720 0. 00:09:54.720 --> 00:09:58.200 What's r double dot? 00:09:58.200 --> 00:10:01.470 How about the r theta dot squared term, 0 or not? 00:10:01.470 --> 00:10:02.290 Nope. 00:10:02.290 --> 00:10:04.243 Let's go on-- z double dot? 00:10:04.243 --> 00:10:04.861 STUDENT: 0. 00:10:04.861 --> 00:10:06.110 PROFESSOR: r theta double dot? 00:10:06.110 --> 00:10:07.082 STUDENT: 0. 00:10:07.082 --> 00:10:08.311 PROFESSOR: r dot theta dot? 00:10:08.311 --> 00:10:08.810 STUDENT: 0 00:10:08.810 --> 00:10:10.540 PROFESSOR: OK, we only end up with one term. 00:10:10.540 --> 00:10:12.770 Actually, I'm going to keep this one for a second-- A 00:10:12.770 --> 00:10:16.030 with respect to O minus. 00:10:16.030 --> 00:10:25.360 And we came up with our R theta dot squared r hat term. 00:10:25.360 --> 00:10:28.390 I leave this in here because I actually don't have to say. 00:10:28.390 --> 00:10:33.000 If I wanted now to do this problem, 00:10:33.000 --> 00:10:35.880 if I asked you to do a problem where I'm doing this, 00:10:35.880 --> 00:10:40.039 and I start accelerating, do you know how to do it? 00:10:40.039 --> 00:10:41.705 There's the answer-- still there, right? 00:10:46.437 --> 00:11:06.155 All right, let's do a quick free body diagram. 00:11:11.660 --> 00:11:18.920 Here's our mass, my master coordinate system out here. 00:11:18.920 --> 00:11:23.735 Let's draw-- let's say it's right here at 90 degrees. 00:11:26.570 --> 00:11:29.300 What are the external forces on the mass in this problem? 00:11:33.196 --> 00:11:34.660 STUDENT: [INAUDIBLE]. 00:11:34.660 --> 00:11:37.120 PROFESSOR: So there's tension in the string, right? 00:11:37.120 --> 00:11:41.700 OK, so now this is a really trivially simple problem. 00:11:41.700 --> 00:11:45.660 So the emphasize here is on the concept. 00:11:45.660 --> 00:11:50.100 So now when you're asked to come up with an equation of motion 00:11:50.100 --> 00:11:55.520 or compute the forces on a mass, use Newton's second law, 00:11:55.520 --> 00:11:58.040 F equals mass times acceleration. 00:11:58.040 --> 00:12:01.870 You now have the complete 3D vector formulation 00:12:01.870 --> 00:12:08.290 for acceleration of a particle in a translating rotating 00:12:08.290 --> 00:12:09.870 coordinate system. 00:12:09.870 --> 00:12:12.780 That's all you need to compute accelerations for lots 00:12:12.780 --> 00:12:14.850 and lots of difficult problems. 00:12:14.850 --> 00:12:18.670 And so if you can write down the acceleration, 00:12:18.670 --> 00:12:23.960 you say it's equal to what? 00:12:23.960 --> 00:12:26.870 Mass-- if you multiply mass times that acceleration, 00:12:26.870 --> 00:12:28.269 what's that equal to? 00:12:28.269 --> 00:12:29.060 STUDENT: The force. 00:12:29.060 --> 00:12:31.393 PROFESSOR: The forces that must be acting on the system. 00:12:31.393 --> 00:12:32.860 And that's the point here. 00:12:32.860 --> 00:12:37.090 So now I want to know the forces on the system, the summation 00:12:37.090 --> 00:12:38.650 of the external forces. 00:12:38.650 --> 00:12:39.940 And then these are vectors. 00:12:39.940 --> 00:12:42.100 And you can do them component by component. 00:12:42.100 --> 00:12:47.010 Some of the external forces in the r hat direction 00:12:47.010 --> 00:12:49.810 must be equal to the mass-- in this case, just 00:12:49.810 --> 00:12:56.360 a particle-- times the acceleration of that particle. 00:12:56.360 --> 00:12:58.050 And in this problem, then that would 00:12:58.050 --> 00:13:01.670 be the mass times the acceleration of A with respect 00:13:01.670 --> 00:13:10.540 to O minus R theta dot squared r hat. 00:13:10.540 --> 00:13:15.690 And this would have to be the r hat component of this thing. 00:13:15.690 --> 00:13:17.990 I said I just want the R component. 00:13:17.990 --> 00:13:20.180 I'd have to figure out if I was running along, 00:13:20.180 --> 00:13:23.110 if I had it in the same direction as r hat. 00:13:23.110 --> 00:13:25.490 What part of that acceleration is in that direction? 00:13:25.490 --> 00:13:26.700 That would come here. 00:13:26.700 --> 00:13:29.040 If there's 0, you just make it 0. 00:13:29.040 --> 00:13:34.610 So let's just let the acceleration of A 00:13:34.610 --> 00:13:36.830 with respect to O be 0. 00:13:36.830 --> 00:13:41.690 Then that says the sum of the forces in the r direction 00:13:41.690 --> 00:13:49.120 is equal to the mass minus mR theta dot squared. 00:13:49.120 --> 00:13:52.210 And if we were to draw a free body diagram, 00:13:52.210 --> 00:13:53.730 we would find out that, ahh, there 00:13:53.730 --> 00:13:55.980 must be a tension on the string pulling 00:13:55.980 --> 00:14:00.120 in on the mass sufficient to give it 00:14:00.120 --> 00:14:03.000 the acceleration that you've computed. 00:14:03.000 --> 00:14:04.460 So every problem, when you're asked 00:14:04.460 --> 00:14:06.790 to compute the force, or the next step up, 00:14:06.790 --> 00:14:09.250 find the equation of motion, the equation of motion 00:14:09.250 --> 00:14:13.115 is just writing this thing out. 00:14:20.250 --> 00:14:30.600 OK, now I want to move on to a more interesting problem. 00:14:55.510 --> 00:15:02.260 All right, I'm going to do this problem. 00:15:02.260 --> 00:15:05.540 So it's a hollow tube. 00:15:05.540 --> 00:15:09.230 And we're going to look at things like, I put a ping pong 00:15:09.230 --> 00:15:11.010 ball in it. 00:15:11.010 --> 00:15:15.345 And if I swing this tube around, the ping pong ball 00:15:15.345 --> 00:15:16.220 is going to come out. 00:15:23.290 --> 00:15:25.710 OK, there must be some forces on that thing 00:15:25.710 --> 00:15:27.405 to cause it to come out. 00:15:27.405 --> 00:15:29.640 There must be some accelerations on them. 00:15:29.640 --> 00:15:31.960 And so I could conceivably have an R, 00:15:31.960 --> 00:15:35.070 a theta double dot acceleration. 00:15:35.070 --> 00:15:39.130 It certainly is going to have theta dot rotation rates. 00:15:39.130 --> 00:15:40.520 The ball is allowed to move. 00:15:40.520 --> 00:15:44.375 So there can be nonzero r dots, r double dots-- 00:15:44.375 --> 00:15:47.390 a lot going on inside of this simple little tube. 00:15:47.390 --> 00:15:49.500 So that's what I want to figure out. 00:15:49.500 --> 00:15:52.990 Let's see if we can come up with a model for this problem. 00:16:08.090 --> 00:16:09.620 So here's my z-axis. 00:16:09.620 --> 00:16:14.980 I have a rotation around it, some theta dot k hat direction. 00:16:14.980 --> 00:16:17.950 Here's my tube. 00:16:17.950 --> 00:16:20.020 It's rotating around. 00:16:20.020 --> 00:16:22.980 So this is sort of a side view, your view of the tube. 00:16:22.980 --> 00:16:26.940 Here's that ping pong ball in there. 00:16:26.940 --> 00:16:35.110 And I'm going to idealize that ping pong 00:16:35.110 --> 00:16:38.970 ball for a minute, a little more general problem. 00:16:38.970 --> 00:16:42.570 Let's say I have kind of a nut on this thing, a disk, 00:16:42.570 --> 00:16:45.280 something hanging onto the outside. 00:16:45.280 --> 00:16:49.380 And I can control the rate, the speed 00:16:49.380 --> 00:16:51.650 at which this thing goes out. 00:16:51.650 --> 00:16:54.330 The ping pong ball, this is going 00:16:54.330 --> 00:16:55.610 to be an application of this. 00:16:55.610 --> 00:16:57.735 But I want to be able to do several other versions, 00:16:57.735 --> 00:17:02.120 like make the speed constant for a second. 00:17:02.120 --> 00:17:06.690 OK, and looking down on this thing, 00:17:06.690 --> 00:17:13.000 top view, here's our inertial frame, 00:17:13.000 --> 00:17:14.579 maybe out here like this. 00:17:14.579 --> 00:17:16.800 Here's my mass. 00:17:16.800 --> 00:17:20.000 So in polar coordinates, here's your theta. 00:17:20.000 --> 00:17:24.192 The r is this. 00:17:24.192 --> 00:17:27.150 This is your r hat. 00:17:27.150 --> 00:17:28.810 This is your theta hat directions. 00:17:33.750 --> 00:17:39.650 I'm going to let the velocity of A with respect to O be 0, 00:17:39.650 --> 00:17:43.380 so there's no translational of this system. 00:17:43.380 --> 00:17:49.510 And z, z dot, z double dot, those are all 0. 00:17:49.510 --> 00:17:53.280 So nothing's happening in the z direction. 00:17:53.280 --> 00:17:57.090 So I want to first compute the velocity of B 00:17:57.090 --> 00:17:59.740 with respect to O. 00:17:59.740 --> 00:18:04.970 And you ought to be able to do that sort of by inspection. 00:18:04.970 --> 00:18:10.830 It comes only from-- oh, I haven't told you enough. 00:18:10.830 --> 00:18:12.790 What do I want to make happen in this problem? 00:18:12.790 --> 00:18:20.993 I want to let r dot-- it's going to be some vr, 00:18:20.993 --> 00:18:21.743 and it's constant. 00:18:24.660 --> 00:18:27.259 I'm not going to go quite to my ping pong shooter here yet. 00:18:27.259 --> 00:18:29.300 I'm going to do a slightly simpler problem first. 00:18:29.300 --> 00:18:30.260 So this is a constant. 00:18:30.260 --> 00:18:32.980 That means r double dot is 0. 00:18:32.980 --> 00:18:34.860 So this thing is just-- let's say 00:18:34.860 --> 00:18:37.490 you had threads on this thing, and it's a screw, 00:18:37.490 --> 00:18:41.950 and it's just moving its way out at a constant rate. 00:18:41.950 --> 00:18:47.980 And I'm going to have constant angular, so theta dot. 00:18:47.980 --> 00:18:51.350 I'll call that cap omega k hat. 00:18:51.350 --> 00:18:55.450 So the angular rate is also constant. 00:18:55.450 --> 00:18:58.070 All right, if that's the case, can you tell me, 00:18:58.070 --> 00:19:04.030 what's v of B with respect to my fixed frame O? 00:19:07.800 --> 00:19:11.480 Well, any time you're not sure, you go back to this formula, 00:19:11.480 --> 00:19:12.560 throw out terms. 00:19:12.560 --> 00:19:14.640 There's no z dot term. 00:19:14.640 --> 00:19:15.670 This is constant. 00:19:15.670 --> 00:19:18.390 This is 0, 0, 0. 00:19:18.390 --> 00:19:19.850 You have this term. 00:19:19.850 --> 00:19:22.560 It's some vr in the r hat direction. 00:19:22.560 --> 00:19:26.520 You have this term, wherever it happens to be in the theta hat 00:19:26.520 --> 00:19:27.020 direction. 00:19:39.372 --> 00:19:43.950 r dot in the r hat direction plus r theta 00:19:43.950 --> 00:19:46.450 dot in the theta hat direction-- OK, 00:19:46.450 --> 00:19:50.850 we need acceleration next, B with respect to O. 00:19:50.850 --> 00:19:53.110 And now you can crank through the terms again. 00:19:53.110 --> 00:19:57.290 This time, the first term is 0. 00:19:57.290 --> 00:19:59.880 The second term is 0, because it's constant. 00:19:59.880 --> 00:20:02.560 The third term is definitely not 0. 00:20:02.560 --> 00:20:03.660 The fourth term is 0. 00:20:03.660 --> 00:20:04.280 Fifth term? 00:20:08.040 --> 00:20:09.270 0. 00:20:09.270 --> 00:20:11.140 This term? 00:20:11.140 --> 00:20:13.020 Not 0. 00:20:13.020 --> 00:20:14.160 Let me just write them. 00:20:18.830 --> 00:20:23.600 So you get a minus r theta dot squared r hat-- that's 00:20:23.600 --> 00:20:30.904 the radial direction term-- plus 2r dot theta dot theta hat. 00:20:30.904 --> 00:20:31.945 That's the accelerations. 00:20:36.000 --> 00:20:38.530 So you have an acceleration now in the r hat direction 00:20:38.530 --> 00:20:40.640 and in the theta hat direction. 00:20:40.640 --> 00:20:42.610 If there's acceleration to those directions, 00:20:42.610 --> 00:20:44.665 there must be forces. 00:20:47.680 --> 00:20:51.720 Newton's second law now says, again, the force 00:20:51.720 --> 00:20:55.200 is the mass times acceleration. 00:20:55.200 --> 00:20:56.930 And this is a vector. 00:20:56.930 --> 00:20:58.390 This is a vector. 00:20:58.390 --> 00:20:59.470 It has two components. 00:20:59.470 --> 00:21:02.585 And the nice thing about these Newton's laws and vectors 00:21:02.585 --> 00:21:05.180 is you can break the problems down into their vector 00:21:05.180 --> 00:21:07.070 components and treat the r direction 00:21:07.070 --> 00:21:09.300 as one equation of motion, and the theta 00:21:09.300 --> 00:21:11.310 direction as a separate one. 00:21:11.310 --> 00:21:15.240 So we might want to draw a free body diagram. 00:21:15.240 --> 00:21:20.260 Here's this block working its way out. 00:21:20.260 --> 00:21:22.660 We know that there's probably some axial force. 00:21:22.660 --> 00:21:24.700 I'm just going to call it T. And there's 00:21:24.700 --> 00:21:27.700 some other unknown force here in the theta hat direction. 00:21:27.700 --> 00:21:30.704 So this is my theta hat. 00:21:30.704 --> 00:21:32.870 I've drawn this just intentionally in the positive r 00:21:32.870 --> 00:21:33.490 hat direction. 00:21:33.490 --> 00:21:35.210 The sign that comes out will tell us 00:21:35.210 --> 00:21:38.490 which direction it really is if you're not certain. 00:21:38.490 --> 00:21:42.250 Just draw it positive, in the positive r hat direction. 00:21:42.250 --> 00:21:45.520 And that's your free body diagram. 00:21:45.520 --> 00:21:48.180 If I wanted to put gravity in there, I might have. 00:21:48.180 --> 00:21:51.200 But we're doing this in the horizontal plane. 00:21:51.200 --> 00:21:52.560 Gravity is in and out this way. 00:21:52.560 --> 00:21:53.670 It's in the z direction. 00:21:53.670 --> 00:21:57.740 And we know it's constrained, can't move. z double dot is 0. 00:21:57.740 --> 00:22:01.710 So there's certainly a support force that picks up the weight. 00:22:01.710 --> 00:22:04.170 But this is in our horizontal plane. 00:22:04.170 --> 00:22:06.170 There's your free body diagram. 00:22:06.170 --> 00:22:09.310 And we can write two equations to solve for these things. 00:22:09.310 --> 00:22:16.050 So the sum of the forces in the r hat direction is T. 00:22:16.050 --> 00:22:19.640 And that must be equal to the mass times the acceleration 00:22:19.640 --> 00:22:27.690 in the r, minus mr theta dot squared in the r hat direction. 00:22:27.690 --> 00:22:31.890 Sure enough, the tension has to pull inwards 00:22:31.890 --> 00:22:33.820 in the minus r hat direction. 00:22:33.820 --> 00:22:39.115 And that's the full result. 00:22:39.115 --> 00:22:44.350 STUDENT: Where are the T and F forces exactly? 00:22:44.350 --> 00:22:45.620 PROFESSOR: Where are they? 00:22:45.620 --> 00:22:53.450 OK, so I'm going to bring the ball to the outside 00:22:53.450 --> 00:22:54.590 where you can see it. 00:22:54.590 --> 00:22:58.260 This thing is going in this direction, horizontal plane, x, 00:22:58.260 --> 00:22:59.380 y plane. 00:22:59.380 --> 00:23:03.520 It's moving its fixed rate out. 00:23:03.520 --> 00:23:05.550 So its speed when it's in here is r 00:23:05.550 --> 00:23:07.370 over 2 omega in that direction. 00:23:07.370 --> 00:23:10.730 And the speed when it's out here is r omega in that direction. 00:23:10.730 --> 00:23:12.630 So clearly it's picking up speed. 00:23:12.630 --> 00:23:17.320 If it's picking up speed, is it picking up kinetic energy? 00:23:17.320 --> 00:23:19.200 Is there work being done on it somehow 00:23:19.200 --> 00:23:20.570 to build up that energy? 00:23:20.570 --> 00:23:22.970 So there must be some forces at play. 00:23:22.970 --> 00:23:26.090 So there's a normal force from the wall 00:23:26.090 --> 00:23:29.670 of this thing pushing this ball sideways to speed it up, 00:23:29.670 --> 00:23:30.250 for sure. 00:23:30.250 --> 00:23:31.650 That's one force. 00:23:31.650 --> 00:23:35.810 And the other force is because I'm not allowing this thing 00:23:35.810 --> 00:23:38.180 just to go freely out. 00:23:38.180 --> 00:23:41.310 I'm constraining it to constant speed out. 00:23:41.310 --> 00:23:43.900 It would really like to go a lot faster than that. 00:23:43.900 --> 00:23:46.600 So what's holding it back? 00:23:46.600 --> 00:23:49.510 At any instance in time, it has centripetal acceleration. 00:23:49.510 --> 00:23:53.290 And what's making it go in the circle 00:23:53.290 --> 00:23:56.060 is a force that is-- in this case, 00:23:56.060 --> 00:24:00.570 if that were a nut with threads, in the threads are applying 00:24:00.570 --> 00:24:03.368 to the nut to keep it from running away. 00:24:03.368 --> 00:24:05.360 STUDENT: In that free body diagram, 00:24:05.360 --> 00:24:08.965 F, doesn't F act similar between theta and r hat? 00:24:08.965 --> 00:24:10.340 And then there's like a component 00:24:10.340 --> 00:24:13.830 of theta hat in there? 00:24:13.830 --> 00:24:15.700 PROFESSOR: Well, I've broken down. 00:24:15.700 --> 00:24:18.080 I've chosen. 00:24:18.080 --> 00:24:21.200 The total force acting on this thing 00:24:21.200 --> 00:24:23.890 is some combination of a force in that direction 00:24:23.890 --> 00:24:25.760 and a combination in the axial direction. 00:24:25.760 --> 00:24:31.810 So it has some net direction that's neither this nor that. 00:24:31.810 --> 00:24:34.735 STUDENT: From that equation, it looks like what you drew, 00:24:34.735 --> 00:24:39.640 F has an r hat component. 00:24:39.640 --> 00:24:42.830 PROFESSOR: So this thing is rotating 00:24:42.830 --> 00:24:45.560 about some center over here. 00:24:45.560 --> 00:24:47.460 So this is theta dot. 00:24:47.460 --> 00:24:50.690 Theta is going in this direction, theta dot. 00:24:50.690 --> 00:24:55.050 And we've chosen a coordinate system that has unit vectors r 00:24:55.050 --> 00:24:56.510 hat and theta hat. 00:24:56.510 --> 00:24:57.560 And so it makes sense. 00:24:57.560 --> 00:24:59.840 We can express the acceleration in terms 00:24:59.840 --> 00:25:01.100 of those two components. 00:25:01.100 --> 00:25:05.270 It makes sense to express the forces in the same direction. 00:25:05.270 --> 00:25:08.620 So I've just arbitrarily said, I have some unknown force 00:25:08.620 --> 00:25:09.690 that's in this direction. 00:25:09.690 --> 00:25:11.814 And I have another unknown force in that direction. 00:25:14.340 --> 00:25:17.230 Then I'm saying, they account for all forces 00:25:17.230 --> 00:25:19.770 in this direction, whatever their source. 00:25:19.770 --> 00:25:23.000 That tells me that the sum of the external forces 00:25:23.000 --> 00:25:29.090 in the theta hat direction in this case is this unknown F. 00:25:29.090 --> 00:25:31.610 But I know from Newton that that's 00:25:31.610 --> 00:25:33.930 got to be equal to the mass times the acceleration 00:25:33.930 --> 00:25:35.380 in that direction. 00:25:35.380 --> 00:25:42.520 In this case, then, that is 2mr dot theta dot theta hat. 00:25:45.630 --> 00:25:50.250 So just from applying the equation 00:25:50.250 --> 00:25:52.580 and applying Newton's second law, 00:25:52.580 --> 00:25:56.359 I can find out what that force must be. 00:25:56.359 --> 00:25:57.900 It would've been a lot more work if I 00:25:57.900 --> 00:26:00.286 had drawn it in some arbitrary in between direction. 00:26:00.286 --> 00:26:02.660 Because then I'd have to break it down into its compounds 00:26:02.660 --> 00:26:03.690 to write this. 00:26:03.690 --> 00:26:07.060 So I've made it as easy as possible for myself. 00:26:07.060 --> 00:26:08.850 So there's a force like this. 00:26:08.850 --> 00:26:10.840 And there is another force like that. 00:26:10.840 --> 00:26:14.550 This one is caused by the centripetal acceleration. 00:26:14.550 --> 00:26:19.820 Or this force causes the centripetal acceleration. 00:26:19.820 --> 00:26:22.080 In order to make something go in a circular path, 00:26:22.080 --> 00:26:24.940 you have to exert a force on it. 00:26:24.940 --> 00:26:26.900 That's the force. 00:26:26.900 --> 00:26:30.160 In order to accelerate something angularly, 00:26:30.160 --> 00:26:31.500 you have to apply a force. 00:26:31.500 --> 00:26:33.760 That comes from the Euler term. 00:26:33.760 --> 00:26:36.830 And this is a curious term. 00:26:36.830 --> 00:26:38.250 This is the Coriolis term. 00:26:41.290 --> 00:26:42.580 So where does it come from? 00:26:45.870 --> 00:26:47.900 That's the crux of the matter here. 00:26:47.900 --> 00:26:48.900 Where does it come from? 00:26:56.600 --> 00:27:01.420 So part of the reading that you need to do now is Chapter 15. 00:27:01.420 --> 00:27:04.700 Chapter 15, most of it is going to be complete review. 00:27:04.700 --> 00:27:07.810 It just says the conservation of a linear momentum, impulse 00:27:07.810 --> 00:27:08.355 and momentum. 00:27:08.355 --> 00:27:10.900 But it also gets into angular momentum. 00:27:10.900 --> 00:27:14.040 So we're going to talk quite a lot about angular momentum. 00:27:17.660 --> 00:27:28.240 And I want to do a very brief little review right now so 00:27:28.240 --> 00:27:29.700 that it applies to this problem. 00:27:32.320 --> 00:27:34.110 So we've come up with an expression 00:27:34.110 --> 00:27:36.025 that the force in the theta direction 00:27:36.025 --> 00:27:43.420 here comes-- there's got to be that, 2mr dot theta dot. 00:27:52.820 --> 00:27:56.400 So here's my point O, and A for that matter. 00:27:56.400 --> 00:28:00.200 But looking down on our problem, here's 00:28:00.200 --> 00:28:03.160 my mass at some instant in time. 00:28:03.160 --> 00:28:09.400 My rotation rate is theta dot-- or actually it's constant. 00:28:09.400 --> 00:28:15.170 So it's k hat like that, cap omega k hat. 00:28:15.170 --> 00:28:20.520 This is my r hat direction, theta hat. 00:28:20.520 --> 00:28:23.710 Now, I'm going to treat this as a particle. 00:28:23.710 --> 00:28:26.530 Not long-- we're going to be talking about the dynamics 00:28:26.530 --> 00:28:27.560 of rigid bodies. 00:28:27.560 --> 00:28:29.510 We're just doing particles for the moment. 00:28:29.510 --> 00:28:31.860 We think of them just as point masses 00:28:31.860 --> 00:28:34.660 and don't deal with their finite extent. 00:28:34.660 --> 00:28:37.410 So we're still thinking of this as a particle. 00:28:37.410 --> 00:28:40.310 And I'm going to write down the definition of the angular 00:28:40.310 --> 00:28:42.390 momentum of a particle. 00:28:42.390 --> 00:28:47.590 This is B out here with respect to my fixed frame here at O. 00:28:47.590 --> 00:28:49.600 And I'm going to use a lowercase h to describe 00:28:49.600 --> 00:28:51.410 angular momentum of particles. 00:28:51.410 --> 00:28:54.640 And I'll use capital H to describe the angular 00:28:54.640 --> 00:28:57.670 momentum of rigid bodies. 00:28:57.670 --> 00:29:00.920 So this is a particle, the definition 00:29:00.920 --> 00:29:03.940 of the angular momentum of a particle with respect 00:29:03.940 --> 00:29:07.166 to a fixed point. 00:29:07.166 --> 00:29:08.540 We're going to come back to that. 00:29:08.540 --> 00:29:10.670 That'll turn out to have some significance. 00:29:10.670 --> 00:29:17.960 The definition of this is it's RBO, the position vector, 00:29:17.960 --> 00:29:25.535 crossed with the linear momentum evaluated in the fixed frame. 00:29:25.535 --> 00:29:27.660 So that's the definition of angular momentum, which 00:29:27.660 --> 00:29:29.470 is you have linear momentum, and it's 00:29:29.470 --> 00:29:32.810 the cross product with the position vector out to it. 00:29:32.810 --> 00:29:38.845 So in this case, the RBO is capital R R hat. 00:29:44.560 --> 00:29:47.300 Well, it's not-- this varies, excuse me. 00:29:47.300 --> 00:29:50.750 So I'm not going to use-- this I use as a constant. 00:29:50.750 --> 00:29:53.090 I better keep it as the variable. 00:29:53.090 --> 00:29:57.140 Pardon that, so it's r, whatever the local position is, 00:29:57.140 --> 00:30:01.970 in the r hat direction crossed with the linear momentum. 00:30:01.970 --> 00:30:06.270 What's the linear momentum of that particle? 00:30:06.270 --> 00:30:09.050 Mass times velocity. 00:30:09.050 --> 00:30:09.983 What's the velocity? 00:30:09.983 --> 00:30:10.858 STUDENT: [INAUDIBLE]. 00:30:17.182 --> 00:30:19.390 PROFESSOR: Theta dot-- where did we write it up here? 00:30:19.390 --> 00:30:31.790 Someplace-- the total velocity is r dot r hat plus r theta 00:30:31.790 --> 00:30:33.300 dot theta hat. 00:30:36.780 --> 00:30:38.640 And we need a mass in here. 00:30:38.640 --> 00:30:44.530 So mass times velocity would be the momentum. 00:30:44.530 --> 00:30:46.670 And we need the cross products of those. 00:30:46.670 --> 00:30:49.310 r hat cross r hat, you get nothing from that. 00:30:49.310 --> 00:30:56.500 And r hat cross theta hat-- positive or negative? 00:30:56.500 --> 00:30:59.040 Positive k, right? 00:30:59.040 --> 00:31:13.280 So this becomes mr squared theta dot in the k hat direction. 00:31:19.200 --> 00:31:21.685 And this one is a constant. 00:31:21.685 --> 00:31:23.491 We'll write this as cap omega. 00:31:27.340 --> 00:31:32.970 So this is my angular momentum of my particle with respect 00:31:32.970 --> 00:31:35.480 to this fixed reference frame. 00:31:35.480 --> 00:31:38.354 And so one final step that you know 00:31:38.354 --> 00:31:43.220 from your previous physics is-- how's torque 00:31:43.220 --> 00:31:44.400 related to angular momentum? 00:31:49.710 --> 00:31:50.505 Take a guess. 00:31:50.505 --> 00:31:52.600 What do you remember? 00:31:52.600 --> 00:32:01.870 Time derivative-- so d by dt of hBO here. 00:32:01.870 --> 00:32:04.380 The time rate of change of this vector 00:32:04.380 --> 00:32:11.190 is the torque with respect to this point, with respect to O. 00:32:11.190 --> 00:32:12.920 So what are constants in this thing? 00:32:12.920 --> 00:32:15.590 We don't have to deal with their derivatives. 00:32:15.590 --> 00:32:22.640 k hat is-- does the direction of the angular momentum change? 00:32:22.640 --> 00:32:23.960 It's upwards. 00:32:23.960 --> 00:32:26.290 The derivative of k, the unit vector k, 00:32:26.290 --> 00:32:28.430 does it change in this problem? 00:32:28.430 --> 00:32:30.930 No, so its time derivative is 0. 00:32:30.930 --> 00:32:32.490 Its time derivative is 0. 00:32:32.490 --> 00:32:33.999 The m time derivative is 0. 00:32:33.999 --> 00:32:36.040 The only thing you have to take the derivative of 00:32:36.040 --> 00:32:37.605 is r, so 2rr dot. 00:32:54.350 --> 00:32:59.380 So the torque is 2mrr dot cap omega. 00:32:59.380 --> 00:33:01.410 And it's all in the k direction. 00:33:01.410 --> 00:33:11.750 And that had better be rBO cross F. Torque comes 00:33:11.750 --> 00:33:14.290 as a force times a moment arm, right? 00:33:14.290 --> 00:33:18.000 And we computed here a force in the theta direction. 00:33:18.000 --> 00:33:20.610 So what would give us a torque in the k direction? 00:33:20.610 --> 00:33:23.090 An r cross a theta. 00:33:23.090 --> 00:33:26.440 r hat cross theta hat gives you a k. 00:33:26.440 --> 00:33:29.800 r cross-- and this force, the part of the force. 00:33:29.800 --> 00:33:32.450 We have two forces, one in the r direction 00:33:32.450 --> 00:33:34.470 and one in the theta direction. 00:33:34.470 --> 00:33:38.540 The cross product, this term up here, the T gives you nothing, 00:33:38.540 --> 00:33:41.620 r cross r hat cross r hat. 00:33:41.620 --> 00:33:46.420 So the only force that matters here is this Coriolis one. 00:33:46.420 --> 00:33:55.800 And so that force is 2mr dot theta dot. 00:33:55.800 --> 00:33:59.280 And we have the r cross. 00:33:59.280 --> 00:34:01.800 We know this comes out in the k direction. 00:34:01.800 --> 00:34:05.150 And we get an r in here when we compute this product. 00:34:05.150 --> 00:34:09.880 And this, to me, looks an awful lot like this. 00:34:09.880 --> 00:34:11.719 Except I missed an r squared. 00:34:11.719 --> 00:34:12.469 How did I do that? 00:34:19.049 --> 00:34:19.924 STUDENT: [INAUDIBLE]. 00:34:25.900 --> 00:34:27.460 PROFESSOR: Oh no, it's not r squared. 00:34:27.460 --> 00:34:29.650 It started off over here as the angular momentum has an r 00:34:29.650 --> 00:34:30.858 squared, took the derivative. 00:34:30.858 --> 00:34:33.109 It dropped down to 2rr dot. 00:34:33.109 --> 00:34:36.150 And that's right. 00:34:36.150 --> 00:34:39.650 So the Coriolis force, what's it got 00:34:39.650 --> 00:34:41.489 to do with the angular momentum? 00:34:46.460 --> 00:34:48.520 That's kind of the point of the exercise here. 00:34:48.520 --> 00:34:57.700 In order for this ping pong ball to be accelerated, 00:34:57.700 --> 00:35:00.760 as it goes slowly out this tube, the angular momentum 00:35:00.760 --> 00:35:02.630 is increasing. 00:35:02.630 --> 00:35:05.030 In order to change angular momentum with time, 00:35:05.030 --> 00:35:06.910 you have to apply a torque. 00:35:06.910 --> 00:35:09.060 The torque that you have to apply 00:35:09.060 --> 00:35:12.700 is 2mrr dot in the k direction. 00:35:12.700 --> 00:35:16.670 And that is r cross the Coriolis force. 00:35:16.670 --> 00:35:19.240 So the Coriolis force, in this case, 00:35:19.240 --> 00:35:21.990 is the force that's necessary to increase the angular 00:35:21.990 --> 00:35:24.630 momentum of a system. 00:35:24.630 --> 00:35:26.760 That's very often the reason-- that's 00:35:26.760 --> 00:35:30.110 what Coriolis force is about. 00:35:30.110 --> 00:35:36.100 So when you shoot an artillery piece on the Earth, 00:35:36.100 --> 00:35:38.410 you've got that projectile going out there. 00:35:38.410 --> 00:35:41.760 It has angular momentum with respect to the Earth. 00:35:41.760 --> 00:35:44.960 And you'll find out that this little term pops up. 00:35:44.960 --> 00:35:47.920 And in fact, the projectile doesn't go straight. 00:35:47.920 --> 00:35:48.590 It curves. 00:35:51.550 --> 00:35:54.020 There's lots of things that because of conservation 00:35:54.020 --> 00:35:59.810 of angular momentum, you end up with this term popping up. 00:35:59.810 --> 00:36:01.810 In this case, angular momentum is not conserved. 00:36:01.810 --> 00:36:02.750 It's increasing. 00:36:02.750 --> 00:36:04.860 So you have this force to make it happen. 00:36:04.860 --> 00:36:06.260 Any questions about this? 00:36:06.260 --> 00:36:08.090 You're going to use this one a lot. 00:36:08.090 --> 00:36:09.548 You're going to work with it a lot. 00:36:17.200 --> 00:36:20.390 So anytime you see changes in angular momentum happening 00:36:20.390 --> 00:36:25.990 in a problem, in these problems with circular motion, 00:36:25.990 --> 00:36:30.060 velocities of parts increasing in radius, 00:36:30.060 --> 00:36:32.860 you'll almost always see this term pop up. 00:36:32.860 --> 00:36:35.200 Any time you see these changes in angular momentum, 00:36:35.200 --> 00:36:38.030 you'll often see the Coriolis term. 00:36:38.030 --> 00:36:46.070 All right, now we're going to do another interesting problem-- 00:36:46.070 --> 00:36:48.040 simple but interesting. 00:36:48.040 --> 00:36:52.790 And that is really to go-- let's go do this problem 00:36:52.790 --> 00:36:55.170 where the thing is really allowed to come freely out. 00:37:02.079 --> 00:37:03.745 All right, you ready to defend yourself? 00:37:12.050 --> 00:37:15.040 A little short stick is pretty effective at throwing candy. 00:37:19.640 --> 00:37:20.950 You got your safety glasses on? 00:37:20.950 --> 00:37:24.400 You want me to see if I can get one up there? 00:37:24.400 --> 00:37:27.600 Aww, also, I obviously haven't practiced this. 00:37:31.210 --> 00:37:32.550 All right, last one. 00:37:38.380 --> 00:37:40.840 Actually, there's two more. 00:37:48.400 --> 00:37:51.300 What makes this work? 00:37:51.300 --> 00:37:52.880 Let's do this problem. 00:37:56.490 --> 00:37:58.505 So let's look at our candy shooter. 00:38:17.870 --> 00:38:20.460 So I'm whipping this thing around. 00:38:20.460 --> 00:38:21.690 Candy is coming out of here. 00:38:21.690 --> 00:38:24.670 It's at B. 00:38:24.670 --> 00:38:28.970 Again, I get no z, not mess with the z part. 00:38:28.970 --> 00:38:31.170 The z part is trivial to usually deal with. 00:38:31.170 --> 00:38:33.636 Because it's totally independent, 00:38:33.636 --> 00:38:34.635 just separate equations. 00:38:34.635 --> 00:38:36.630 It doesn't complicate things much at all, 00:38:36.630 --> 00:38:38.542 even when you have z. 00:38:38.542 --> 00:38:42.340 Now, in this case, the velocity-- and here's 00:38:42.340 --> 00:38:45.230 my O frame here. 00:38:45.230 --> 00:38:47.110 The velocity B with respect to O, 00:38:47.110 --> 00:38:52.520 well, now it's got-- I'm not going to move. 00:38:52.520 --> 00:38:55.080 I'm standing still when I do it. 00:38:55.080 --> 00:38:57.660 So the first term is 0. 00:38:57.660 --> 00:39:02.350 It's got an r term, r dot in the r hat direction. 00:39:02.350 --> 00:39:08.570 And it has an r theta dot theta hat term. 00:39:13.810 --> 00:39:17.330 And these can now-- this might be changing. 00:39:17.330 --> 00:39:19.380 This might have a time derivative, r double dot. 00:39:19.380 --> 00:39:20.130 It certainly does. 00:39:20.130 --> 00:39:22.000 It's accelerating coming out of that tube. 00:39:22.000 --> 00:39:25.670 And my theta, angular motion, it can be accelerating, too. 00:39:25.670 --> 00:39:28.985 So we're going to have to deal with those. 00:39:28.985 --> 00:39:30.360 I need to know the accelerations. 00:39:36.100 --> 00:39:39.060 Now, I haven't emphasized it till now, 00:39:39.060 --> 00:39:42.990 but I find it conceptually useful to-- when 00:39:42.990 --> 00:39:45.750 you work with polar coordinates, you 00:39:45.750 --> 00:39:48.810 can have this ability to aggregate 00:39:48.810 --> 00:39:57.910 the terms in these two component directions. 00:39:57.910 --> 00:39:59.240 So you have this. 00:39:59.240 --> 00:40:01.640 All the r terms go together. 00:40:01.640 --> 00:40:06.870 And we've let the z double dot term-- 00:40:06.870 --> 00:40:08.270 it's just separate by itself. 00:40:08.270 --> 00:40:10.380 It drops out easily. 00:40:10.380 --> 00:40:12.040 And you have the theta hat term. 00:40:12.040 --> 00:40:25.890 And that's the r theta double dot plus 2r dot theta dot. 00:40:25.890 --> 00:40:27.900 And these are in the theta hat direction. 00:40:27.900 --> 00:40:30.530 There's your Coriolis term, your Euler acceleration 00:40:30.530 --> 00:40:33.500 term, centripetal term. 00:40:33.500 --> 00:40:34.039 Yeah? 00:40:34.039 --> 00:40:34.914 STUDENT: [INAUDIBLE]. 00:40:38.964 --> 00:40:40.630 PROFESSOR: Well, I don't know if I ought 00:40:40.630 --> 00:40:42.692 to tell you secrets about me. 00:40:42.692 --> 00:40:45.025 Because it's going to give you an advantage on the quiz. 00:40:48.300 --> 00:40:50.840 But I've almost never, ever been known 00:40:50.840 --> 00:40:55.630 to ask a question that says, "derive." 00:40:55.630 --> 00:40:59.920 But I'll sure ask you concept questions. 00:40:59.920 --> 00:41:01.930 I really want you to understand the principles. 00:41:01.930 --> 00:41:04.625 I don't get real hung up on having you do 00:41:04.625 --> 00:41:07.029 the grungy grind it out things. 00:41:07.029 --> 00:41:08.570 Do I want you to remember the formula 00:41:08.570 --> 00:41:13.130 for how to take the derivative of a vector in rotating frame? 00:41:13.130 --> 00:41:15.150 Yeah, that's where these have come from. 00:41:18.250 --> 00:41:21.830 You had better remember this. 00:41:21.830 --> 00:41:24.850 These two formulas, the velocity formula and this, 00:41:24.850 --> 00:41:27.820 the acceleration formula, are just core to this course. 00:41:27.820 --> 00:41:29.670 Now, the way quizzes are done-- first quiz, 00:41:29.670 --> 00:41:31.900 you come in, one sheet of paper. 00:41:31.900 --> 00:41:35.580 What had better be on your paper? 00:41:35.580 --> 00:41:37.030 OK? 00:41:37.030 --> 00:41:39.470 And second quiz, two sheets, final, three sheets, 00:41:39.470 --> 00:41:40.870 that kind of thing. 00:41:40.870 --> 00:41:43.290 But conceptually, you've got forces in the r, 00:41:43.290 --> 00:41:47.000 forces in the z, accelerations in r theta 00:41:47.000 --> 00:41:49.455 and z, forces r theta and z. 00:41:52.210 --> 00:41:53.890 And for these planar motion problems, 00:41:53.890 --> 00:41:57.190 this one is sure easy to use. 00:41:57.190 --> 00:42:00.980 So let's think about this problem. 00:42:00.980 --> 00:42:06.650 I'm going to let this be frictionless just 00:42:06.650 --> 00:42:09.410 to make it easy. 00:42:09.410 --> 00:42:12.130 All right, so what possible forces act 00:42:12.130 --> 00:42:14.870 on the hunk of candy? 00:42:14.870 --> 00:42:17.210 Let's do a free body diagram of the hunk of candy 00:42:17.210 --> 00:42:19.680 coming out of here. 00:42:19.680 --> 00:42:22.010 What are the forces? 00:42:22.010 --> 00:42:23.640 And let's keep it planar. 00:42:23.640 --> 00:42:25.080 We can get gravity into this. 00:42:25.080 --> 00:42:26.770 But let's just do it in a plane. 00:42:26.770 --> 00:42:29.500 So I'm just going horizontal and slinging this thing. 00:42:29.500 --> 00:42:31.160 Gravity's in the z direction, and I've 00:42:31.160 --> 00:42:32.430 constrained it in the z. 00:42:32.430 --> 00:42:37.140 So it's something supporting the gravity in the tube. 00:42:37.140 --> 00:42:40.540 So definitely there's an mg on this thing downwards. 00:42:40.540 --> 00:42:42.990 But it's in the z direction, and we're not 00:42:42.990 --> 00:42:45.200 letting it move in the z. 00:42:45.200 --> 00:42:49.290 What about the horizontal, this direction? 00:42:49.290 --> 00:42:53.750 The r-- this is my r hat direction. 00:42:53.750 --> 00:42:56.960 This is my theta hat direction. 00:42:56.960 --> 00:42:59.516 Whoops, not either-- theta is going this way. 00:42:59.516 --> 00:43:03.260 So I've drawn this as a side view. 00:43:03.260 --> 00:43:06.430 What's the force in the r hat direction? 00:43:06.430 --> 00:43:07.305 STUDENT: [INAUDIBLE]. 00:43:10.600 --> 00:43:12.350 PROFESSOR: OK, in order to do the problem, 00:43:12.350 --> 00:43:14.980 you have to figure that out. 00:43:14.980 --> 00:43:15.850 What are the forces? 00:43:19.030 --> 00:43:22.150 What are the source of forces in the r direction? 00:43:22.150 --> 00:43:24.016 Here's the r direction. 00:43:24.016 --> 00:43:24.890 STUDENT: [INAUDIBLE]. 00:43:24.890 --> 00:43:25.723 PROFESSOR: Say what? 00:43:28.230 --> 00:43:29.140 Come on, you guys. 00:43:29.140 --> 00:43:31.158 Somebody be-- yeah. 00:43:31.158 --> 00:43:33.650 STUDENT: [INAUDIBLE]. 00:43:33.650 --> 00:43:34.860 PROFESSOR: There are not any. 00:43:34.860 --> 00:43:35.776 Is that what you said? 00:43:35.776 --> 00:43:36.710 There aren't any. 00:43:36.710 --> 00:43:38.025 And why's that? 00:43:38.025 --> 00:43:38.900 STUDENT: [INAUDIBLE]. 00:43:38.900 --> 00:43:41.780 PROFESSOR: Right, so there's no forces in the r direction. 00:43:41.780 --> 00:43:44.980 So there's no forces on this thing in the r. 00:43:44.980 --> 00:43:48.260 And so then this is a side view. 00:43:48.260 --> 00:43:50.260 We could do the top view looking down. 00:43:50.260 --> 00:43:53.260 Top view, you've got your x, y. 00:43:53.260 --> 00:43:56.580 You also have your-- here's the ball. 00:43:56.580 --> 00:43:58.310 Here's the r hat. 00:43:58.310 --> 00:44:00.040 Here's the theta hat. 00:44:00.040 --> 00:44:04.800 Now, free body diagram in the top view-- 00:44:04.800 --> 00:44:10.937 well, there's some force here probably. 00:44:10.937 --> 00:44:14.290 STUDENT: I was going to ask that the fact that we 00:44:14.290 --> 00:44:16.685 have no forces in the r hat direction, but we do 00:44:16.685 --> 00:44:18.529 have acceleration. 00:44:18.529 --> 00:44:19.445 PROFESSOR: Absolutely. 00:44:19.445 --> 00:44:22.030 She's commenting that we do have accelerations 00:44:22.030 --> 00:44:23.240 in the r direction, right? 00:44:23.240 --> 00:44:24.490 STUDENT: And we have no force. 00:44:24.490 --> 00:44:25.910 PROFESSOR: And no force. 00:44:25.910 --> 00:44:28.765 So that's the conundrum of this problem. 00:44:28.765 --> 00:44:30.140 That's the point of this problem. 00:44:30.140 --> 00:44:30.931 So let me continue. 00:44:34.070 --> 00:44:36.830 Free body diagram-- I'm looking down on it. 00:44:36.830 --> 00:44:38.080 I'm allowing for some force. 00:44:38.080 --> 00:44:42.590 It's the normal force that comes from the pipe exerting 00:44:42.590 --> 00:44:44.420 the force on the candy. 00:44:44.420 --> 00:44:47.410 And since it's frictionless, it can only be normal to the pipe. 00:44:47.410 --> 00:44:54.270 OK, so there's the free body diagram in the top view. 00:44:54.270 --> 00:44:56.466 So we can write two equations. 00:44:56.466 --> 00:44:58.510 We can write three questions-- this one 00:44:58.510 --> 00:45:01.286 equal to 0, this one in the r hat direction, 00:45:01.286 --> 00:45:02.785 this one in the theta hat direction. 00:45:22.600 --> 00:45:27.350 Sum of the forces r hat direction must be 0. 00:45:27.350 --> 00:45:28.667 And we have a mass. 00:45:28.667 --> 00:45:29.666 We have an acceleration. 00:45:40.170 --> 00:45:42.120 Solve for r double dot. 00:45:50.860 --> 00:45:58.930 Remarkable-- there's no force in the r hat direction. 00:45:58.930 --> 00:46:05.220 The position of the object, the r 00:46:05.220 --> 00:46:08.490 coordinate, it has a velocity. 00:46:08.490 --> 00:46:10.250 It has an acceleration. 00:46:14.380 --> 00:46:18.045 But the total acceleration in the r hat directions 00:46:18.045 --> 00:46:19.590 are actually 0. 00:46:19.590 --> 00:46:25.790 The rate of change of the velocity 00:46:25.790 --> 00:46:28.860 of this thing in the radial direction, r double dot, 00:46:28.860 --> 00:46:30.870 is nonzero. 00:46:30.870 --> 00:46:32.151 But there are no forces on it. 00:46:37.340 --> 00:46:42.150 I have pondered another way to explain this. 00:46:42.150 --> 00:46:43.390 I'm still thinking about it. 00:46:43.390 --> 00:46:45.010 And you think about this, too. 00:46:45.010 --> 00:46:48.300 How do you explain this in the absence of forces? 00:46:48.300 --> 00:46:49.960 And it's partly where the concept 00:46:49.960 --> 00:46:53.760 comes from of fictitious forces, that centrifugal force is 00:46:53.760 --> 00:46:54.580 a force. 00:46:54.580 --> 00:46:55.140 It's not. 00:46:55.140 --> 00:46:56.600 It's an acceleration. 00:46:56.600 --> 00:47:01.330 This is just saying that-- let's go back to this problem. 00:47:01.330 --> 00:47:05.170 In order to make something go in a circle, 00:47:05.170 --> 00:47:08.200 you have to put a force on it to cause 00:47:08.200 --> 00:47:11.280 the centripetal acceleration. 00:47:11.280 --> 00:47:16.090 You have to allow the thing to go out if you're not 00:47:16.090 --> 00:47:18.190 forcing it to go in a circle. 00:47:18.190 --> 00:47:20.940 You have to allow it to go out at that rate in order 00:47:20.940 --> 00:47:25.020 for there to be no centripetal accelerations on the object. 00:47:25.020 --> 00:47:25.521 Yeah? 00:47:25.521 --> 00:47:27.603 STUDENT: So I was going to say, when you start it, 00:47:27.603 --> 00:47:29.026 you push it in the y direction. 00:47:29.026 --> 00:47:29.930 So that's a force. 00:47:29.930 --> 00:47:32.620 It's not in the r hat, but it's in the y. 00:47:32.620 --> 00:47:33.830 PROFESSOR: To get it started. 00:47:33.830 --> 00:47:35.958 STUDENT: Right, and there's no force opposing it 00:47:35.958 --> 00:47:38.090 in that direction [INAUDIBLE]. 00:47:38.090 --> 00:47:45.990 PROFESSOR: Does it experience centripetal acceleration? 00:47:45.990 --> 00:47:47.550 So there's no rotation. 00:47:47.550 --> 00:47:50.125 Does it experience centripetal acceleration? 00:47:53.100 --> 00:47:54.560 What do you think? 00:47:54.560 --> 00:47:58.450 Yeah, because it goes through curved motion. 00:47:58.450 --> 00:48:00.360 At any instance in time when it's doing that, 00:48:00.360 --> 00:48:02.279 there is a radius of curvature. 00:48:02.279 --> 00:48:04.070 You can at that instant in time think of it 00:48:04.070 --> 00:48:07.050 as being in a circular path. 00:48:07.050 --> 00:48:09.840 And sometimes it's just really easy 00:48:09.840 --> 00:48:13.510 to do these kind of problems with normal and tangential 00:48:13.510 --> 00:48:14.970 coordinates. 00:48:14.970 --> 00:48:23.670 So I'm looking down on the x, y plane. 00:48:23.670 --> 00:48:25.330 Gravity is into the Earth. 00:48:25.330 --> 00:48:27.790 So I'm looking down on a vehicle, a car. 00:48:27.790 --> 00:48:31.940 And that car, the guy is kind of drunk. 00:48:31.940 --> 00:48:35.100 He's going down the road like this. 00:48:35.100 --> 00:48:36.440 So here's my y. 00:48:36.440 --> 00:48:37.770 Here's my x. 00:48:37.770 --> 00:48:49.270 And y equals some A sine 2 pi over the wavelength, 00:48:49.270 --> 00:48:58.150 2 pi over lambda, times x at some A sine kx. 00:48:58.150 --> 00:49:00.380 And as he drives down-- if you're in a car, 00:49:00.380 --> 00:49:03.490 and you're doing that, you get thrown side to side in the car. 00:49:03.490 --> 00:49:04.997 So you are being accelerated. 00:49:04.997 --> 00:49:06.580 And so we want to be able to calculate 00:49:06.580 --> 00:49:09.850 the acceleration due to the fact that you're going 00:49:09.850 --> 00:49:12.480 down and doing a curved path. 00:49:12.480 --> 00:49:16.390 And we deal with these things sometimes 00:49:16.390 --> 00:49:19.260 with a convenient little set of coordinates 00:49:19.260 --> 00:49:23.810 that are our normal and tangential unit 00:49:23.810 --> 00:49:30.140 vectors, u normal and u tangential, 00:49:30.140 --> 00:49:31.350 at any instant in time. 00:49:34.610 --> 00:49:38.990 And we know that if this is along the path, 00:49:38.990 --> 00:49:41.700 at any instant in time you're right here, 00:49:41.700 --> 00:49:45.070 what direction is your velocity? 00:49:45.070 --> 00:49:52.170 Just definition of velocity-- tangent to the path, right? 00:49:52.170 --> 00:49:54.170 So at any instant in time, the velocity 00:49:54.170 --> 00:49:58.000 has got to be tangent to the path at that moment. 00:49:58.000 --> 00:50:09.170 So velocity, the vector, has a magnitude and a unit vector 00:50:09.170 --> 00:50:12.606 uT here I'll call it, tangent. 00:50:12.606 --> 00:50:15.020 That's all there is to it. 00:50:15.020 --> 00:50:20.320 And the acceleration of this thing 00:50:20.320 --> 00:50:22.435 is your time derivative of this. 00:50:26.330 --> 00:50:34.000 And that's going to give you a v dot uT hat plus a v. 00:50:34.000 --> 00:50:39.290 And now you need a time derivative of this guy. 00:50:39.290 --> 00:50:41.904 But this is a unit length vector. 00:50:41.904 --> 00:50:43.320 You can plug it into that equation 00:50:43.320 --> 00:50:46.700 for the derivative of a rotating vector 00:50:46.700 --> 00:50:48.440 and calculate what this should be. 00:50:48.440 --> 00:50:50.210 You could also just draw it out. 00:50:50.210 --> 00:50:53.060 So I'll draw this for you. 00:50:58.880 --> 00:51:01.750 How are we doing on time? 00:51:01.750 --> 00:51:04.070 I should just be able to finish this. 00:51:11.500 --> 00:51:16.500 Here's my unit vector in the tangential direction. 00:51:16.500 --> 00:51:23.320 As I'm going around this curve, this 00:51:23.320 --> 00:51:27.650 is my tangential direction. 00:51:27.650 --> 00:51:29.280 There's some instant I have a radius. 00:51:29.280 --> 00:51:31.050 We call that rho. 00:51:31.050 --> 00:51:33.655 And in the little time, delta t, I 00:51:33.655 --> 00:51:38.300 go through an angle delta theta in delta t. 00:51:38.300 --> 00:51:42.070 And this is my uT vector here. 00:51:42.070 --> 00:51:46.120 It changes by a little bit. 00:51:46.120 --> 00:51:53.290 That's the change in the uT unit vector in this time, delta t. 00:51:53.290 --> 00:51:55.870 And it goes perpendicular. 00:51:55.870 --> 00:51:58.652 And it goes in the positive un direction. 00:52:01.700 --> 00:52:06.050 So delta-- what's the easiest way to write this one? 00:52:12.030 --> 00:52:26.600 Delta uT equals some theta dot delta t-- that's 00:52:26.600 --> 00:52:35.390 the angle-- times the length of the unit vector, 1. 00:52:35.390 --> 00:52:38.410 That's the distance it goes, so 1. 00:52:38.410 --> 00:52:42.990 So r omega, 1 theta dot is the distance 00:52:42.990 --> 00:52:48.500 that this unit vector goes through in delta t. 00:52:48.500 --> 00:52:54.940 And the direction it goes in is u normal hat. 00:52:54.940 --> 00:53:01.820 So delta uT over delta t limit as t 00:53:01.820 --> 00:53:11.280 goes to 0, you get theta dot un, just like before. 00:53:11.280 --> 00:53:13.700 So the time derivative of this unit vector 00:53:13.700 --> 00:53:17.080 in the tangential direction is just theta dot 00:53:17.080 --> 00:53:19.770 in the normal direction. 00:53:19.770 --> 00:53:24.090 And then from that, we can very quickly 00:53:24.090 --> 00:53:27.630 derive the rest of this acceleration. 00:53:27.630 --> 00:53:30.820 The acceleration then is that plus this. 00:53:30.820 --> 00:53:41.119 We now know an expression for-- this is my uT dot term. 00:53:41.119 --> 00:53:42.410 I'm going to plug that in here. 00:53:42.410 --> 00:53:45.510 This then, we need an expression for v. What's v? 00:53:48.770 --> 00:53:52.000 Well, at that instant in time, it has some radius rho. 00:53:52.000 --> 00:53:55.100 It has an angular velocity theta dot. 00:53:55.100 --> 00:53:57.930 So rho theta dot would be the v here. 00:54:07.980 --> 00:54:13.240 So I'm looking for an expression for vuT dot. 00:54:18.300 --> 00:54:24.928 So that's v theta dot un. 00:54:24.928 --> 00:54:28.550 But theta dot is v over rho. 00:54:31.150 --> 00:54:37.210 v squared-- sorry about this-- over rho un. 00:54:37.210 --> 00:54:45.160 So this guy up here, this is v dot uT plus v squared-- 00:54:45.160 --> 00:54:47.150 I'll rewrite it-- over rho un. 00:54:50.082 --> 00:54:51.540 So if you're speeding up, if you're 00:54:51.540 --> 00:54:54.250 going from 30 miles an hour to 40 miles an hour, 00:54:54.250 --> 00:54:55.520 that's your tangential. 00:54:55.520 --> 00:54:57.360 That's your speed along the path. 00:54:57.360 --> 00:54:58.901 That's this term. 00:54:58.901 --> 00:55:00.650 But because you're going around the curve, 00:55:00.650 --> 00:55:02.720 you have an acceleration of v squared over rho. 00:55:02.720 --> 00:55:05.010 You've run into this before in physics. 00:55:05.010 --> 00:55:07.760 This is where it comes from. 00:55:07.760 --> 00:55:12.490 This is a centripetal acceleration like term. 00:55:12.490 --> 00:55:15.210 If you replace v with rho theta dot, 00:55:15.210 --> 00:55:17.557 you'd get r theta dot squared. 00:55:17.557 --> 00:55:19.890 So you can either put it in terms of v squared over rho, 00:55:19.890 --> 00:55:22.820 or you can put it in terms of rho theta dot squared. 00:55:22.820 --> 00:55:28.350 And rho theta dot squared sounds a lot like my acceleration term 00:55:28.350 --> 00:55:28.890 right here. 00:55:32.280 --> 00:55:35.500 So with that simple little formula, 00:55:35.500 --> 00:55:38.580 you can do-- you need one other thing. 00:55:41.160 --> 00:55:44.960 And you just go look it up in the book. 00:55:44.960 --> 00:55:48.680 There is an expression from calculus for the radius 00:55:48.680 --> 00:55:52.570 of curvature of a path. 00:55:52.570 --> 00:55:56.480 And it has first and second derivatives of y with respect 00:55:56.480 --> 00:55:56.980 to x. 00:55:56.980 --> 00:56:00.710 So you need a dy/dx and a d2y dx squared. 00:56:00.710 --> 00:56:03.650 From a sine function, you can calculate that. 00:56:03.650 --> 00:56:05.960 So you calculate rho from a formula 00:56:05.960 --> 00:56:07.910 that's in the book for calculating 00:56:07.910 --> 00:56:09.240 radius of curvature. 00:56:09.240 --> 00:56:11.100 And then you're done, all right? 00:56:11.100 --> 00:56:15.520 So see you on Thursday.