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 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.260 at ocw.mit.edu. 00:00:20.872 --> 00:00:22.330 PROFESSOR: So I'm going to give you 00:00:22.330 --> 00:00:30.130 a quick example of what I think is a good way to do solutions. 00:00:30.130 --> 00:00:31.405 Our approach to solutions. 00:00:45.090 --> 00:00:45.980 State the problem. 00:00:48.910 --> 00:00:51.030 I'm going to give you a little formulaic here. 00:00:51.030 --> 00:00:51.720 Draw figures. 00:01:00.870 --> 00:01:03.440 On the first day I said I like to think of dynamics problem. 00:01:03.440 --> 00:01:06.500 I break them down into three categories. 00:01:06.500 --> 00:01:08.185 One is to describe the motion. 00:01:14.120 --> 00:01:15.810 What does describing the motion mean? 00:01:15.810 --> 00:01:21.670 Well how many-- the number of degrees of 00:01:21.670 --> 00:01:23.790 freedom in the problem. 00:01:23.790 --> 00:01:30.450 That implies the number of equations of motion 00:01:30.450 --> 00:01:33.360 that you're going to need to solve that problem. 00:01:33.360 --> 00:01:35.710 So you got to identify the number of degrees of freedom. 00:01:35.710 --> 00:01:41.370 It also then tells you the number of coordinates you need. 00:01:44.310 --> 00:01:46.730 So this is all part of describing the motion. 00:01:46.730 --> 00:01:50.040 It's figuring out how many coordinates, assigning them. 00:01:58.400 --> 00:02:01.860 So assigning the coordinates. 00:02:01.860 --> 00:02:05.520 And then finally, essentially all underneath this you 00:02:05.520 --> 00:02:06.896 essentially do the kinematics. 00:02:21.240 --> 00:02:27.097 And that's the velocities, accelerations, so forth. 00:02:27.097 --> 00:02:29.055 So once you have-- you've explained the motion. 00:02:34.420 --> 00:02:35.960 And this I guess is four. 00:02:35.960 --> 00:02:43.155 Explain the correct physical laws. 00:02:47.060 --> 00:02:48.460 You know how they apply. 00:02:48.460 --> 00:02:50.280 f equals ma. 00:02:50.280 --> 00:02:52.925 Newton's first, second, third law of conservation of momentum 00:02:52.925 --> 00:02:56.750 or whatever you want-- you think is the appropriate thing. 00:02:56.750 --> 00:02:59.840 So explain what the physical laws are and apply them. 00:03:02.620 --> 00:03:05.100 And finally, do the math. 00:03:10.287 --> 00:03:12.120 So if you could break problems down that way 00:03:12.120 --> 00:03:14.450 it'll give you a nice, logical flow. 00:03:14.450 --> 00:03:16.980 So I'm going to give you a bit of an example problem. 00:03:27.140 --> 00:03:29.640 And I'm also going to kind of pose a brain teaser to you 00:03:29.640 --> 00:03:31.060 at the end of this problem today. 00:03:31.060 --> 00:03:33.660 I want to give you something to think about. 00:03:33.660 --> 00:03:37.710 So I'm going to draw my problem. 00:03:37.710 --> 00:03:43.130 This is a block on an incline. 00:03:46.840 --> 00:03:50.820 It's got some scales to measure your weight. 00:03:50.820 --> 00:03:55.510 And you're standing on this thing. 00:03:55.510 --> 00:03:56.695 Riding it down the incline. 00:04:00.000 --> 00:04:02.870 And the first question about this 00:04:02.870 --> 00:04:08.490 is to find the position as a function of time. 00:04:26.534 --> 00:04:27.450 So that's the problem. 00:04:27.450 --> 00:04:28.980 That's part a. 00:04:32.650 --> 00:04:34.670 So state the problem, find the position. 00:04:45.024 --> 00:04:46.190 So how-- well what do we do? 00:04:46.190 --> 00:04:48.690 Well draw figures, I've started with that. 00:04:48.690 --> 00:04:50.240 Next, describe the motion. 00:04:50.240 --> 00:04:56.140 I need a free-- We have to figure out 00:04:56.140 --> 00:04:59.360 how many degrees of freedom this problem has. 00:04:59.360 --> 00:05:01.030 I'm going to just declare no rotation. 00:05:01.030 --> 00:05:04.180 Going to treat it as a particle. 00:05:04.180 --> 00:05:06.871 So a particle has how many, generally how many degrees of 00:05:06.871 --> 00:05:07.370 freedom? 00:05:07.370 --> 00:05:09.190 How many coordinates to completely describe 00:05:09.190 --> 00:05:10.226 where it's at? 00:05:10.226 --> 00:05:11.040 AUDIENCE: Three. 00:05:11.040 --> 00:05:12.240 PROFESSOR: Three. 00:05:12.240 --> 00:05:14.880 So I may need as many as three coordinates 00:05:14.880 --> 00:05:17.180 to describe the motion of this thing. 00:05:17.180 --> 00:05:19.780 And if I really doing complete equations of motion 00:05:19.780 --> 00:05:21.330 I need three equations of motion. 00:05:21.330 --> 00:05:25.560 So this is two, three-- describing the motion I 00:05:25.560 --> 00:05:28.576 have three degrees of freedom. 00:05:28.576 --> 00:05:30.075 I'm going to need three coordinates. 00:05:36.820 --> 00:05:42.200 So in this case here's my picture 00:05:42.200 --> 00:05:45.560 I'm going just to set up a Cartesian coordinate system 00:05:45.560 --> 00:05:49.970 aligned in a helpful way. 00:05:49.970 --> 00:05:53.390 X, y, z coming out. 00:05:53.390 --> 00:05:56.910 And this is my fixed inertial reference frame. 00:05:56.910 --> 00:05:58.820 And here's my center of mass. 00:05:58.820 --> 00:06:02.050 And basically my coordinates are describing the position 00:06:02.050 --> 00:06:03.130 of the center of mass. 00:06:19.140 --> 00:06:22.230 So that's pretty much the describing the motion, 00:06:22.230 --> 00:06:23.340 what I need for now. 00:06:23.340 --> 00:06:25.250 We'll get to the velocities and accelerations 00:06:25.250 --> 00:06:27.666 when we get to the math part. 00:06:27.666 --> 00:06:28.415 Apply the physics. 00:06:36.570 --> 00:06:38.380 I'm going to use Newton's second law. 00:06:48.570 --> 00:06:55.260 Sum of the external forces, mass times the acceleration. 00:06:55.260 --> 00:06:59.584 That's the physical law I'm going to apply. 00:06:59.584 --> 00:07:00.625 Draw a free body diagram. 00:07:16.800 --> 00:07:18.870 And I'm just going to consider the block 00:07:18.870 --> 00:07:22.709 and the person this is the whole collection, it's one thing. 00:07:22.709 --> 00:07:24.500 I'm just not keep drawing the person on it, 00:07:24.500 --> 00:07:26.840 but-- so here's my object, including 00:07:26.840 --> 00:07:29.540 the weight of the person. 00:07:29.540 --> 00:07:35.410 And it's going to have-- here's its center of mass. 00:07:35.410 --> 00:07:39.040 Obviously in mg, gravitational force, 00:07:39.040 --> 00:07:42.610 it's going to have a normal force. 00:07:42.610 --> 00:07:44.492 Going to have a friction force and how do I 00:07:44.492 --> 00:07:46.950 figure out what the friction-- which direction the friction 00:07:46.950 --> 00:07:47.645 is? 00:07:47.645 --> 00:07:51.310 I assume motion down the hill. 00:07:51.310 --> 00:07:53.070 Friction will oppose it. 00:07:53.070 --> 00:07:55.040 I draw in the arrows in the direction 00:07:55.040 --> 00:07:57.590 I expect the forces to act. 00:07:57.590 --> 00:07:58.965 Then I'll use the sign convention 00:07:58.965 --> 00:08:01.340 that the arrows tell me what signs. 00:08:01.340 --> 00:08:06.780 So since my x-coordinate is down the hill. 00:08:06.780 --> 00:08:08.080 This is y. 00:08:08.080 --> 00:08:11.320 This is x, friction x, in the minus x direction. 00:08:17.820 --> 00:08:23.900 And I haven't-- I've left out a key piece of information. 00:08:23.900 --> 00:08:26.710 Got to have the angle of the slope. 00:08:26.710 --> 00:08:37.270 And once you have the angle of the slope that's theta, 00:08:37.270 --> 00:08:38.840 this is theta. 00:08:38.840 --> 00:08:42.500 And I'm going to need to-- this is in a direction of one 00:08:42.500 --> 00:08:44.120 of my coordinates and so is this, 00:08:44.120 --> 00:08:47.120 but I need to break the gravitational piece 00:08:47.120 --> 00:08:51.520 into components lined up with my coordinates. 00:08:51.520 --> 00:08:55.420 And so you have a theta here as well. 00:08:55.420 --> 00:08:57.445 And now I can write my equation to motion. 00:09:02.450 --> 00:09:05.350 And the nice thing about vectors is 00:09:05.350 --> 00:09:08.690 that when you have three equations of motion, 00:09:08.690 --> 00:09:12.950 three coordinates, each that are components of vectors 00:09:12.950 --> 00:09:14.770 in the x, y and z direction, it gives us 00:09:14.770 --> 00:09:17.130 three equations immediately. 00:09:17.130 --> 00:09:19.590 So for example, the summation of the forces 00:09:19.590 --> 00:09:22.230 in this problem in the z-direction, 00:09:22.230 --> 00:09:25.770 the external forces are sums to? 00:09:25.770 --> 00:09:26.469 AUDIENCE: 0. 00:09:26.469 --> 00:09:27.010 PROFESSOR: 0. 00:09:27.010 --> 00:09:29.620 OK so we get a trivial solution out of that. 00:09:29.620 --> 00:09:34.340 And we don't have to go much further. 00:09:34.340 --> 00:09:36.760 Summation of the forces in the y-direction 00:09:36.760 --> 00:09:38.180 gives us some useful information. 00:09:42.770 --> 00:09:50.860 And then the y-directed forces I have an n and a minus mg 00:09:50.860 --> 00:09:53.910 and I think it's cosine theta. 00:09:59.320 --> 00:10:01.580 Which tells me immediately what n is. 00:10:01.580 --> 00:10:09.630 So from statics I get to-- and from what we know about models 00:10:09.630 --> 00:10:16.870 of friction then we know that we can model the friction as mu 00:10:16.870 --> 00:10:25.120 times n for mu mg cosine theta. 00:10:25.120 --> 00:10:27.726 So from the statics I learn a bunch of things 00:10:27.726 --> 00:10:29.225 that I need to know for the problem. 00:10:34.760 --> 00:10:43.389 So now I get to the real heart of the problem, 00:10:43.389 --> 00:10:45.430 writing my equation of motion in the x-direction. 00:10:56.370 --> 00:11:00.580 And the forces in the x-direction mg 00:11:00.580 --> 00:11:02.100 sine theta down the hill. 00:11:02.100 --> 00:11:04.920 So it's positive. 00:11:04.920 --> 00:11:08.540 Minus the friction is up the hill. 00:11:08.540 --> 00:11:09.500 So I get mg. 00:11:18.264 --> 00:11:34.280 I'm mixing up my m's here but there's no other m so-- this 00:11:34.280 --> 00:11:40.280 basically says, that x double dot, the m's all cancel out. 00:11:40.280 --> 00:11:52.470 That x double dot is g sine theta minus mu g cosine theta. 00:11:52.470 --> 00:11:57.140 And that just happens to be a pretty simple to solve, 00:11:57.140 --> 00:11:58.920 ordinary differential equation. 00:11:58.920 --> 00:12:01.650 This is an equation in which the acceleration in the problem 00:12:01.650 --> 00:12:02.290 is constant. 00:12:02.290 --> 00:12:04.000 The data's not changing. 00:12:04.000 --> 00:12:05.530 None of these things are changing 00:12:05.530 --> 00:12:07.497 and so you can just solve this one. 00:12:07.497 --> 00:12:09.205 Now we're to the third part, do the math. 00:12:16.110 --> 00:12:18.550 This one you can just integrate. 00:12:18.550 --> 00:12:21.525 And so you find out that well, x dot then 00:12:21.525 --> 00:12:25.370 and I'm just going to call this c some constant. 00:12:25.370 --> 00:12:31.340 So x dot ct plus an initial velocity, if it had one. 00:12:31.340 --> 00:12:34.280 And x of t, what you're looking for. 00:12:45.120 --> 00:12:48.370 And now so v0 and x0 are just your initial conditions. 00:12:48.370 --> 00:12:51.410 More than likely 0 if you set it up cleverly. 00:12:51.410 --> 00:12:53.890 So that's just modeling quickly what 00:12:53.890 --> 00:12:56.490 I think a good way to lay out a problem is. 00:12:56.490 --> 00:13:00.230 Describing the motion, explaining the physics, 00:13:00.230 --> 00:13:05.580 doing the math, drawing good pictures, stating the problem. 00:13:05.580 --> 00:13:07.809 All right so now the brain teaser 00:13:07.809 --> 00:13:09.100 that I want you to think about. 00:13:33.360 --> 00:13:35.725 So here's the mass of the block plus the scales. 00:13:39.760 --> 00:13:42.290 And here's the-- here you're standing on it. 00:13:45.800 --> 00:13:47.190 So mass of the person. 00:13:52.550 --> 00:13:53.930 You're riding this down the hill. 00:14:10.630 --> 00:14:11.530 So this is part b. 00:14:22.706 --> 00:14:24.205 Think about that when in the shower. 00:14:27.400 --> 00:14:30.610 If you've got a really simple way to do it, 00:14:30.610 --> 00:14:32.110 great write it up. 00:14:32.110 --> 00:14:34.920 It's not terribly hard and it's mostly-- I'll give you a hint. 00:14:34.920 --> 00:14:39.500 Thinking in terms of free body diagram helps a lot. 00:14:39.500 --> 00:14:41.930 And we'll come back to this kind of fun problem. 00:14:56.440 --> 00:14:58.890 OK, want to go-- that was part one. 00:14:58.890 --> 00:15:03.210 I want to go onto this recapping the center of mass quickly. 00:15:03.210 --> 00:15:11.519 We learned a couple of important things. 00:15:11.519 --> 00:15:13.310 We got-- we talked about the center of mass 00:15:13.310 --> 00:15:16.730 because we were just talking about Newton's three laws. 00:15:16.730 --> 00:15:18.894 From looking at the first law, found 00:15:18.894 --> 00:15:20.810 that it's useful in determining whether or not 00:15:20.810 --> 00:15:22.530 you're in a inertial frame, we used it. 00:15:22.530 --> 00:15:24.320 Second law we've just applied it. 00:15:24.320 --> 00:15:26.960 We used to do-- get equations of motion. 00:15:26.960 --> 00:15:31.730 Third law was about-- we used it when we're 00:15:31.730 --> 00:15:33.150 thinking about center of mass. 00:15:33.150 --> 00:15:36.040 We used it to define what the center of mass was. 00:15:36.040 --> 00:15:40.730 So we said the total mass of a system, 00:15:40.730 --> 00:15:41.890 I better draw my picture. 00:15:41.890 --> 00:15:43.245 Here's my system of particles. 00:15:45.940 --> 00:15:50.910 M1 with position vectors. 00:15:50.910 --> 00:15:53.420 So a whole mess of particles with their position vectors. 00:15:53.420 --> 00:15:56.270 This is ri with respect to O for example. 00:15:56.270 --> 00:15:59.480 This is my O x, y, z frame. 00:15:59.480 --> 00:16:03.680 We said that this total mass of the particles 00:16:03.680 --> 00:16:11.520 somewhere out here there's a center of mass with a position 00:16:11.520 --> 00:16:22.300 vector rg with respect to O. So the definition of my center 00:16:22.300 --> 00:16:29.140 of mass is this is equal to the summation of the m, i, r, i, o 00:16:29.140 --> 00:16:33.170 and these are position vectors. 00:16:33.170 --> 00:16:35.760 So that's the definition of my center of mass. 00:16:35.760 --> 00:16:43.215 If I take two time derivatives of that we arrived at mt rg 00:16:43.215 --> 00:16:46.760 with the respect to O double dot. 00:16:46.760 --> 00:16:53.500 Summation over i of my m, i, r, i, o double dots. 00:16:57.040 --> 00:17:01.730 And then importantly that's the summation 00:17:01.730 --> 00:17:07.883 of all of the external forces on each 00:17:07.883 --> 00:17:09.869 of these-- each one of these by itself 00:17:09.869 --> 00:17:12.780 satisfies Newton's law, second law, which 00:17:12.780 --> 00:17:14.690 he wrote about particles. 00:17:14.690 --> 00:17:16.105 Each one has a summation. 00:17:16.105 --> 00:17:17.980 I've summed these and I'm going to sum these, 00:17:17.980 --> 00:17:23.920 but each one has a summation of internal forces acting on it. 00:17:23.920 --> 00:17:27.130 These I call the f, i, j's. 00:17:27.130 --> 00:17:29.900 And we learn-- something about the third law 00:17:29.900 --> 00:17:31.530 tells us about that summation. 00:17:31.530 --> 00:17:34.390 The third law tells us what? 00:17:34.390 --> 00:17:38.650 That goes to 0 and that was that's the really powerful 00:17:38.650 --> 00:17:42.530 piece of the third law that we make great use of. 00:17:42.530 --> 00:17:46.190 Because this now essentially allows us to say, 00:17:46.190 --> 00:17:49.060 that the summation of the external 00:17:49.060 --> 00:17:53.110 forces on an assembly of particles, 00:17:53.110 --> 00:18:06.140 on a system of particles, is equal to the total mass 00:18:06.140 --> 00:18:11.820 times the acceleration of the center of mass. 00:18:11.820 --> 00:18:17.080 And that-- what that does in one stroke 00:18:17.080 --> 00:18:22.400 takes you from Newton, who's laws applied to particles 00:18:22.400 --> 00:18:27.487 and allows you to apply Newton's second law to rigid bodies. 00:18:27.487 --> 00:18:29.070 Because rigid bodies can be thought up 00:18:29.070 --> 00:18:33.250 a bunch of particles, which are represented 00:18:33.250 --> 00:18:36.520 in that simple equation. 00:18:36.520 --> 00:18:39.430 And that gets you from particles to rigid bodies. 00:18:39.430 --> 00:18:42.190 And we all know from physics that a summation 00:18:42.190 --> 00:18:44.200 of the external force on this thing 00:18:44.200 --> 00:18:45.834 is the mass times the acceleration 00:18:45.834 --> 00:18:46.750 of the center of mass. 00:18:50.000 --> 00:18:53.730 So that's actually quite an important powerful law. 00:18:53.730 --> 00:18:57.060 It provides this for us. 00:18:57.060 --> 00:19:01.900 Now I said I wanted to give you a quick, very 00:19:01.900 --> 00:19:04.970 useful application of thinking about center of mass. 00:19:04.970 --> 00:19:08.870 I showed you the other day, I had my carbon fiber tube here. 00:19:08.870 --> 00:19:11.120 Showed you that trick for finding the center of mass 00:19:11.120 --> 00:19:12.703 just by sliding your fingers along it. 00:19:12.703 --> 00:19:14.730 And you end up at the center of mass. 00:19:14.730 --> 00:19:17.190 Well as a practical matter other things-- 00:19:17.190 --> 00:19:19.900 you really want to be able find center of mass. 00:19:19.900 --> 00:19:23.290 This is a glider, a sailplane called an LS8. 00:19:23.290 --> 00:19:26.330 I happen-- I'm a glider flight instructor. 00:19:26.330 --> 00:19:29.660 Been flying gliders for 35 years or something like that. 00:19:29.660 --> 00:19:34.280 And I flew a glider of this type just recently. 00:19:34.280 --> 00:19:38.980 That machine has a 49 to one let's call it 50 to one 00:19:38.980 --> 00:19:41.710 to make it easy, glide ratio. 00:19:41.710 --> 00:19:45.030 Means if you're a mile above the ground in still air, 00:19:45.030 --> 00:19:47.850 you will go 50 miles before you touch the ground. 00:19:47.850 --> 00:19:50.840 So they're really amazing high performance machines. 00:19:50.840 --> 00:19:53.930 One of the things about all aircraft that you actually 00:19:53.930 --> 00:19:56.810 need to know is, you really need to know where 00:19:56.810 --> 00:19:59.630 the center of mass of it is. 00:19:59.630 --> 00:20:02.140 And if the center of mass is in the wrong place 00:20:02.140 --> 00:20:04.780 the plane will not fly properly. 00:20:04.780 --> 00:20:07.475 And so you can't just go throw on 50 pounds of lead 00:20:07.475 --> 00:20:09.300 in the tail of that plane and expect 00:20:09.300 --> 00:20:12.160 to survive the next flight. 00:20:12.160 --> 00:20:14.250 So you have to know where the center of mass 00:20:14.250 --> 00:20:16.560 is, and in fact, you want the center of mass 00:20:16.560 --> 00:20:21.040 about 25% of the-- if the wing is this wide from front 00:20:21.040 --> 00:20:22.890 to back it's called the cord. 00:20:22.890 --> 00:20:25.600 About 25% back from the leading edge 00:20:25.600 --> 00:20:30.560 is about where the center of lift of a wing is. 00:20:30.560 --> 00:20:33.110 And you want your center of mass also 00:20:33.110 --> 00:20:35.350 called the center of gravity in these situations, 00:20:35.350 --> 00:20:37.890 you want it to be pretty close to the center of lift 00:20:37.890 --> 00:20:41.130 so that their balance in the plane flies nicely. 00:20:41.130 --> 00:20:52.780 So is there a simple way to find the center of mass of something 00:20:52.780 --> 00:20:55.380 like a sailplane? 00:20:55.380 --> 00:20:58.060 So I'm going to drew-- this is exactly how you do it. 00:20:58.060 --> 00:21:01.490 So I'll draw a quick picture of my sailplane here. 00:21:11.302 --> 00:21:13.760 You usually have a little skid or a tail wheel on the back. 00:21:13.760 --> 00:21:17.210 And to find the center of mass you just set them on scales. 00:21:21.270 --> 00:21:24.410 You weigh it. 00:21:24.410 --> 00:21:26.340 You pick a coordinate system. 00:21:26.340 --> 00:21:29.200 Doesn't matter where it is, as long as it's fixed. 00:21:29.200 --> 00:21:32.960 The easiest place is right at the nose of the sailplane. 00:21:32.960 --> 00:21:37.180 So we'll make this x, make this y. 00:21:37.180 --> 00:21:39.080 You take it and you can take a tape measure 00:21:39.080 --> 00:21:45.180 and measure the distance from your reference point 00:21:45.180 --> 00:21:48.200 to the position where the wheel sits on the scales. 00:21:48.200 --> 00:21:50.720 We'll call that L1 here. 00:21:50.720 --> 00:21:53.820 And that's typically about five feet. 00:21:53.820 --> 00:21:58.230 And back here you have another position 00:21:58.230 --> 00:22:02.350 to where you have your second set of scales, that's L2. 00:22:02.350 --> 00:22:05.465 And a typical sailplane that's about 15 feet. 00:22:14.420 --> 00:22:16.247 Apply Newton's law. 00:22:16.247 --> 00:22:17.580 This thing's not going anywhere. 00:22:17.580 --> 00:22:19.910 Sum of the forces in the vertical direction 00:22:19.910 --> 00:22:23.710 is equal to 0. 00:22:23.710 --> 00:22:25.390 Free body diagram. 00:22:25.390 --> 00:22:30.990 Well you have somewhere about here, where the wing is, 00:22:30.990 --> 00:22:33.170 this is your center of mass. 00:22:33.170 --> 00:22:38.880 And you have m total times g downwards there. 00:22:38.880 --> 00:22:42.715 You have two weights pushing on the sailplane. 00:22:42.715 --> 00:22:46.920 A W2 pushing up, holding up the tail. 00:22:46.920 --> 00:22:51.040 And a W1 holding up the main gear. 00:22:51.040 --> 00:22:58.250 And so from the sum of the forces in the y-direction, 00:22:58.250 --> 00:22:59.960 that had better be 0. 00:22:59.960 --> 00:23:14.390 So you know that W1 plus W2 minus mtg is 0. 00:23:14.390 --> 00:23:18.260 And so you find out that the total weight of the sailplane 00:23:18.260 --> 00:23:23.874 is no surprise, the sum of the two weights on the scales. 00:23:23.874 --> 00:23:24.868 AUDIENCE: [INAUDIBLE]. 00:23:27.850 --> 00:23:29.530 PROFESSOR: Yeah. 00:23:29.530 --> 00:23:33.960 Let's do this, m, t, g. 00:23:33.960 --> 00:23:35.480 So the total weight times gravity 00:23:35.480 --> 00:23:38.820 is just the sum of the two readings on the scales. 00:23:38.820 --> 00:23:43.200 And the second piece that you need to do to do this problem 00:23:43.200 --> 00:23:45.440 is, you can have an equation that says, 00:23:45.440 --> 00:23:48.360 the sum of the external torques with respect 00:23:48.360 --> 00:23:52.930 to you're-- through a fixed point O, 00:23:52.930 --> 00:23:56.520 is equal to the mass moment of inertia times the angular 00:23:56.520 --> 00:23:58.530 acceleration, oftentimes written as alpha. 00:23:58.530 --> 00:24:03.560 In this case, that's going to be 0, it's going nowhere. 00:24:03.560 --> 00:24:06.300 So what are the external torques with respect to this point? 00:24:06.300 --> 00:24:08.220 Well we have a right-handed coordinate system. 00:24:08.220 --> 00:24:10.485 You have W1 up, times L1. 00:24:15.080 --> 00:24:23.210 W2 up, times L2 minus W1 plus W2, 00:24:23.210 --> 00:24:29.750 which is the total weight of the sailplane times 00:24:29.750 --> 00:24:35.760 rg, the location of the center of mass. 00:24:35.760 --> 00:24:37.770 That's this distance that we're looking for. 00:24:46.570 --> 00:24:48.860 So we know everything here, except rg. 00:24:48.860 --> 00:24:52.690 So we can solve for rg with respect to O 00:24:52.690 --> 00:25:03.680 and it's simply W1 L1 plus W2 L2 over W1 plus W2. 00:25:03.680 --> 00:25:15.720 And if you run the numbers, typically W1 600 pounds, 00:25:15.720 --> 00:25:23.560 W2 the numbers I've done here is-- might be 40 pounds. 00:25:26.940 --> 00:25:29.590 And I've already said five feet and 15 feet. 00:25:29.590 --> 00:25:31.580 And you work the answer and you come up 00:25:31.580 --> 00:25:40.830 with 5.62 feet, which puts this-- here is the wheel 00:25:40.830 --> 00:25:44.650 and you're a little bit aft of the main gear. 00:25:44.650 --> 00:25:47.240 And that's where you, basically where you want to be. 00:25:47.240 --> 00:25:53.247 That's a real practical use of knowing about centers of mass 00:25:53.247 --> 00:25:54.330 and how to calculate them. 00:26:00.970 --> 00:26:06.080 So that's the second item I wanted to talk about today. 00:26:06.080 --> 00:26:08.370 Essentially a recap of the center of mass. 00:26:08.370 --> 00:26:11.480 And now I want to move on to talking 00:26:11.480 --> 00:26:18.290 about a serious introduction to-- we've 00:26:18.290 --> 00:26:21.380 had the introduction, velocities and accelerations. 00:26:21.380 --> 00:26:24.510 We have to have a way of writing down 00:26:24.510 --> 00:26:35.170 the acceleration of a mass, a point, a dog in a rotating, 00:26:35.170 --> 00:26:38.740 translating, reference frame with the possibility 00:26:38.740 --> 00:26:41.670 that in addition to that, the dog's moving. 00:26:41.670 --> 00:26:44.230 So we want to have equations-- we want to have the ability 00:26:44.230 --> 00:26:47.310 to write down expressions for the velocity 00:26:47.310 --> 00:26:54.110 and acceleration of a mass moving in a translating, 00:26:54.110 --> 00:26:57.690 rotating, reference frame. 00:26:57.690 --> 00:26:58.680 So we've started this. 00:26:58.680 --> 00:27:02.080 We did pretty much did velocities to begin with. 00:27:31.740 --> 00:27:33.080 So here's my inertial frame. 00:27:41.030 --> 00:27:43.800 Call it O or O x, y, z. 00:27:43.800 --> 00:27:46.470 Here's my rigid body out there. 00:27:51.350 --> 00:27:57.200 It has a point a something else, b might be the dog. 00:27:57.200 --> 00:28:00.810 And we've described the position of this 00:28:00.810 --> 00:28:04.940 as the position of this point a, with respect to O. 00:28:04.940 --> 00:28:10.640 And at this point we're going to locate a reference frame 00:28:10.640 --> 00:28:13.700 attached to the rigid body. 00:28:13.700 --> 00:28:15.570 And so it's going to be called a, 00:28:15.570 --> 00:28:19.530 and I'll call it x prime, y prime, z prime. 00:28:19.530 --> 00:28:21.350 It's attached to the rigid body, it 00:28:21.350 --> 00:28:24.230 rotates with the rigid body and its attached 00:28:24.230 --> 00:28:26.340 at some fixed point. 00:28:26.340 --> 00:28:27.860 Now what would oftentimes would be 00:28:27.860 --> 00:28:30.950 a smart choice for that fixed point at point A? 00:28:35.144 --> 00:28:36.076 AUDIENCE: [INAUDIBLE]. 00:28:36.076 --> 00:28:38.492 PROFESSOR: If you're going to write an equation expressing 00:28:38.492 --> 00:28:40.280 the motion of this. 00:28:40.280 --> 00:28:41.832 Where would you make point a? 00:28:41.832 --> 00:28:42.759 [INTERPOSING VOICES] 00:28:42.759 --> 00:28:43.800 AUDIENCE: Center of mass. 00:28:43.800 --> 00:28:45.350 PROFESSOR: Center of mass. 00:28:45.350 --> 00:28:48.320 So very, very often, especially when objects are free 00:28:48.320 --> 00:28:50.820 floating around out there you're going to make smart choices 00:28:50.820 --> 00:28:52.770 and you're going to put this coordinate system 00:28:52.770 --> 00:28:53.936 right on the center of mass. 00:28:53.936 --> 00:28:56.040 But it doesn't have to be, but it can be. 00:28:59.420 --> 00:29:04.390 So we were interested in knowing things 00:29:04.390 --> 00:29:09.590 about the motion of this point in our inertial reference 00:29:09.590 --> 00:29:15.080 frame, in terms of positions of our coordinate system. 00:29:15.080 --> 00:29:23.190 And then also this vector here rd, with respect to a. 00:29:23.190 --> 00:29:29.590 Now last time we came up with expressions for the velocity 00:29:29.590 --> 00:29:35.285 of b with respect to O. 00:29:35.285 --> 00:29:39.320 We said in general it's the velocity of your-- where 00:29:39.320 --> 00:29:41.320 your coordinate system's located. 00:29:41.320 --> 00:29:46.260 The translating-- the velocity of the translating frame 00:29:46.260 --> 00:29:55.780 plus the derivative of rba, time derivative 00:29:55.780 --> 00:30:00.430 of the position as seen from, if you were sitting at a. 00:30:03.360 --> 00:30:06.520 And another way to say that, or this 00:30:06.520 --> 00:30:12.420 is a derivative taken with the rotation rate momentarily set 00:30:12.420 --> 00:30:14.490 equal to 0. 00:30:14.490 --> 00:30:16.370 Another way to think of it. 00:30:16.370 --> 00:30:20.180 Plus a piece that comes from rotation. 00:30:20.180 --> 00:30:24.150 So the rotation with respect to the fixed frame, these 00:30:24.150 --> 00:30:30.010 are all vectors, the rotation with respect 00:30:30.010 --> 00:30:36.720 to the fixed frame, cross product with r, b, a. 00:30:36.720 --> 00:30:38.460 And this-- and we said this is actually 00:30:38.460 --> 00:30:41.590 a general formula for the derivative of-- this piece is 00:30:41.590 --> 00:30:48.840 the derivative of a vector in a frame, in a fixed frame. 00:30:48.840 --> 00:30:51.970 You have two pieces, the derivative as 00:30:51.970 --> 00:30:55.197 seen without rotation plus the contribution that 00:30:55.197 --> 00:30:56.030 comes from rotation. 00:31:05.430 --> 00:31:08.920 When I did this center of mass thing a second ago, 00:31:08.920 --> 00:31:12.350 I just kind of quickly wrote down two time derivatives 00:31:12.350 --> 00:31:13.820 of the position vector. 00:31:13.820 --> 00:31:18.310 There's no omega cross O's in there right? 00:31:18.310 --> 00:31:20.960 Why could I do that? 00:31:20.960 --> 00:31:24.520 This is actually kind of an important distinct point. 00:31:24.520 --> 00:31:29.530 I could do that they didn't say very specifically when I did it 00:31:29.530 --> 00:31:31.920 was an assumption I was making. 00:31:31.920 --> 00:31:34.610 Except for perhaps they drew it. 00:31:34.610 --> 00:31:39.430 This was done in a Cartesian coordinate system. 00:31:39.430 --> 00:31:41.250 And my coordinates were x, y and z 00:31:41.250 --> 00:31:47.471 and the unit vectors were i, j, k and do they move? 00:31:47.471 --> 00:31:47.970 No. 00:31:47.970 --> 00:31:49.178 What's their time derivative? 00:31:49.178 --> 00:31:50.220 [INTERPOSING VOICES] 00:31:50.220 --> 00:31:52.761 PROFESSOR: When you don't-- when the inner vectors don't have 00:31:52.761 --> 00:31:54.811 time derivatives you don't get these terms. 00:31:54.811 --> 00:31:57.060 This is the only term that contributes so I could just 00:31:57.060 --> 00:31:58.390 write that equation. 00:31:58.390 --> 00:32:02.350 But we now have a reference frame attached to a body 00:32:02.350 --> 00:32:04.540 and this reference frame is rotating. 00:32:04.540 --> 00:32:07.940 And that means that the direction of the unit vectors 00:32:07.940 --> 00:32:11.710 attach-- the unit vector attached to x-prime here 00:32:11.710 --> 00:32:13.520 is moving, it's rotating. 00:32:13.520 --> 00:32:15.700 And it's going to have a time derivative. 00:32:15.700 --> 00:32:17.855 So we have to-- and that is given 00:32:17.855 --> 00:32:19.480 and you take those time derivatives you 00:32:19.480 --> 00:32:20.920 get this second piece. 00:32:24.140 --> 00:32:26.330 I'm going to give you the answer in advance. 00:32:26.330 --> 00:32:32.460 The acceleration of b with respect to O I'm 00:32:32.460 --> 00:32:37.460 going to give you the full 3D equation. 00:32:37.460 --> 00:32:41.240 Then we'll go back and see a bit where it comes from. 00:32:41.240 --> 00:32:45.660 So here's-- it's the time derivative of that velocity 00:32:45.660 --> 00:32:55.376 expression with respect to time taken in the inertial frame O, 00:32:55.376 --> 00:32:55.875 x, y, z. 00:33:02.310 --> 00:33:04.911 And am I going to have enough room to get this on? 00:33:04.911 --> 00:33:05.535 It'll be close. 00:33:19.090 --> 00:33:22.280 All right this has several pieces. 00:33:22.280 --> 00:33:28.070 It's got a contribution of the acceleration of a with respect 00:33:28.070 --> 00:33:30.551 to O. That's just the acceleration of this point. 00:33:30.551 --> 00:33:33.050 Has nothing to do with rotation, so it's just a straight out 00:33:33.050 --> 00:33:37.550 acceleration of my translating frame with respect to O. 00:33:37.550 --> 00:33:40.490 That's the first piece. 00:33:40.490 --> 00:33:44.690 The second piece is related-- is the derivative of this guy that 00:33:44.690 --> 00:33:46.620 comes from the derivative of this. 00:33:46.620 --> 00:33:54.010 It's the acceleration of b with respect to a as seen 00:33:54.010 --> 00:33:54.895 in this a frame. 00:34:00.000 --> 00:34:01.500 If you read the Williams book, it's 00:34:01.500 --> 00:34:03.940 called the relative acceleration. 00:34:03.940 --> 00:34:06.732 It's relative to the-- if you were sitting at point A, 00:34:06.732 --> 00:34:08.565 it's what you would see as the acceleration. 00:34:12.770 --> 00:34:25.239 Plus 2 omega cross velocity of b with respect 00:34:25.239 --> 00:34:36.850 to a as seen from a plus omega dot, the derivative 00:34:36.850 --> 00:34:57.070 of the rotation rate, cross rba plus omega cross, 00:34:57.070 --> 00:35:03.590 omega cross r, b, a. 00:35:07.536 --> 00:35:09.110 Kind of daunting right? 00:35:09.110 --> 00:35:11.250 A little messy. 00:35:11.250 --> 00:35:19.740 Basically one, two, three, four, five different terms. 00:35:19.740 --> 00:35:24.130 And you're going to-- and they all have names and meanings. 00:35:24.130 --> 00:35:27.370 And one of the things that will really help you 00:35:27.370 --> 00:35:30.420 is to get familiar, you really need 00:35:30.420 --> 00:35:36.480 to be familiar with the meaning of each one of the terms. 00:35:36.480 --> 00:35:38.690 And it's not terribly difficult. 00:35:38.690 --> 00:35:43.240 This one, just the acceleration of the translating frame. 00:35:43.240 --> 00:35:44.850 So if it's a merry-go-round sitting 00:35:44.850 --> 00:35:47.580 on a train and the train's heading down the track, 00:35:47.580 --> 00:35:50.320 its acceleration of the train. 00:35:50.320 --> 00:35:52.370 Rotating frame is attached to the merry-go-round. 00:35:54.990 --> 00:35:58.300 And if you've got the dog on the merry-go-round 00:35:58.300 --> 00:36:02.990 this is then the acceleration of the dog relative to this, 00:36:02.990 --> 00:36:04.460 the merry-go-round. 00:36:04.460 --> 00:36:07.040 This position of the coordinate system 00:36:07.040 --> 00:36:11.060 attached to the merry-go-round has no rotation in it. 00:36:11.060 --> 00:36:17.110 This is the velocity of that point, the dog, 00:36:17.110 --> 00:36:19.850 as seen from the A frame. 00:36:19.850 --> 00:36:22.100 Again, you have no sense of rotation. 00:36:22.100 --> 00:36:24.250 Rotation is not a part of this. 00:36:24.250 --> 00:36:27.050 Cross product with the rotation rate. 00:36:30.430 --> 00:36:31.780 This is the accelerate. 00:36:31.780 --> 00:36:35.810 This is the angular acceleration cross product with rba. 00:36:35.810 --> 00:36:38.910 Now that's a term-- what does that mean? 00:36:38.910 --> 00:36:44.400 I'm swinging a baseball bat and I'm accelerating this thing. 00:36:44.400 --> 00:36:48.540 Idealize it as just something on a radius accelerating. 00:36:48.540 --> 00:36:52.030 The acceleration of a point out here 00:36:52.030 --> 00:36:55.520 is the radius times the angular acceleration. 00:36:55.520 --> 00:36:57.300 So that's all this term is. 00:36:57.300 --> 00:36:59.870 And it's called the Euler acceleration. 00:36:59.870 --> 00:37:01.702 But it's just simply r theta double dot. 00:37:04.866 --> 00:37:08.190 This, if you multiply it out and just think about units, 00:37:08.190 --> 00:37:10.790 this ends up looking like r omega squared. 00:37:10.790 --> 00:37:13.040 Have you run into that before? 00:37:13.040 --> 00:37:15.290 What's that? 00:37:15.290 --> 00:37:16.618 Common language. 00:37:16.618 --> 00:37:18.630 AUDIENCE: [INAUDIBLE]. 00:37:18.630 --> 00:37:21.010 PROFESSOR: That's as a centrifugal-- centripetal, 00:37:21.010 --> 00:37:23.060 this is centripetal acceleration. 00:37:23.060 --> 00:37:25.600 So this is the centripetal acceleration term, 00:37:25.600 --> 00:37:28.060 that's the Euler acceleration term, 00:37:28.060 --> 00:37:30.025 this is the local acceleration. 00:37:30.025 --> 00:37:32.100 This is the acceleration of your frame. 00:37:32.100 --> 00:37:33.910 This is the strange one. 00:37:33.910 --> 00:37:35.685 This is the Coriolis acceleration. 00:37:40.390 --> 00:37:42.520 And we'll get familiar with it too. 00:37:42.520 --> 00:37:48.520 So that's the full blown 3D acceleration equation. 00:37:48.520 --> 00:37:50.700 And by the way the vector-- the velocity one 00:37:50.700 --> 00:37:52.730 is also perfect 3D. 00:37:52.730 --> 00:38:01.170 Now in this course we won't do much in the way of 3D dynamics 00:38:01.170 --> 00:38:01.860 problems. 00:38:01.860 --> 00:38:02.793 Yes. 00:38:02.793 --> 00:38:08.070 AUDIENCE: Does the point b on the rigid plane move? 00:38:08.070 --> 00:38:10.195 PROFESSOR: Does the-- it may. 00:38:10.195 --> 00:38:19.450 It could be this is the-- an asteroid out there in space 00:38:19.450 --> 00:38:22.540 and you've got-- this is home base 00:38:22.540 --> 00:38:26.580 and that's a guy out there in a space suit running. 00:38:29.650 --> 00:38:31.910 So we want to be able to describe 00:38:31.910 --> 00:38:38.500 the acceleration of that guy as seen from a fixed reference 00:38:38.500 --> 00:38:40.710 frame. 00:38:40.710 --> 00:38:43.720 Now why would we want to know that acceleration? 00:38:43.720 --> 00:38:47.180 Why do we want to know it in a fixed frame? 00:38:47.180 --> 00:38:53.480 Well if you want to calculate the forces on the person. 00:38:53.480 --> 00:38:57.200 Well how much-- what's he have to do with his feet 00:38:57.200 --> 00:38:59.600 to brace himself or whatever? 00:38:59.600 --> 00:39:01.570 What are the actual forces? 00:39:01.570 --> 00:39:04.970 You have to know the acceleration on the person. 00:39:04.970 --> 00:39:08.030 But Newton's laws, in order to say f equals ma, 00:39:08.030 --> 00:39:13.690 Newton's laws have to be applied in inertial reference frames. 00:39:13.690 --> 00:39:16.590 Is this thing out there doing this an a inertial reference 00:39:16.590 --> 00:39:17.571 frame? 00:39:17.571 --> 00:39:18.070 No. 00:39:18.070 --> 00:39:20.140 So you can't calculate the forces 00:39:20.140 --> 00:39:23.530 without having some idea of this inertial frame. 00:39:23.530 --> 00:39:26.750 So this is the way of getting the acceleration 00:39:26.750 --> 00:39:33.690 on-- at a location on a moving, rotating body with respect 00:39:33.690 --> 00:39:36.410 to an inertial frame. 00:39:36.410 --> 00:39:40.160 And with all the terms present. 00:39:40.160 --> 00:39:45.250 Now most discourse has generally has addressed problems 00:39:45.250 --> 00:39:49.740 which are in most textbooks address only planar motion 00:39:49.740 --> 00:39:51.270 problems. 00:39:51.270 --> 00:39:55.530 Planar motion basically means that we can find 00:39:55.530 --> 00:39:59.280 the translations to a plane. 00:39:59.280 --> 00:40:04.860 So imagine an x, and a y, and a z upwards coordinate system 00:40:04.860 --> 00:40:06.450 attached to the top this table. 00:40:06.450 --> 00:40:09.370 And I only allow motions that are around the table. 00:40:09.370 --> 00:40:12.860 And I only allow motions that have a single axis rotation. 00:40:12.860 --> 00:40:14.720 And that's lined up with z. 00:40:14.720 --> 00:40:17.210 Those are essentially planar motion problems. 00:40:17.210 --> 00:40:20.650 And most courses in dynamics, it lists [INAUDIBLE]. 00:40:20.650 --> 00:40:22.290 That's as far as they get. 00:40:22.290 --> 00:40:24.610 And the most of the problems that you'll do 00:40:24.610 --> 00:40:27.670 will be planar motion problems. 00:40:27.670 --> 00:40:33.970 But that equation reduces to the planar motion problem as well. 00:40:33.970 --> 00:40:36.270 We will do a little bit of 3D, because there's 00:40:36.270 --> 00:40:37.941 a class of problems that I really 00:40:37.941 --> 00:40:40.440 think it's important for you to understand that just come up 00:40:40.440 --> 00:40:43.997 all the time that require a little 3D. 00:40:43.997 --> 00:40:45.830 And as you want to have some things going on 00:40:45.830 --> 00:40:47.240 out of the plane, but we'll still 00:40:47.240 --> 00:40:53.420 confine the axis of rotation to a single direction. 00:40:53.420 --> 00:40:53.920 Yeah. 00:40:53.920 --> 00:40:58.330 AUDIENCE: [INAUDIBLE] in a planar, 00:40:58.330 --> 00:41:02.740 but then would you have three to view the freedom? [INAUDIBLE]. 00:41:02.740 --> 00:41:04.430 PROFESSOR: That's a great question 00:41:04.430 --> 00:41:05.880 she said if you had a dog running 00:41:05.880 --> 00:41:08.630 on the merry-go-round how many degrees of freedom do you have? 00:41:08.630 --> 00:41:13.510 So in general rigid bodies, each rigid body, 00:41:13.510 --> 00:41:16.100 each independent rigid body has-- 00:41:16.100 --> 00:41:18.580 you have to describe its location of its center of mass 00:41:18.580 --> 00:41:20.270 and that takes how many coordinates? 00:41:20.270 --> 00:41:23.340 How many coordinates I'll call them. 00:41:23.340 --> 00:41:25.172 AUDIENCE: [INAUDIBLE]. 00:41:25.172 --> 00:41:28.660 PROFESSOR: Well in general three. 00:41:28.660 --> 00:41:35.227 And it can now rotate around three different axes. 00:41:35.227 --> 00:41:36.310 And that takes three more. 00:41:36.310 --> 00:41:39.740 So rigid bodies have six degrees of freedom. 00:41:39.740 --> 00:41:42.620 And any problem when you go to address the problem you 00:41:42.620 --> 00:41:45.460 essentially for a rigid body you start with six. 00:41:45.460 --> 00:41:47.270 And you start applying constraints 00:41:47.270 --> 00:41:50.850 to reduce it down to the number of remaining 00:41:50.850 --> 00:41:52.810 degrees of freedom. 00:41:52.810 --> 00:41:55.380 So if it's confined to a plane and no z-motion 00:41:55.380 --> 00:41:58.250 is allowed one constraint. 00:41:58.250 --> 00:42:01.470 If it is on a plane and it's only 00:42:01.470 --> 00:42:04.640 allowed to rotate about the z-axis that 00:42:04.640 --> 00:42:08.920 means you've constrained its rotation in around y and x. 00:42:08.920 --> 00:42:10.640 So that's two more. 00:42:10.640 --> 00:42:16.560 And so now you're down to three degrees of freedom left, xy 00:42:16.560 --> 00:42:19.890 and a rotation about the z-axis. 00:42:19.890 --> 00:42:21.680 So planar motion problems generally 00:42:21.680 --> 00:42:26.450 have three degrees of freedom. 00:42:26.450 --> 00:42:28.230 But instant-- let's just say we're 00:42:28.230 --> 00:42:30.020 just interested in just something that 00:42:30.020 --> 00:42:33.240 rotates and doesn't translate. 00:42:33.240 --> 00:42:35.610 How many degrees of freedom does that have then? 00:42:35.610 --> 00:42:36.770 Just one. 00:42:36.770 --> 00:42:40.259 x and y are forced not to-- no motion, two more 00:42:40.259 --> 00:42:41.550 constraints you're down to one. 00:42:41.550 --> 00:42:45.712 So lots of problems we do are in fact single degree 00:42:45.712 --> 00:42:46.545 of freedom problems. 00:42:57.190 --> 00:43:01.640 So to do planar motion problems we oftentimes 00:43:01.640 --> 00:43:04.010 use polar coordinates. 00:43:04.010 --> 00:43:08.420 So I'm going to introduce r theta. 00:43:08.420 --> 00:43:10.980 And I'm actually going to call it cylindrical coordinates. 00:43:21.300 --> 00:43:26.298 And cylindrical coordinates then you have an r, a theta and z. 00:43:34.910 --> 00:43:37.900 And let's think about well let's see, I have a demo, 00:43:37.900 --> 00:43:40.540 a little demo here. 00:43:40.540 --> 00:43:45.450 So here's a problem with a single axis of rotation. 00:43:48.740 --> 00:43:54.570 And it's a-- there is a mass out here and just this is a rhyme. 00:43:54.570 --> 00:43:58.260 And so think of this think of this mass 00:43:58.260 --> 00:44:03.350 out here as being a bug walking out this rod. 00:44:03.350 --> 00:44:08.032 And the rod, this thing goes round and round. 00:44:08.032 --> 00:44:09.580 It's not a merry-go-round but it's 00:44:09.580 --> 00:44:12.660 a merry-go-round with a gang plank on it. 00:44:12.660 --> 00:44:14.035 It's going up at an angle and you 00:44:14.035 --> 00:44:16.118 can walk the gang plank while the merry-go-round's 00:44:16.118 --> 00:44:16.690 going around. 00:44:16.690 --> 00:44:19.330 So that's what we got here. 00:44:19.330 --> 00:44:23.540 So this is actually allowed to change position of this mass. 00:44:23.540 --> 00:44:28.900 So how would I describe that with cylindrical coordinates? 00:44:28.900 --> 00:44:34.900 Let me so it's going to take-- one would be a side view. 00:44:38.160 --> 00:44:40.930 So you see your vertical axis here 00:44:40.930 --> 00:44:43.135 and have to have a bearing to hold it in place. 00:44:46.790 --> 00:44:53.970 Here's the arm, here's the bug walking out the arm, 00:44:53.970 --> 00:44:55.480 has some rotation rate. 00:44:58.200 --> 00:45:13.440 Theta dot this is the z-axis and the position of this point 00:45:13.440 --> 00:45:22.280 is described by a r vector, in the r hat 00:45:22.280 --> 00:45:23.800 direction, which I think is the unit 00:45:23.800 --> 00:45:25.630 vectors you're used to using. 00:45:25.630 --> 00:45:30.375 And then this is the z-component in the k hat direction. 00:45:32.920 --> 00:45:45.730 And this vector here would be r of v with respect to what shall 00:45:45.730 --> 00:45:46.230 I call it? 00:45:46.230 --> 00:45:55.800 I'll make this a and over here someplace 00:45:55.800 --> 00:46:07.400 I have a fixed inertial let's see I got to be careful here. 00:46:07.400 --> 00:46:13.620 I want that to be z then I have a y going out here x, y, z 00:46:13.620 --> 00:46:15.180 pointing upwards. 00:46:15.180 --> 00:46:17.480 And my-- this fixed inertial system the unit 00:46:17.480 --> 00:46:25.840 vectors here this would be i hat, j hat and k hat. 00:46:25.840 --> 00:46:30.920 But these-- this rotating system with its unit 00:46:30.920 --> 00:46:35.360 vector little k and this vector they're the same, 00:46:35.360 --> 00:46:38.020 they're parallel. 00:46:38.020 --> 00:46:42.620 But looking down on this, this is my polar coordinate system. 00:46:42.620 --> 00:46:45.560 Now I'm going to look at my top view. 00:46:45.560 --> 00:46:46.870 I will see a projection. 00:46:49.700 --> 00:46:54.340 I'll just see the r, this is rr hat, 00:46:54.340 --> 00:46:57.840 this is my point B. This is theta. 00:47:01.600 --> 00:47:03.640 And I have a unit vector. 00:47:03.640 --> 00:47:08.190 So the unit vector r hat is something-- the unit long 00:47:08.190 --> 00:47:10.870 pointing in this direction. 00:47:10.870 --> 00:47:14.320 And the unit vector in this direction is theta hat. 00:47:14.320 --> 00:47:18.430 And it's perpendicular to that radius. 00:47:18.430 --> 00:47:20.330 So now I have my three unit vectors. 00:47:20.330 --> 00:47:24.530 One pointing in the direction of r, here's 00:47:24.530 --> 00:47:29.920 also my unit vector is just to make sure there's no confusion. 00:47:29.920 --> 00:47:32.218 This unit vector is in this direction. 00:47:35.360 --> 00:47:36.940 K is in that direction. 00:47:36.940 --> 00:47:39.050 Theta is in that direction. 00:47:44.450 --> 00:47:47.080 And over here you still have your-- now here's 00:47:47.080 --> 00:47:54.570 my x, y, z out of the board inertial frame. 00:47:54.570 --> 00:47:59.166 And this-- my inertial frame this might be r, b, o. 00:48:02.770 --> 00:48:04.180 So in my inertial frame. 00:48:04.180 --> 00:48:07.085 I want to know what's going on here. 00:48:07.085 --> 00:48:08.960 I want to be able to calculate the velocities 00:48:08.960 --> 00:48:09.876 and the accelerations. 00:48:16.970 --> 00:48:21.730 So the notation here gets-- can get a little confusing. 00:48:21.730 --> 00:48:26.010 The rbo notation that I've been using all along, 00:48:26.010 --> 00:48:30.880 that's the motion-- that's the position vector describing 00:48:30.880 --> 00:48:34.650 that point in my inertial frame. 00:48:34.650 --> 00:48:39.310 And my-- just lowercase r here, no scrub scripts or anything, 00:48:39.310 --> 00:48:44.850 that's just going to-- that's my polar coordinate r theta 00:48:44.850 --> 00:48:47.500 and z that happened to be in this case, 00:48:47.500 --> 00:48:50.120 this is a rotating frame. 00:48:50.120 --> 00:48:52.190 This is a rotating frame. 00:48:52.190 --> 00:48:58.870 It's the center of this coordinate system's at a. 00:48:58.870 --> 00:49:00.720 But this thing rotates. 00:49:00.720 --> 00:49:03.400 So this is a pretty simplified version 00:49:03.400 --> 00:49:07.760 of this general problem. 00:49:07.760 --> 00:49:11.250 Now because it's simplified, you can actually-- 00:49:11.250 --> 00:49:12.680 it's a lot easier to use. 00:49:12.680 --> 00:49:14.240 Also has some real limitations. 00:49:14.240 --> 00:49:17.230 You can only going just-- there's limited things 00:49:17.230 --> 00:49:18.500 that you can describe with it. 00:49:42.970 --> 00:49:48.070 So let's start by describing velocities in cylindrical 00:49:48.070 --> 00:49:48.570 coordinates. 00:50:23.290 --> 00:50:27.800 Remember this rba is the length of this guy here. 00:50:27.800 --> 00:50:31.710 And it's made up of rr hat zk hat. 00:50:59.900 --> 00:51:01.585 So to express the velocity we have 00:51:01.585 --> 00:51:03.710 to take a time derivative of this r, b, a 00:51:03.710 --> 00:51:07.210 and I'm going to express it in terms of r theta nz. 00:51:07.210 --> 00:51:09.190 And to get acceleration I have to take two time 00:51:09.190 --> 00:51:12.210 derivatives of this, but this is going to be expressed 00:51:12.210 --> 00:51:14.880 in my cylindrical coordinates. 00:51:14.880 --> 00:51:17.420 This is where I'm going. 00:51:17.420 --> 00:51:24.660 And lots of problems-- many, many of these problems 00:51:24.660 --> 00:51:27.240 have fixed axes of rotations and this velocity 00:51:27.240 --> 00:51:29.540 and this acceleration are zero. 00:51:29.540 --> 00:51:31.080 You just drop it out. 00:51:31.080 --> 00:51:34.000 I'm going to do that just to keep this-- make 00:51:34.000 --> 00:51:35.250 this problem a little simpler. 00:51:35.250 --> 00:51:37.890 So we can just focus on these terms. 00:51:37.890 --> 00:51:45.970 So let's just let the there be no translational of this frame. 00:51:45.970 --> 00:51:48.900 And that says that Va with respect 00:51:48.900 --> 00:51:53.679 to O the acceleration of A with respect to O over zero. 00:51:53.679 --> 00:51:55.470 So I want you to just focus on these terms. 00:51:55.470 --> 00:51:57.160 I don't lose anything, I can put these back in 00:51:57.160 --> 00:51:58.200 later if I need them. 00:51:58.200 --> 00:52:00.157 I just don't want to keep carrying them along. 00:52:30.530 --> 00:52:33.220 So I have my-- remember my side view. 00:52:37.440 --> 00:52:47.140 This is r, r hat zk hat that's my point. 00:52:47.140 --> 00:52:48.210 And my top view. 00:53:06.970 --> 00:53:12.990 This is my projection just looking down on it 00:53:12.990 --> 00:53:15.580 what I see is the length r. 00:53:15.580 --> 00:53:20.929 And what I see in my unit vector going that way r direction 00:53:20.929 --> 00:53:21.970 and theta that direction. 00:53:25.290 --> 00:53:32.870 And this is x and the i and a y with a j hat 00:53:32.870 --> 00:53:36.640 vector looking down on it. 00:53:36.640 --> 00:53:42.660 My rotation rate mega with respect to my inertial frame, 00:53:42.660 --> 00:53:46.100 is sum theta dot k hat. 00:53:50.850 --> 00:53:54.350 All right so now let's find the velocity of b with respect 00:53:54.350 --> 00:53:59.640 to O. Well it's 0, no translation, plus-- 00:53:59.640 --> 00:54:06.850 and now I need a time derivative of rb with respect to a. 00:54:06.850 --> 00:54:20.402 But this is then r, b, a is r, r hat plus z k hat. 00:54:20.402 --> 00:54:22.675 And I need the time derivative of that. 00:54:26.360 --> 00:54:41.980 So I get an r dot r hat plus an r r hat dot plus a z dot k. 00:54:41.980 --> 00:54:44.840 So this is the product of two things. 00:54:44.840 --> 00:54:46.760 They're both time dependent. 00:54:46.760 --> 00:54:51.380 So I have to get two pieces, k does not change in direction. 00:54:51.380 --> 00:54:53.330 So it has no time derivative. 00:54:53.330 --> 00:54:55.570 So I only have a z dot k. 00:54:55.570 --> 00:54:57.300 So this is a result of doing this, 00:54:57.300 --> 00:54:59.740 but I now have to figure out what 00:54:59.740 --> 00:55:03.205 is the time derivative of the unit vector in the r direction. 00:55:17.910 --> 00:55:21.260 So when I told-- when we worked out this formula the other day 00:55:21.260 --> 00:55:24.640 for the time derivative of a rotating vector, 00:55:24.640 --> 00:55:27.300 I mostly did it, it was kind of an intuitive argument. 00:55:27.300 --> 00:55:28.820 So on this one occasion I'm going 00:55:28.820 --> 00:55:32.130 to give you an example of actually figuring out 00:55:32.130 --> 00:55:35.390 what the derivative of this rotating vector is. 00:55:35.390 --> 00:55:39.120 And if you go read that kinematics handout 00:55:39.120 --> 00:55:42.920 and it does this in kind of full blown form for-- in general. 00:55:42.920 --> 00:55:45.780 So I'm just going to do it as one example. 00:55:45.780 --> 00:55:51.880 So here's our looking down on this, the projection on the xy 00:55:51.880 --> 00:55:55.100 plane, here's our r-vector. 00:55:55.100 --> 00:55:58.060 And here's this unit vector and it 00:55:58.060 --> 00:56:02.900 starts from-- I have a unit vector starting from a it's 00:56:02.900 --> 00:56:06.370 unit-- it's one long. 00:56:06.370 --> 00:56:08.930 And this is r hat. 00:56:08.930 --> 00:56:15.900 And it's of unit length and it's in this particular direction. 00:56:15.900 --> 00:56:21.050 Now in a little bit it time delta t, it moves. 00:56:21.050 --> 00:56:23.490 It moves to here. 00:56:23.490 --> 00:56:33.480 So this is delta r hat and what direction does it move? 00:56:33.480 --> 00:56:34.920 AUDIENCE: [INAUDIBLE]. 00:56:34.920 --> 00:56:38.830 PROFESSOR: Yeah, it moves in the-- moves in the theta hat 00:56:38.830 --> 00:56:41.090 direction. 00:56:41.090 --> 00:56:45.470 And the amount that it moves is the rotation rate, 00:56:45.470 --> 00:56:49.050 theta dot, delta t. 00:56:52.190 --> 00:56:59.640 So delta r hat, if I solve for this, delta t. 00:56:59.640 --> 00:57:04.930 And this is in the theta hat direction, 00:57:04.930 --> 00:57:09.185 is theta dot theta hat. 00:57:13.280 --> 00:57:17.810 So this is the limit as t, delta t 00:57:17.810 --> 00:57:24.770 goes to 0 you get the derivative of r hat with respect to time. 00:57:24.770 --> 00:57:27.870 Its direction is in the theta hat direction 00:57:27.870 --> 00:57:31.980 and its magnitude is theta dot. 00:57:31.980 --> 00:57:35.740 So that's the time derivative of the unit vector r hat. 00:57:39.390 --> 00:57:40.210 Yeah. 00:57:40.210 --> 00:57:43.556 AUDIENCE: How does that work with units? 00:57:43.556 --> 00:57:47.700 PROFESSOR: How does it work with units? 00:57:47.700 --> 00:57:55.640 What's left out of here is that this is unit length 00:57:55.640 --> 00:57:57.426 and has dimensions. 00:57:57.426 --> 00:58:02.270 It's unit length, one whatever unit system you're working. 00:58:02.270 --> 00:58:06.440 So that is implicitly in here. 00:58:06.440 --> 00:58:11.900 It's one meter theta dot and that theta dot, the delta t, 00:58:11.900 --> 00:58:12.880 the times go away. 00:58:12.880 --> 00:58:17.940 You're left with one meter times the magnitude of theta dot. 00:58:17.940 --> 00:58:19.630 So the distance it actually moves 00:58:19.630 --> 00:58:25.370 is r theta, the r delta theta, delta theta is theta dot delta 00:58:25.370 --> 00:58:27.645 t and the radius happens to be 1. 00:58:31.100 --> 00:58:34.830 So whatever unit system you're working in it's a unit vector. 00:58:34.830 --> 00:58:36.130 Has unit length. 00:58:36.130 --> 00:58:38.330 So its units are buried right there. 00:58:38.330 --> 00:58:40.890 Good question. 00:58:40.890 --> 00:58:45.120 OK so now we know what this is. 00:58:45.120 --> 00:58:50.670 So now we can come back finish our description of the velocity 00:58:50.670 --> 00:58:58.820 of b with respect to a then is r dot r hat 00:58:58.820 --> 00:59:10.795 plus z dot k hat plus theta dot times 00:59:10.795 --> 00:59:17.050 so r times the derivative of the unit vector r, which we just 00:59:17.050 --> 00:59:21.292 figured out is theta dot theta hat times r. 00:59:21.292 --> 00:59:25.620 R theta dot theta hat. 00:59:25.620 --> 00:59:28.620 So that's my velocity of b with respect to a. 00:59:28.620 --> 00:59:31.640 My velocity of B with respect to O all you have to add in 00:59:31.640 --> 00:59:33.720 is the velocity of A with perspective to O, 00:59:33.720 --> 00:59:35.580 which we've let be 0 for now. 00:59:35.580 --> 00:59:40.080 So for the moment this is also d with respect to O. 00:59:40.080 --> 00:59:42.910 But this is the general piece of the velocity 00:59:42.910 --> 00:59:46.294 of b with respect to a in polar cylindrical coordinates. 00:59:50.170 --> 00:59:54.050 Now we could have-- so I've actually worked it out, just 00:59:54.050 --> 00:59:57.090 shown you, just drew the picture and figured out the derivative. 00:59:57.090 --> 00:59:59.850 We could have used that magic formula. 00:59:59.850 --> 01:00:03.920 The formula for the derivative of a vector in a rotating 01:00:03.920 --> 01:00:04.420 frame. 01:00:15.720 --> 01:00:18.330 So I'll just do that quickly to remind you 01:00:18.330 --> 01:00:20.020 how we could have done this. 01:00:20.020 --> 01:00:27.620 rba with respect to time as seen in the O frame. 01:00:30.570 --> 01:00:36.990 Is the partial derivative of rba with respect 01:00:36.990 --> 01:00:47.577 to time as seen in the rotating frame, plus omega cross r, b, 01:00:47.577 --> 01:00:48.076 a. 01:00:55.370 --> 01:01:02.030 This term is that and that. 01:01:04.840 --> 01:01:08.760 The derivative of this rba as seen 01:01:08.760 --> 01:01:11.440 from inside of the rotating frame, 01:01:11.440 --> 01:01:17.090 is just the change in length, this is rba here from the side. 01:01:17.090 --> 01:01:21.389 So the change in length of that vector, the derivative 01:01:21.389 --> 01:01:22.680 of-- the time derivative of it. 01:01:22.680 --> 01:01:28.290 It's the vector sum of the r dot plus z dot. 01:01:28.290 --> 01:01:36.130 So this piece comes from this and this. 01:01:36.130 --> 01:01:42.840 And this piece should-- this one here, it better be this. 01:01:42.840 --> 01:01:44.650 Well this is-- let's figure it out. 01:01:44.650 --> 01:01:49.140 This is omega in the k hat direction, 01:01:49.140 --> 01:01:59.520 cross and rba is r r hat plus z k hat. 01:01:59.520 --> 01:02:03.610 k cross k is 0. 01:02:03.610 --> 01:02:09.460 k cross r theta hat. 01:02:09.460 --> 01:02:15.530 K cross-- k hat cross r hat is theta hat positive. 01:02:15.530 --> 01:02:21.770 r omega theta hat, same thing as r theta dot theta hat. 01:02:21.770 --> 01:02:25.310 So we could have just applied this formula for the derivative 01:02:25.310 --> 01:02:29.730 of a rotating vector and we would 01:02:29.730 --> 01:02:30.854 have gotten the same thing. 01:02:41.230 --> 01:02:43.347 OK just ran out of boards. 01:03:20.180 --> 01:03:22.460 Now a quick little exercise you could do on your own 01:03:22.460 --> 01:03:25.510 is, we're going to need to be able to calculate 01:03:25.510 --> 01:03:28.092 the derivative of theta hat. 01:03:28.092 --> 01:03:29.800 Well just plug it in that little formula. 01:03:36.650 --> 01:03:41.250 And the first term you'll find out the derivative of the theta 01:03:41.250 --> 01:03:48.800 hat, the length doesn't change in time, it's a unit vector. 01:03:48.800 --> 01:03:50.920 So you only have the second piece. 01:03:50.920 --> 01:03:54.800 So it's sum omega cross theta and you're going 01:03:54.800 --> 01:03:57.859 to get minus theta dot r hat. 01:04:17.940 --> 01:04:21.140 So I really want to get here. 01:04:21.140 --> 01:04:23.480 The acceleration of b and O. That's 01:04:23.480 --> 01:04:28.420 the real-- that's the single piece we really 01:04:28.420 --> 01:04:30.210 need to finish the kinematics. 01:04:30.210 --> 01:04:32.080 So we can do most any problems. 01:04:32.080 --> 01:04:35.100 Got to be able to describe the acceleration of a point 01:04:35.100 --> 01:04:37.770 and translating rotating frame. 01:04:37.770 --> 01:04:41.960 And that's going to be the acceleration of a with respect 01:04:41.960 --> 01:04:53.970 to O, plus a time derivative of the velocity of b with respect 01:04:53.970 --> 01:04:55.250 to O. 01:04:55.250 --> 01:04:58.790 We've calculated the velocity, we 01:04:58.790 --> 01:05:05.300 need to be able to essentially carry out this derivative. 01:05:05.300 --> 01:05:10.250 Two time derivatives of the r, b, a, or a single time 01:05:10.250 --> 01:05:12.870 derivative of bva. 01:05:12.870 --> 01:05:18.150 Well we just computed the velocity 01:05:18.150 --> 01:05:20.310 in this-- of this rotating frame and this 01:05:20.310 --> 01:05:22.150 is our final expression. 01:05:22.150 --> 01:05:26.860 So we need to compute the time derivative of that. 01:05:26.860 --> 01:05:32.650 I just-- so it's going to look like r 01:05:32.650 --> 01:05:43.520 dot r hat plus over here a term r theta dot theta hat 01:05:43.520 --> 01:05:50.250 and pardon me for doing this I think 01:05:50.250 --> 01:05:51.740 it'll be cleaner in the end. 01:05:51.740 --> 01:05:55.180 I'm going to start with my z dot k, keep it over here, 01:05:55.180 --> 01:06:01.500 plus r dot r hat plus this term. 01:06:04.440 --> 01:06:06.500 And this is going to take up a lot room. 01:06:06.500 --> 01:06:08.600 Just spread out this way so it-- 01:06:17.000 --> 01:06:19.410 So let's just-- I'm going to write down where this comes 01:06:19.410 --> 01:06:21.860 out, this is a little tedious, but then you'll 01:06:21.860 --> 01:06:26.190 have seen it once hopefully believe that it really works. 01:06:31.410 --> 01:06:34.350 So these terms, this first term here just 01:06:34.350 --> 01:06:36.650 gives you z double dot. 01:06:36.650 --> 01:06:37.770 So let's write her down. 01:06:37.770 --> 01:06:44.870 So just z double dot k hat time derivative of this, 01:06:44.870 --> 01:06:50.780 plus an r double dot r hat, but now I 01:06:50.780 --> 01:06:53.740 have to take-- do it, flip it and do the other side of it. 01:06:53.740 --> 01:07:05.460 So I get my-- how do I want to do this? 01:07:15.909 --> 01:07:16.659 Yeah, I like this. 01:07:26.730 --> 01:07:29.826 So this term kept-- leads to this. 01:07:29.826 --> 01:07:35.270 This term brings you to here. 01:07:35.270 --> 01:07:50.410 This term, you get r dot times theta dot theta hat plus an r 01:07:50.410 --> 01:07:56.660 and now you need to take a time derivative of theta dot theta 01:07:56.660 --> 01:07:58.550 hat. 01:07:58.550 --> 01:07:59.810 So that's going to expand. 01:07:59.810 --> 01:08:02.080 So this brings you to here. 01:08:28.680 --> 01:08:33.399 Now we've done this derivative, so we can put it in. 01:08:33.399 --> 01:08:59.930 So this gives us this term over here, 01:08:59.930 --> 01:09:01.449 so let's keep adding these up. 01:09:11.520 --> 01:09:15.319 Notice this gives me an r dot theta dot theta hat. 01:09:15.319 --> 01:09:20.399 This gives me an r dot theta dot theta hat. 01:09:23.050 --> 01:09:24.330 Two identical terms. 01:09:28.680 --> 01:09:38.649 Now this term gives me an r theta double dot theta hat. 01:09:38.649 --> 01:09:40.340 And now this-- now we need to take 01:09:40.340 --> 01:09:43.020 the derivative of theta hat and that gives you 01:09:43.020 --> 01:09:46.515 minus theta dot r hat. 01:09:46.515 --> 01:09:59.980 So you get a minus r but multiplied by theta dot again 01:09:59.980 --> 01:10:04.220 squared r hat. 01:10:04.220 --> 01:10:05.820 So I think we're about there. 01:10:05.820 --> 01:10:07.820 We're going to start collecting things together. 01:10:23.220 --> 01:10:30.930 Now we have a two r dot theta dot 01:10:30.930 --> 01:10:46.750 theta hat plus an r theta double dot theta hat, minus r theta 01:10:46.750 --> 01:10:49.780 dot squared r half. 01:10:49.780 --> 01:10:51.880 So these things, these derivatives, 01:10:51.880 --> 01:10:55.730 just all kind of flowed down and led to more terms. 01:10:58.760 --> 01:11:06.140 But now if we compare-- that we get one, two, three, four, 01:11:06.140 --> 01:11:08.960 five, I clump these together. 01:11:08.960 --> 01:11:14.250 This is the change in length of the r vector. 01:11:14.250 --> 01:11:19.500 Stretch in the position has a z component and a r component. 01:11:19.500 --> 01:11:23.630 This is the movement of the coordinate system if it moves. 01:11:23.630 --> 01:11:26.900 This is the Coriolis term. 01:11:26.900 --> 01:11:30.470 This is the Euler acceleration and this is 01:11:30.470 --> 01:11:33.350 the centripetal acceleration. 01:11:33.350 --> 01:11:36.040 So this is the-- what happens when 01:11:36.040 --> 01:11:41.700 you start with that messy vector thing 01:11:41.700 --> 01:11:47.320 and apply it, restrict it to a cylindrical coordinate problem, 01:11:47.320 --> 01:11:50.400 which is basically planar motion. 01:11:50.400 --> 01:11:53.470 But you allow some things in the z direction only. 01:11:53.470 --> 01:11:58.090 So polar coordinates is a more limited form of that, 01:11:58.090 --> 01:12:01.290 but it's-- every term comes back, 01:12:01.290 --> 01:12:03.270 every term is still in it. 01:12:03.270 --> 01:12:07.950 So acceleration of the moving coordinate 01:12:07.950 --> 01:12:11.080 system, change of the length of the position 01:12:11.080 --> 01:12:14.580 vector in the moving coordinate system, 01:12:14.580 --> 01:12:19.610 the Coriolis term the Euler acceleration term that angular 01:12:19.610 --> 01:12:25.490 acceleration speedup and finally the centripetal term. 01:12:25.490 --> 01:12:29.030 Now the way you go about solving problems, 01:12:29.030 --> 01:12:32.490 usins-- doing problems in polar coordinates. 01:12:32.490 --> 01:12:36.110 So now you're asked to find an equation of motion. 01:12:36.110 --> 01:12:40.560 This is an expression for the acceleration 01:12:40.560 --> 01:12:42.770 of whatever it is you're trying to describe 01:12:42.770 --> 01:12:46.370 in an inertial frame. 01:12:46.370 --> 01:12:51.690 So that when you say-- you can now say f equals ma. 01:12:51.690 --> 01:12:53.700 If you know the-- sometimes you're 01:12:53.700 --> 01:12:55.550 given the forces in problems and your asked 01:12:55.550 --> 01:12:59.370 to find the accelerations. 01:12:59.370 --> 01:13:01.140 But what if you're given the acceleration 01:13:01.140 --> 01:13:03.610 and you're asked to find the force? 01:13:03.610 --> 01:13:06.566 All right, I give you the simplest problem of this kind. 01:13:09.810 --> 01:13:13.030 What's the tension in the string? 01:13:13.030 --> 01:13:17.517 Well if I know that-- so I just say, 01:13:17.517 --> 01:13:19.850 the way you do these problems is how many things can you 01:13:19.850 --> 01:13:20.480 eliminate? 01:13:20.480 --> 01:13:22.710 Well I wasn't moving, this term goes away. 01:13:22.710 --> 01:13:23.930 That's zero. 01:13:23.930 --> 01:13:26.370 Z wasn't involved, it's not changing, 01:13:26.370 --> 01:13:30.259 it's just constant angular rotation, that term is 0. 01:13:30.259 --> 01:13:32.300 What's the change in length of the string while I 01:13:32.300 --> 01:13:33.660 was doing it? 01:13:33.660 --> 01:13:36.410 Now if that term's 0 we're getting easier all time. 01:13:36.410 --> 01:13:41.860 How far-- how fast was the length getting longer? 01:13:41.860 --> 01:13:43.180 That terms gone away. 01:13:43.180 --> 01:13:46.432 Was I speeding up, or slowing down, or constant speed? 01:13:46.432 --> 01:13:47.890 Well we'll say it's constant speed, 01:13:47.890 --> 01:13:50.060 ooh this problem's getting easier all the time. 01:13:50.060 --> 01:13:51.930 I'm down to one term. 01:13:51.930 --> 01:13:53.090 f equals ma. 01:14:03.730 --> 01:14:10.960 Minus r theta dot squared r hat. 01:14:10.960 --> 01:14:15.750 So the force that I must have been applying to the string 01:14:15.750 --> 01:14:18.690 was in the minus r hat direction and had 01:14:18.690 --> 01:14:22.380 magnitude mr omega squared. 01:14:25.400 --> 01:14:27.010 So that's actually all there is to it. 01:14:27.010 --> 01:14:29.830 We're using polar coordinates and cylindrical coordinates 01:14:29.830 --> 01:14:33.210 to do second law problems. 01:14:33.210 --> 01:14:35.410 So there's a couple of problems that you're 01:14:35.410 --> 01:14:37.770 doing these kind of things on the homework 01:14:37.770 --> 01:14:40.430 set that's being put out today. 01:14:40.430 --> 01:14:44.330 So give them a try. 01:14:44.330 --> 01:14:45.827 Have a good weekend.