WEBVTT 00:00:00.100 --> 00:00:02.450 The following content is provided under a Creative 00:00:02.450 --> 00:00:03.830 Commons license. 00:00:03.830 --> 00:00:06.070 Your support will help MIT OpenCourseWare 00:00:06.070 --> 00:00:10.170 continue to offer high quality educational resources for free. 00:00:10.170 --> 00:00:12.710 To make a donation or to view additional materials 00:00:12.710 --> 00:00:16.620 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.620 --> 00:00:17.325 at ocw.mit.edu. 00:00:21.156 --> 00:00:23.670 PROFESSOR: In a minute, we're going 00:00:23.670 --> 00:00:26.430 to go through this exercise of key concepts for the week. 00:00:26.430 --> 00:00:30.220 But for you early birds, I'll offer, 00:00:30.220 --> 00:00:34.070 and we may get to some of this. 00:00:34.070 --> 00:00:37.130 Is there just something from the weak or the problem set? 00:00:37.130 --> 00:00:39.820 Just some topic that's still bugging you 00:00:39.820 --> 00:00:41.420 that you just don't get and you'd 00:00:41.420 --> 00:00:44.170 like me to put on a little list and I might get to it? 00:00:47.544 --> 00:00:50.436 AUDIENCE: So if we solve for velocity, 00:00:50.436 --> 00:00:51.882 is it enough to just take the time 00:00:51.882 --> 00:00:53.810 derivative of that velocity. 00:00:53.810 --> 00:00:55.410 PROFESSOR: Yes. 00:00:55.410 --> 00:01:00.770 So if you have all of the terms, the full velocity 00:01:00.770 --> 00:01:05.110 expression in vector notation, and you 00:01:05.110 --> 00:01:08.520 want to know the acceleration of that point, 00:01:08.520 --> 00:01:12.560 just take the derivatives taking care of all the rotating unit 00:01:12.560 --> 00:01:13.929 vectors and all that stuff. 00:01:13.929 --> 00:01:14.429 Yeah? 00:01:14.429 --> 00:01:16.720 AUDIENCE: When you take the derivative of the velocity, 00:01:16.720 --> 00:01:20.057 where does the Coriolis term drop from? 00:01:20.057 --> 00:01:21.790 PROFESSOR: Where does it fall out of? 00:01:21.790 --> 00:01:24.380 AUDIENCE: Yeah. 00:01:24.380 --> 00:01:26.410 PROFESSOR: I can't answer it in words. 00:01:26.410 --> 00:01:28.070 I can't remember exactly. 00:01:28.070 --> 00:01:32.610 But Coriolis term-- curiously, it 00:01:32.610 --> 00:01:36.370 comes from two different places. 00:01:36.370 --> 00:01:42.380 It's two omega v rel kind of term. 00:01:42.380 --> 00:01:45.720 And actually, it comes from two different places 00:01:45.720 --> 00:01:48.916 when you grind out the derivatives of the velocity. 00:01:48.916 --> 00:01:49.862 AUDIENCE: [INAUDIBLE]. 00:01:54.592 --> 00:02:01.930 PROFESSOR: So in terms of using the general-- so I remember. 00:02:01.930 --> 00:02:03.980 Now what I want you to do and, partly, 00:02:03.980 --> 00:02:07.820 what this session is going to do is 00:02:07.820 --> 00:02:09.930 get you to use the acceleration formulas enough 00:02:09.930 --> 00:02:12.380 that you have those just committed to memory. 00:02:12.380 --> 00:02:13.620 Now we're going to let you have crib sheets when 00:02:13.620 --> 00:02:14.450 you go into quizzes. 00:02:14.450 --> 00:02:16.658 And I would always have the velocity and acceleration 00:02:16.658 --> 00:02:21.320 formulas in polar coordinates and in full vector form. 00:02:21.320 --> 00:02:23.260 If you identify each of the terms correctly, 00:02:23.260 --> 00:02:26.880 that formula will work just fine. 00:02:26.880 --> 00:02:29.020 Getting the right components of rotation rates 00:02:29.020 --> 00:02:30.560 can be a little tricky, as you might 00:02:30.560 --> 00:02:34.920 have found in that homework problem from this week. 00:02:34.920 --> 00:02:39.440 In that time derivative of a rotating vector, 00:02:39.440 --> 00:02:41.360 the omega cross the vector. 00:02:41.360 --> 00:02:43.250 What omega is that? 00:02:43.250 --> 00:02:44.390 That's tricky. 00:02:44.390 --> 00:02:47.170 That you have to be careful with. 00:02:47.170 --> 00:02:48.537 But really remember. 00:02:48.537 --> 00:02:50.870 If you remember the velocity and acceleration equations, 00:02:50.870 --> 00:02:52.050 you can trust them. 00:02:52.050 --> 00:02:54.490 They will work. 00:02:54.490 --> 00:02:58.270 So OK, we can start now. 00:02:58.270 --> 00:03:02.015 And I'll go back to here. 00:03:06.330 --> 00:03:10.440 So take that minute, like we did last time. 00:03:10.440 --> 00:03:12.700 Write down on a piece of paper two, three, four. 00:03:12.700 --> 00:03:15.200 What you think are the key concepts 00:03:15.200 --> 00:03:18.030 that were important in the past week. 00:03:18.030 --> 00:03:22.290 By important, you need to know them to do a good job 00:03:22.290 --> 00:03:24.340 on the quiz coming up. 00:03:24.340 --> 00:03:26.750 Important concepts. 00:03:26.750 --> 00:03:29.546 OK, let's make a list. 00:03:29.546 --> 00:03:31.212 AUDIENCE: Did you say polar coordinates? 00:03:31.212 --> 00:03:33.296 PROFESSOR: Polar coordinates. 00:03:33.296 --> 00:03:36.340 And I'm going to generalize that to call 00:03:36.340 --> 00:03:47.050 it choosing coordinate systems. 00:03:50.220 --> 00:03:52.030 There's an art to that. 00:03:52.030 --> 00:03:54.639 So when you to use them. 00:03:54.639 --> 00:03:55.930 It's [INAUDIBLE] of that, yeah. 00:03:55.930 --> 00:03:58.305 AUDIENCE: The equation for some of the parts? 00:03:58.305 --> 00:03:59.400 PROFESSOR: Oh, yeah. 00:03:59.400 --> 00:04:01.720 The torque equation. 00:04:01.720 --> 00:04:08.897 Some of the external torques vectors is dh/dt. 00:04:13.470 --> 00:04:17.144 And torques are always with respect to some point. 00:04:17.144 --> 00:04:20.690 So dh/dt. 00:04:20.690 --> 00:04:24.680 Angular momentum is with respect to some point. 00:04:24.680 --> 00:04:27.220 And there's a second term to this, though. 00:04:27.220 --> 00:04:33.580 And it is velocity of the point in the inertial frame 00:04:33.580 --> 00:04:37.570 cross the linear momentum in the inertial frame. 00:04:40.110 --> 00:04:42.020 That one, you haven't had to use much yet. 00:04:42.020 --> 00:04:47.561 But that's one of the two really important Newton law kind 00:04:47.561 --> 00:04:49.060 of things that we use in the course. 00:04:49.060 --> 00:04:51.350 Sum of forces equals mass times acceleration. 00:04:51.350 --> 00:04:53.800 Sum of torques equals that. 00:04:53.800 --> 00:04:55.574 OK, I have another one. 00:04:55.574 --> 00:04:57.562 AUDIENCE: Derivative of rotating unit vectors. 00:04:57.562 --> 00:04:59.550 PROFESSOR: Derivative rotating vectors. 00:05:15.480 --> 00:05:19.260 And taking derivatives rotating vectors always boils down to, 00:05:19.260 --> 00:05:22.040 eventually, it's down to you got to do the unit vector. 00:05:22.040 --> 00:05:23.130 So that's part of it. 00:05:23.130 --> 00:05:24.046 How about another one? 00:05:27.160 --> 00:05:29.080 AUDIENCE: [INAUDIBLE]. 00:05:29.080 --> 00:05:34.020 PROFESSOR: OK, so I'll generalize that to, 00:05:34.020 --> 00:05:45.740 essentially, being able to find equations of motion. 00:05:45.740 --> 00:06:05.410 So from sum of forces and-- all right. 00:06:05.410 --> 00:06:08.580 OK, anything else? 00:06:15.460 --> 00:06:26.615 I'll tell you one really important one that you've been. 00:06:26.615 --> 00:06:29.930 Maybe you think we did it last week. 00:06:29.930 --> 00:06:36.620 I'd say being able to find accelerations 00:06:36.620 --> 00:06:38.500 and moving coordinate systems. 00:06:38.500 --> 00:06:40.410 We did mostly velocities last week. 00:06:40.410 --> 00:06:43.650 But finding velocities and accelerations 00:06:43.650 --> 00:06:45.555 in rotating and moving frames. 00:07:05.240 --> 00:07:07.540 OK, so that's a pretty good list. 00:07:07.540 --> 00:07:10.370 [INAUDIBLE] ask me what I thought the important things 00:07:10.370 --> 00:07:11.350 from the week were? 00:07:15.090 --> 00:07:17.065 That hits the important stuff. 00:07:17.065 --> 00:07:17.565 OK. 00:07:22.890 --> 00:07:26.140 What we were talking about as people were arriving is 00:07:26.140 --> 00:07:28.410 and I might not get this today if we run out of time. 00:07:28.410 --> 00:07:30.243 But does anybody just got a burning question 00:07:30.243 --> 00:07:33.902 about some concept that just didn't work out for you 00:07:33.902 --> 00:07:36.310 or a problem set that came up? 00:07:36.310 --> 00:07:36.810 Yeah? 00:07:36.810 --> 00:07:39.792 AUDIENCE: The derivatives for [INAUDIBLE] vectors 00:07:39.792 --> 00:07:43.271 and [INAUDIBLE]. 00:07:43.271 --> 00:07:47.831 PROFESSOR: So that's really the derivative of rotation rate 00:07:47.831 --> 00:07:48.330 vector. 00:07:48.330 --> 00:07:48.430 AUDIENCE: Yeah. 00:07:48.430 --> 00:07:49.840 PROFESSOR: That was, kind of, a nasty problem 00:07:49.840 --> 00:07:51.280 the first time you hit that. 00:07:51.280 --> 00:07:54.300 That was new, right? 00:07:54.300 --> 00:07:55.510 Let me write that one down. 00:08:06.849 --> 00:08:08.920 And rotating frames and all that. 00:08:08.920 --> 00:08:09.420 I get it. 00:08:09.420 --> 00:08:10.747 Does somebody else have another one? 00:08:10.747 --> 00:08:12.496 AUDIENCE: [INAUDIBLE] of that. [INAUDIBLE] 00:08:12.496 --> 00:08:15.494 just like which omega is it when you're looking [INAUDIBLE]. 00:08:15.494 --> 00:08:16.202 PROFESSOR: Right. 00:08:16.202 --> 00:08:17.150 Oh, yeah. 00:08:17.150 --> 00:08:18.100 Right. 00:08:18.100 --> 00:08:22.300 OK, anybody else a burning question? 00:08:22.300 --> 00:08:24.160 I think I'm actually going to start there. 00:08:24.160 --> 00:08:28.040 We're going to spend just a minute on this one. 00:08:28.040 --> 00:08:37.039 And as an example problem yesterday in the lecture, 00:08:37.039 --> 00:08:38.450 I basically did the same. 00:08:38.450 --> 00:08:40.970 I had a rotor here. 00:08:40.970 --> 00:08:43.789 It could have a disk on it just to make it clear. 00:08:43.789 --> 00:08:47.360 So this thing is rotating at some omega 2. 00:08:47.360 --> 00:08:53.790 And it was on a merry go round going at some omega 1. 00:08:53.790 --> 00:08:56.400 And I'll pick at coordinate system here. 00:08:56.400 --> 00:09:05.020 I think it was yxz that I wrote in the lecture notes. 00:09:05.020 --> 00:09:06.440 And that's the rotating one. 00:09:06.440 --> 00:09:09.660 So you have an o and an a here. 00:09:09.660 --> 00:09:17.875 And you have a fixed frame that but the y 00:09:17.875 --> 00:09:19.980 ones, the little ones, attach to the platform. 00:09:19.980 --> 00:09:21.330 It's going around. 00:09:21.330 --> 00:09:22.410 OK? 00:09:22.410 --> 00:09:29.060 So the rotation rate of the platform here is what? 00:09:34.370 --> 00:09:38.670 In magnitude and unit vector. 00:09:38.670 --> 00:09:40.470 Pardon? 00:09:40.470 --> 00:09:43.750 Mega 1 k hat. 00:09:43.750 --> 00:09:47.760 And does it matter which k because they're parallel. 00:09:47.760 --> 00:09:49.460 So the little k and the big k. 00:09:49.460 --> 00:10:01.190 OK, what's the rotation rate of this shift, 00:10:01.190 --> 00:10:03.740 in terms of a magnitude and a unit vector? 00:10:03.740 --> 00:10:07.324 AUDIENCE: Omega 2 lower case y dot. 00:10:07.324 --> 00:10:08.282 PROFESSOR: OK, omega 2. 00:10:08.282 --> 00:10:14.310 And I'll call it at j1 here. 00:10:14.310 --> 00:10:16.710 It's hard to distinguish my handwriting on the board 00:10:16.710 --> 00:10:19.050 between uppers lowers. 00:10:19.050 --> 00:10:21.590 So with z1's, x1's. 00:10:21.590 --> 00:10:24.565 OK, so it's in the J1 direction, right? 00:10:24.565 --> 00:10:26.440 And I've just intentionally made it positive. 00:10:26.440 --> 00:10:28.810 If it were in the other direction, have a minus there. 00:10:28.810 --> 00:10:37.290 OK, so now at the end, the problem 00:10:37.290 --> 00:10:43.520 asked for the time derivative of the rotation 00:10:43.520 --> 00:10:50.320 rate of the spinning. 00:10:50.320 --> 00:10:52.510 The problem that in the homework, 00:10:52.510 --> 00:10:55.000 this thing was actually inclined, basically. 00:10:59.990 --> 00:11:09.090 It's easy to just say the total rotation rate of the shaft. 00:11:09.090 --> 00:11:10.590 Let's get that settled first. 00:11:10.590 --> 00:11:15.770 In the fixed inertial frame, the total rotation rate of this 00:11:15.770 --> 00:11:17.102 is what? 00:11:17.102 --> 00:11:19.026 AUDIENCE: [INAUDIBLE]. 00:11:19.026 --> 00:11:21.200 PROFESSOR: Yeah, and you can sum rotation rates. 00:11:21.200 --> 00:11:30.100 So it's omega 1 k hat plus omega 2 j1. 00:11:30.100 --> 00:11:35.540 And I want to take the time derivative of this omega chafed 00:11:35.540 --> 00:11:36.040 total. 00:11:39.150 --> 00:11:43.430 So it's d by dt of these things. 00:11:43.430 --> 00:11:45.210 So I can start to do that. 00:11:45.210 --> 00:11:48.990 So d by by dt is, in this case, is k change a direction. 00:11:48.990 --> 00:11:50.260 No problem. 00:11:50.260 --> 00:11:52.610 Could that change? 00:11:52.610 --> 00:11:54.230 Could it? 00:11:54.230 --> 00:11:55.284 Yeah, sure. 00:11:55.284 --> 00:11:56.950 I mean, it could be accelerating, right. 00:11:56.950 --> 00:11:58.990 If it is, then you get a term here 00:11:58.990 --> 00:12:04.530 that comes from this one that would be an omega 1 dot k hat 00:12:04.530 --> 00:12:05.030 plus. 00:12:05.030 --> 00:12:08.036 And now we need to take the derivative of this piece. 00:12:08.036 --> 00:12:09.012 All right. 00:12:11.940 --> 00:12:14.600 And that's now a rotating vector. 00:12:14.600 --> 00:12:18.040 And we've said that the derivative of any rotating 00:12:18.040 --> 00:12:28.180 vector dq dt is the derivative of t as if the rotation is 0. 00:12:28.180 --> 00:12:32.200 It's its magnitude increasing, not its direction changing. 00:12:32.200 --> 00:12:35.670 Plus omega cross q. 00:12:35.670 --> 00:12:38.790 And I've intentionally left sub-scripts and server scripts 00:12:38.790 --> 00:12:40.540 off here because that's the issue here. 00:12:40.540 --> 00:12:44.270 Which ones go where? 00:12:44.270 --> 00:12:47.650 All right, so this is what we need to apply 00:12:47.650 --> 00:12:50.430 to taking that derivative. 00:12:50.430 --> 00:12:54.790 We took this derivative and found this term 00:12:54.790 --> 00:12:59.370 was what gave us the omega 1.k. 00:12:59.370 --> 00:13:04.687 And when taking derivative of this, what was the answer here? 00:13:04.687 --> 00:13:05.312 AUDIENCE: Zero. 00:13:05.312 --> 00:13:08.649 PROFESSOR: Zero because the rotation rate is omega 1. 00:13:08.649 --> 00:13:09.440 You're crossing it. 00:13:09.440 --> 00:13:11.785 Well omega 1k crossed with omega 1k. 00:13:11.785 --> 00:13:12.760 That goes with zero. 00:13:12.760 --> 00:13:14.520 There's no change in direction. 00:13:14.520 --> 00:13:16.680 So we applied this formula once to this. 00:13:16.680 --> 00:13:19.140 Now we're going to apply this formula here. 00:13:19.140 --> 00:13:22.910 So I'll do it this way. 00:13:22.910 --> 00:13:24.170 So this is term one. 00:13:29.640 --> 00:13:32.250 Term one gave us this. 00:13:32.250 --> 00:13:34.560 Term two is going to give us this. 00:13:34.560 --> 00:13:37.820 So take this. 00:13:37.820 --> 00:13:42.270 Take the derivative of-- where we go here-- this guy. 00:13:44.890 --> 00:13:46.430 Now when I say this is the-- it's 00:13:46.430 --> 00:13:49.520 the partial derivative of this thing, 00:13:49.520 --> 00:13:53.990 ignoring the contribution from the change of direction. 00:13:53.990 --> 00:13:56.610 So it's as if some rotation were zero. 00:13:56.610 --> 00:13:57.785 Which rotation? 00:13:57.785 --> 00:13:58.284 Right. 00:13:58.284 --> 00:13:59.188 AUDIENCE: Omega 1? 00:13:59.188 --> 00:14:01.090 PROFESSOR: Omega 1. 00:14:01.090 --> 00:14:02.350 Good. 00:14:02.350 --> 00:14:07.538 So what is the answer to this piece of that derivative? 00:14:07.538 --> 00:14:08.442 AUDIENCE: Omega 2. 00:14:08.442 --> 00:14:14.980 PROFESSOR: Omega 2 dot what direction? 00:14:14.980 --> 00:14:16.110 Same direction, right. 00:14:16.110 --> 00:14:18.300 Now we want to apply this piece. 00:14:18.300 --> 00:14:22.050 So now the question comes down plus some omega 00:14:22.050 --> 00:14:23.622 cross with the q. 00:14:23.622 --> 00:14:25.600 And what's the q in this case? 00:14:30.710 --> 00:14:33.270 Two. 00:14:33.270 --> 00:14:36.620 So this is omega 2 j1 hat. 00:14:36.620 --> 00:14:38.560 That's coming from here. 00:14:38.560 --> 00:14:40.410 That's what this piece is. 00:14:40.410 --> 00:14:43.200 So now we need what's crossed with it. 00:14:43.200 --> 00:14:45.042 AUDIENCE: [INAUDIBLE]. 00:14:45.042 --> 00:14:45.625 PROFESSOR: Hm? 00:14:45.625 --> 00:14:47.080 AUDIENCE: Omega [INAUDIBLE]. 00:14:57.415 --> 00:15:03.290 PROFESSOR: I'm trying to think of a clear way to help you. 00:15:03.290 --> 00:15:07.780 What is causing its direction to change? 00:15:07.780 --> 00:15:10.500 AUDIENCE: [INAUDIBLE]. 00:15:10.500 --> 00:15:11.470 PROFESSOR: Right. 00:15:11.470 --> 00:15:13.130 That's the one you care about. 00:15:13.130 --> 00:15:15.500 What's causing its direction to change? 00:15:15.500 --> 00:15:18.320 What's causing this thing to come up? 00:15:18.320 --> 00:15:20.920 And its direction is changing because the platforms 00:15:20.920 --> 00:15:22.040 are moving. 00:15:22.040 --> 00:15:25.110 So it's omega 1. 00:15:25.110 --> 00:15:27.740 K cross j. 00:15:27.740 --> 00:15:31.323 So k cross j gives you? 00:15:31.323 --> 00:15:33.207 AUDIENCE: [INAUDIBLE]. 00:15:33.207 --> 00:15:39.320 PROFESSOR: K cross j negative i 1 in the rotating frame, right. 00:15:39.320 --> 00:15:40.650 So this whole thing. 00:15:40.650 --> 00:15:50.180 Omega 1 dot k plus omega 2 dot j1 plus. 00:15:50.180 --> 00:15:51.680 No, this is a minus, actually. 00:15:51.680 --> 00:15:55.160 We said, minus omega 1. 00:15:55.160 --> 00:15:56.320 Omega 2. 00:15:56.320 --> 00:15:58.690 K cross j is i 1 hat. 00:16:02.402 --> 00:16:03.330 OK? 00:16:03.330 --> 00:16:07.360 And finally, how do you deal with the fact 00:16:07.360 --> 00:16:11.760 if this rotor had been on an inclined angle? 00:16:11.760 --> 00:16:15.740 Some phi, which now this is the exact problem 00:16:15.740 --> 00:16:18.570 that was on homework. 00:16:18.570 --> 00:16:23.630 So propose a way of attacking that. 00:16:23.630 --> 00:16:26.984 Changes this problem just a little bit. 00:16:26.984 --> 00:16:29.604 AUDIENCE: Can you change the reference frame [INAUDIBLE]? 00:16:29.604 --> 00:16:31.770 PROFESSOR: No, you don't change the reference frame. 00:16:31.770 --> 00:16:33.330 You still have a coordinate system. 00:16:35.914 --> 00:16:37.330 Work with the coordinates you got. 00:16:37.330 --> 00:16:40.510 AUDIENCE: [INAUDIBLE] to its components? 00:16:40.510 --> 00:16:44.672 PROFESSOR: In which reference frame? 00:16:44.672 --> 00:16:45.604 AUDIENCE: [INAUDIBLE]. 00:16:45.604 --> 00:16:47.090 PROFESSOR: Yeah, the moving [? one. ?] 00:16:47.090 --> 00:16:48.798 That makes it easy because this thing now 00:16:48.798 --> 00:16:53.260 has a vector omega 2 like that. 00:16:53.260 --> 00:16:55.310 And you can break it up into a piece like that 00:16:55.310 --> 00:16:56.660 and a piece like that. 00:16:56.660 --> 00:16:59.590 So now this is phi. 00:16:59.590 --> 00:17:03.480 Omega 2 cosine phi is the side, right. 00:17:03.480 --> 00:17:04.900 Omega 2 sine phi is that side. 00:17:04.900 --> 00:17:07.369 What direction is it? 00:17:07.369 --> 00:17:09.069 What's its unit vector direction? 00:17:09.069 --> 00:17:10.020 This piece. 00:17:10.020 --> 00:17:10.740 AUDIENCE: K? 00:17:10.740 --> 00:17:12.099 PROFESSOR: K, OK. 00:17:12.099 --> 00:17:18.191 And we're going to end up crossing it with this term, 00:17:18.191 --> 00:17:18.690 right. 00:17:18.690 --> 00:17:21.540 So k cross k. 00:17:21.540 --> 00:17:22.160 Nothing. 00:17:22.160 --> 00:17:25.250 So in fact, this bit doesn't change in direction 00:17:25.250 --> 00:17:26.020 as it's rotating. 00:17:26.020 --> 00:17:26.690 Does it? 00:17:26.690 --> 00:17:29.660 So it isn't going to contribute to that second term. 00:17:29.660 --> 00:17:33.070 Only piece you have to be concerned with is this one. 00:17:33.070 --> 00:17:38.350 So down in here in this part of it, all you would do 00:17:38.350 --> 00:17:41.370 is say omega 1 k cross. 00:17:41.370 --> 00:17:46.440 And now you just put in the q full omega 00:17:46.440 --> 00:17:54.630 2, which is omega 2 sine k plus omega 2 cosine. 00:17:54.630 --> 00:17:56.410 Sorry about the writing here. 00:17:56.410 --> 00:17:58.534 And do the cross product. 00:17:58.534 --> 00:17:59.950 You find out that term disappears. 00:17:59.950 --> 00:18:04.830 And you get an omega 1 omega 2 cosine. 00:18:04.830 --> 00:18:08.480 And this term here is still the answer. 00:18:08.480 --> 00:18:12.560 But you end up with a cosine phi in here 00:18:12.560 --> 00:18:15.330 because it's just the component of omega 2 that's 00:18:15.330 --> 00:18:18.730 in the j1 direction now that contributes. 00:18:18.730 --> 00:18:21.140 OK, great. 00:18:21.140 --> 00:18:22.260 Good. 00:18:22.260 --> 00:18:22.800 That help? 00:18:26.590 --> 00:18:27.570 I'll leave this. 00:18:27.570 --> 00:18:29.320 Let's get on to the problem for the day. 00:18:29.320 --> 00:18:39.180 The problem for the day is you've got a rotating arm. 00:18:39.180 --> 00:18:41.230 You have a mass sitting on it. 00:18:41.230 --> 00:18:46.190 The arm is rotating at some rotation rate omega. 00:18:46.190 --> 00:18:52.220 You might have an acceleration omega dot theta double dot. 00:18:56.000 --> 00:18:56.950 This mask can slide. 00:18:59.490 --> 00:19:08.190 So the problem is this thing is-- I'll start here. 00:19:08.190 --> 00:19:10.430 So it's rotating like this. 00:19:10.430 --> 00:19:13.670 And eventually, the thing begins to slide. 00:19:13.670 --> 00:19:15.590 And in fact, if I do it, this actually 00:19:15.590 --> 00:19:19.580 works a little better here because I have 00:19:19.580 --> 00:19:20.950 a little tiny groove in this. 00:19:20.950 --> 00:19:22.030 It helps a little bit. 00:19:22.030 --> 00:19:23.320 So it's rotating like that. 00:19:23.320 --> 00:19:24.570 It eventually begins to slide. 00:19:24.570 --> 00:19:28.630 And if the rotation rate is slow, it slides down. 00:19:28.630 --> 00:19:33.710 If I made the rotation rate fast enough, it slides up. 00:19:33.710 --> 00:19:37.740 And the angle at which it happens clearly 00:19:37.740 --> 00:19:40.920 depends on rotation rates, accelerations, the angle, 00:19:40.920 --> 00:19:43.570 friction coefficients, and things of that sort. 00:19:43.570 --> 00:19:47.520 So what I want you to start with is here's, 00:19:47.520 --> 00:19:50.490 basically, a drawing of it. 00:19:50.490 --> 00:19:57.060 And I don't want to give you any excess information. 00:19:57.060 --> 00:20:00.710 But what I want you to do is each, by yourselves, 00:20:00.710 --> 00:20:03.180 take just a minute, and draw a free body 00:20:03.180 --> 00:20:05.126 diagram of this problem. 00:20:05.126 --> 00:20:07.500 And then, after that, I'm going to have you get in groups 00:20:07.500 --> 00:20:08.280 and improve on it. 00:20:08.280 --> 00:20:09.270 But start yourself. 00:20:09.270 --> 00:20:12.990 Draw a free body diagram that will describe 00:20:12.990 --> 00:20:17.755 this mask on this thing, plank. 00:20:17.755 --> 00:20:25.537 OK, we've got a sketch with a little free body diagram on it. 00:20:25.537 --> 00:20:27.120 What I want you to do now, we're going 00:20:27.120 --> 00:20:29.630 to do this two or three times today. 00:20:29.630 --> 00:20:31.640 Get in groups of four five here. 00:20:31.640 --> 00:20:36.040 Compare notes, and come up with a group free body diagram. 00:20:36.040 --> 00:20:38.795 So you're pretty close. 00:20:42.490 --> 00:20:43.100 Let me look. 00:20:43.100 --> 00:20:44.890 Where was your guys solution again? 00:20:50.940 --> 00:20:53.890 There is maybe about as good as it gets. 00:20:57.090 --> 00:21:02.590 So here's our mass. 00:21:02.590 --> 00:21:06.630 I think everybody had an mg. 00:21:06.630 --> 00:21:12.340 Almost everybody had a normal force. 00:21:12.340 --> 00:21:14.120 Everybody had a friction force. 00:21:14.120 --> 00:21:16.770 Some discussion about what direction the friction force 00:21:16.770 --> 00:21:18.330 ought to be in. 00:21:18.330 --> 00:21:21.980 I'll draw it uphill here. 00:21:21.980 --> 00:21:26.100 And I'll call it ft for tangential, here. 00:21:30.120 --> 00:21:32.900 So some had about that much. 00:21:32.900 --> 00:21:36.690 There's a very important missing piece, which 00:21:36.690 --> 00:21:39.340 actually several groups had. 00:21:39.340 --> 00:21:41.670 What is that? 00:21:41.670 --> 00:21:45.281 Can anybody guess what I'm-- three of your groups had 00:21:45.281 --> 00:21:47.280 something on here that I don't have up here yet. 00:21:49.870 --> 00:21:52.756 I hear coordinates. 00:21:52.756 --> 00:21:54.130 This one, you can almost get away 00:21:54.130 --> 00:21:56.254 with not putting it in the coordinate system first. 00:21:56.254 --> 00:21:59.330 But you, usually, need to have a coordinate system in order 00:21:59.330 --> 00:22:01.455 to determine the direction of the forces. 00:22:01.455 --> 00:22:02.560 OK? 00:22:02.560 --> 00:22:04.660 And we'll get to that in a second. 00:22:04.660 --> 00:22:09.505 So everybody that I saw had written down, I saw, 00:22:09.505 --> 00:22:12.380 a number of the little x, y, z frames sitting out here 00:22:12.380 --> 00:22:14.160 like that. 00:22:14.160 --> 00:22:16.270 Maybe one was lined up. 00:22:16.270 --> 00:22:16.820 Not sure. 00:22:16.820 --> 00:22:17.460 Most not. 00:22:17.460 --> 00:22:19.780 Most like this. 00:22:19.780 --> 00:22:22.720 Is polar coordinates a good choice for this problem? 00:22:22.720 --> 00:22:24.076 Why not? 00:22:24.076 --> 00:22:25.022 AUDIENCE: [INAUDIBLE]. 00:22:28.806 --> 00:22:32.840 PROFESSOR: So the question is whether or not r theta and z 00:22:32.840 --> 00:22:35.920 can completely describe the motion in the problem. 00:22:35.920 --> 00:22:37.560 AUDIENCE: You have tangential forces, 00:22:37.560 --> 00:22:39.634 so it's better if you use rectangular. 00:22:42.479 --> 00:22:43.895 PROFESSOR: The radius is changing. 00:22:43.895 --> 00:22:46.490 But that's known as r dot. 00:22:46.490 --> 00:22:48.790 r dot can handle the motion to this this way. 00:22:48.790 --> 00:22:51.880 r double dot can handle acceleration this way. 00:22:51.880 --> 00:22:54.320 Theta dot can handle this motion. 00:22:54.320 --> 00:22:57.690 Theta double dot can handle that acceleration. 00:22:57.690 --> 00:23:00.350 Actually, is there any z motion or z-forces? 00:23:00.350 --> 00:23:01.000 No. 00:23:01.000 --> 00:23:02.875 Actually, the polar coordinates will actually 00:23:02.875 --> 00:23:04.550 work here just fine. 00:23:04.550 --> 00:23:08.430 So you definitely have to pick a coordinate system. 00:23:08.430 --> 00:23:13.490 So we're going to have an that system and theta hat 00:23:13.490 --> 00:23:16.850 in this direction because in this problem, 00:23:16.850 --> 00:23:23.910 whether you use x and y as a rotating frame here or r theta, 00:23:23.910 --> 00:23:29.050 you want to pick your coordinate system so that these things 00:23:29.050 --> 00:23:30.620 break down easily. 00:23:30.620 --> 00:23:32.950 In this case, the normal equation 00:23:32.950 --> 00:23:34.680 will be in the theta hat direction. 00:23:34.680 --> 00:23:39.400 The sliding direction will be in just one unit vector direction. 00:23:39.400 --> 00:23:42.280 So you don't have multiple components in this direction. 00:23:42.280 --> 00:23:45.490 If you use that coordinate system to describe this motion, 00:23:45.490 --> 00:23:46.660 you have x and y terms. 00:23:46.660 --> 00:23:48.500 It makes it harder. 00:23:48.500 --> 00:23:50.610 So this is a pretty good coordinate system to use. 00:23:50.610 --> 00:23:53.060 So that's a really important missing piece 00:23:53.060 --> 00:23:56.460 is the coordinate system, if you don't have it. 00:23:56.460 --> 00:24:03.400 And as an aside about free body diagrams-- free body diagrams, 00:24:03.400 --> 00:24:09.870 except my d is turned around-- how 00:24:09.870 --> 00:24:13.375 do you know that the friction force is in that direction. 00:24:13.375 --> 00:24:14.770 You had to assume something. 00:24:18.126 --> 00:24:20.500 This is such a trivial problem you can kind of figure out 00:24:20.500 --> 00:24:20.832 in your head. 00:24:20.832 --> 00:24:21.970 I gave the demonstration. 00:24:21.970 --> 00:24:22.690 It slid down. 00:24:22.690 --> 00:24:25.460 But also, I said , you do it fast enough, it goes up. 00:24:25.460 --> 00:24:26.840 So you don't really know. 00:24:26.840 --> 00:24:31.530 So you have to have made an assumption about it. 00:24:31.530 --> 00:24:45.590 So the general rule is to assign or assume 00:24:45.590 --> 00:24:59.170 positive values for all motions. 00:24:59.170 --> 00:25:08.105 And then, you deduce the direction of forces. 00:25:11.660 --> 00:25:15.420 And I'll take another quick aside here 00:25:15.420 --> 00:25:19.070 to illustrate this in a problem that works better 00:25:19.070 --> 00:25:23.530 than the current one for this. 00:25:23.530 --> 00:25:27.330 An obvious coordinate system for this little mass on rollers 00:25:27.330 --> 00:25:28.170 inertial frame. 00:25:28.170 --> 00:25:30.820 Here's an x. 00:25:30.820 --> 00:25:35.360 And I asked you to do a free body diagram of that. 00:25:35.360 --> 00:25:37.390 Well you would show me a normal force. 00:25:37.390 --> 00:25:39.510 You'd show me an mg. 00:25:39.510 --> 00:25:41.290 But now I ask you to tell me the direction 00:25:41.290 --> 00:25:42.530 to draw the spring force. 00:25:46.640 --> 00:25:51.450 So if you adopt the rule that for each body 00:25:51.450 --> 00:25:55.770 you're working with, you assume that it has positive x, 00:25:55.770 --> 00:26:00.000 positive x dot, positive y, positive y dot 00:26:00.000 --> 00:26:03.139 then you can deduce what direction of the forces 00:26:03.139 --> 00:26:04.180 would result [INAUDIBLE]. 00:26:04.180 --> 00:26:08.900 So in this case, the only motion that generates a force up here 00:26:08.900 --> 00:26:10.966 is what? 00:26:10.966 --> 00:26:12.900 AUDIENCE: [INAUDIBLE]. 00:26:12.900 --> 00:26:15.990 PROFESSOR: Is x. x dot. x double dot. 00:26:15.990 --> 00:26:19.890 The only motion that will generate a force 00:26:19.890 --> 00:26:23.470 is displacement because it generates a force where? 00:26:23.470 --> 00:26:24.464 In the spring. 00:26:24.464 --> 00:26:25.880 If they add a dash [? pod ?] here, 00:26:25.880 --> 00:26:28.570 them velocity generates a force, too. 00:26:28.570 --> 00:26:34.840 But displacement, if you assume the displacement is positive, 00:26:34.840 --> 00:26:37.670 then which direction is the spring force? 00:26:37.670 --> 00:26:41.100 OK, so then you have spring force this way. 00:26:41.100 --> 00:26:43.691 And that's value kx. 00:26:43.691 --> 00:26:45.690 And when you write it in the equation of motion, 00:26:45.690 --> 00:26:47.750 you put a minus sign to account for the direction 00:26:47.750 --> 00:26:49.780 of that error. 00:26:49.780 --> 00:26:51.130 OK. 00:26:51.130 --> 00:26:55.280 So in this problem we kind of-- and if you ever 00:26:55.280 --> 00:26:57.376 do it wrong, if you pick this direction wrong, 00:26:57.376 --> 00:26:58.500 what happens in the answer? 00:26:58.500 --> 00:27:00.500 Will you still get the right answer? 00:27:00.500 --> 00:27:03.020 If you're consistent, you'll get another negative sign 00:27:03.020 --> 00:27:04.370 popping up to fix it. 00:27:04.370 --> 00:27:04.870 OK. 00:27:07.510 --> 00:27:09.060 Next. 00:27:09.060 --> 00:27:12.060 So I think we're in good shape now. 00:27:12.060 --> 00:27:14.010 I'm going to draw the friction force uphill 00:27:14.010 --> 00:27:16.610 because I'm going to assume this thing's going to slide down. 00:27:16.610 --> 00:27:20.070 And I'm going to simplify the problem a little bit for you. 00:27:20.070 --> 00:27:21.450 And I'm going to say that there's 00:27:21.450 --> 00:27:24.260 no angular acceleration. 00:27:24.260 --> 00:27:27.250 Constant angular rate. 00:27:27.250 --> 00:27:31.820 The distance that the mass starts off up the slope 00:27:31.820 --> 00:27:33.190 is at one and a half feet. 00:27:35.840 --> 00:27:37.440 So it's going up like this. 00:27:37.440 --> 00:27:40.290 And it, eventually, reaches 50 degrees. 00:27:40.290 --> 00:27:44.210 And at 50 degrees, it begins to slide downhill. 00:27:44.210 --> 00:27:47.660 So it's going up a constant rate and starts to slide. 00:27:47.660 --> 00:27:50.160 So there must be some friction coefficient 00:27:50.160 --> 00:27:54.430 that provides a system with a property such 00:27:54.430 --> 00:27:56.540 that it slides at 50 degrees. 00:27:56.540 --> 00:28:01.220 So the problem here is define the coefficient of friction. 00:28:01.220 --> 00:28:03.840 So now you're in your groups. 00:28:03.840 --> 00:28:05.070 Solve the problem. 00:28:05.070 --> 00:28:07.840 And what I really want you to do is come up with an expression. 00:28:07.840 --> 00:28:13.562 mu equals in variables and constants. 00:28:13.562 --> 00:28:15.270 Don't bother plugging in numbers in that. 00:28:18.825 --> 00:28:20.950 Well if you're comfortable using polar coordinates, 00:28:20.950 --> 00:28:22.074 that'd be the way to do it. 00:28:22.074 --> 00:28:23.890 But you need to write some equations now. 00:28:23.890 --> 00:28:26.800 And I suggest you need to think in terms of equations 00:28:26.800 --> 00:28:27.700 [? in ?] motion. 00:28:27.700 --> 00:28:33.420 I think it's time to come back together here and work on this. 00:28:37.170 --> 00:28:41.040 Two, three, or four groups are, sort of, struggling with this. 00:28:41.040 --> 00:28:45.340 And it's because you're not really 00:28:45.340 --> 00:28:51.440 going to first principles and doing the problem step by step. 00:28:51.440 --> 00:28:53.530 You're, sort of, jumping to the answer 00:28:53.530 --> 00:28:57.350 because it's a trivial, simple, Mickey Mouse problem that you 00:28:57.350 --> 00:29:00.030 did in high school. 00:29:00.030 --> 00:29:03.674 So I'm, kind of, pounding on you a little bit. 00:29:03.674 --> 00:29:05.590 We give you a simple problem so we can do them 00:29:05.590 --> 00:29:07.330 in a short period of time. 00:29:07.330 --> 00:29:09.980 But you need to learn to do them in the rigorous way 00:29:09.980 --> 00:29:14.030 so that you learn the real fundamental stuff that you 00:29:14.030 --> 00:29:16.860 have to know. 00:29:16.860 --> 00:29:19.750 So none of you-- you all are sort of thinking about, 00:29:19.750 --> 00:29:21.250 well, we got sum of the forces here. 00:29:21.250 --> 00:29:23.080 We've got an acceleration there. 00:29:23.080 --> 00:29:26.180 But nobody is just sitting down and saying 00:29:26.180 --> 00:29:31.370 that some of the forces is equal to the mass 00:29:31.370 --> 00:29:34.490 times the acceleration and working the problem out. 00:29:37.460 --> 00:29:38.550 OK, bad dog. 00:29:41.100 --> 00:29:44.960 Just because this problem is almost the statics problem 00:29:44.960 --> 00:29:48.870 doesn't mean you-- when things are statics problem, 00:29:48.870 --> 00:29:50.620 it just means that acceleration goes to 0. 00:29:50.620 --> 00:29:52.270 And the sum of the forces is now 0. 00:29:52.270 --> 00:29:54.350 And you solve the problem. 00:29:54.350 --> 00:29:59.410 Start with Newtons, in this case, it's Newtons second law. 00:29:59.410 --> 00:30:02.640 It helps to know an expression for acceleration, 00:30:02.640 --> 00:30:04.410 so you don't have to grind it out. 00:30:04.410 --> 00:30:08.510 So polar coordinates works pretty well in this problem. 00:30:08.510 --> 00:30:14.480 And I want you to commit to memory two acceleration 00:30:14.480 --> 00:30:15.500 equations. 00:30:15.500 --> 00:30:19.430 One in polar coordinates, and the one the general vector 1. 00:30:19.430 --> 00:30:22.640 So in polar coordinates, the one that I memorize 00:30:22.640 --> 00:30:32.180 is a with respect to o plus r double dot minus r theta 00:30:32.180 --> 00:30:44.060 dot squared r hat plus r theta double dot plus 2 omega r 00:30:44.060 --> 00:30:48.040 dot theta hat. 00:30:48.040 --> 00:30:53.500 Coriolis, Eulerian centripetal, and your linear acceleration. 00:30:53.500 --> 00:30:55.780 What's this term? 00:30:55.780 --> 00:30:58.938 What's it doing there? 00:30:58.938 --> 00:31:00.730 AUDIENCE: [INAUDIBLE]. 00:31:00.730 --> 00:31:04.780 PROFESSOR: Yeah, but the reason I remember it is, 00:31:04.780 --> 00:31:09.230 in this course, we break down every dynamics problem 00:31:09.230 --> 00:31:12.710 we're confronted with into sub problems that 00:31:12.710 --> 00:31:17.240 can be solved as the sum of a translation 00:31:17.240 --> 00:31:20.210 of a body plus a rotation. 00:31:20.210 --> 00:31:22.720 So this expression allows you to write down 00:31:22.720 --> 00:31:27.090 the acceleration of a point on a body, which is 00:31:27.090 --> 00:31:31.080 both translating and rotating. 00:31:31.080 --> 00:31:32.800 What this term accounts for. 00:31:36.750 --> 00:31:38.770 The acceleration contributed by what? 00:31:42.290 --> 00:31:45.430 So I've got this general-- let's see. 00:31:45.430 --> 00:31:48.570 I got a merry go round that's on wheel 00:31:48.570 --> 00:31:49.970 and is rolling along here. 00:31:49.970 --> 00:31:51.235 And I have an inertial frame. 00:31:54.400 --> 00:31:55.970 And I got a point a here. 00:31:55.970 --> 00:31:59.460 But now I've also got a system in here, 00:31:59.460 --> 00:32:04.880 which I'm describing with an r theta connected to this. 00:32:04.880 --> 00:32:07.530 Thing can have translational acceleration. 00:32:07.530 --> 00:32:09.182 That's that term. 00:32:09.182 --> 00:32:10.890 This problem doesn't happen to have that. 00:32:10.890 --> 00:32:11.880 It goes to 0. 00:32:11.880 --> 00:32:14.090 In this problem, that's a 0. 00:32:14.090 --> 00:32:21.160 In this problem, we can now apply this equation 00:32:21.160 --> 00:32:22.510 to that problem. 00:32:22.510 --> 00:32:26.665 How many sub equations are we going to get? 00:32:30.212 --> 00:32:30.920 And why no three. 00:32:30.920 --> 00:32:32.735 AUDIENCE: But I said two. 00:32:32.735 --> 00:32:34.270 PROFESSOR: I know you said two. 00:32:34.270 --> 00:32:35.978 But I'm asking why not three. 00:32:35.978 --> 00:32:37.730 AUDIENCE: Because [INAUDIBLE] theta. 00:32:37.730 --> 00:32:42.670 PROFESSOR: And a z, which you need because it describes theta 00:32:42.670 --> 00:32:44.100 dot. 00:32:44.100 --> 00:32:46.372 So you have a z direction. 00:32:46.372 --> 00:32:48.830 But is there any forces in this problem in the z direction? 00:32:48.830 --> 00:32:50.710 Are there any accelerations in the z direction? 00:32:50.710 --> 00:32:51.950 No, so you get a trivial answer there. 00:32:51.950 --> 00:32:53.533 So you could just write that one down. 00:32:53.533 --> 00:32:55.100 Mass times acceleration equals zero. 00:32:55.100 --> 00:32:56.016 And there's no forces. 00:32:56.016 --> 00:32:57.810 So there's three possible equations. 00:32:57.810 --> 00:33:00.590 Only two of them are meaningful in this problem. 00:33:00.590 --> 00:33:04.240 And we have one that we can summon the that component 00:33:04.240 --> 00:33:06.210 and one in the theta hat. 00:33:06.210 --> 00:33:11.200 So the sum of the forces in the theta hat direction 00:33:11.200 --> 00:33:13.445 for this problem are what? 00:33:17.620 --> 00:33:21.170 From your free body diagram. 00:33:21.170 --> 00:33:22.295 AUDIENCE: The normal force? 00:33:22.295 --> 00:33:24.520 PROFESSOR: The normal in the positive theta 00:33:24.520 --> 00:33:26.218 hat direction and? 00:33:26.218 --> 00:33:28.658 AUDIENCE: [INAUDIBLE]. 00:33:28.658 --> 00:33:32.930 PROFESSOR: Minus mg. 00:33:32.930 --> 00:33:35.840 And I think it's [? percosen ?] theta. 00:33:35.840 --> 00:33:37.200 And that's your theta hat. 00:33:37.200 --> 00:33:42.580 And what is the acceleration in that direction? 00:33:42.580 --> 00:33:44.920 Well you go to acceleration formula. 00:33:44.920 --> 00:33:46.550 And now it's inspect the terms. 00:33:46.550 --> 00:33:51.920 This one is 0 because the platform is fixed. 00:33:51.920 --> 00:33:53.850 r double dot? 00:33:53.850 --> 00:33:56.730 0 because we're just waiting. 00:33:56.730 --> 00:33:59.170 It's just trying to calculate when it begins to slide. 00:33:59.170 --> 00:34:01.490 r theta dot squared. 00:34:01.490 --> 00:34:03.010 Not 0. 00:34:03.010 --> 00:34:06.172 r theta double dot. 00:34:06.172 --> 00:34:06.880 Well it might be. 00:34:06.880 --> 00:34:11.389 But I said omega dots 0 constant. 00:34:11.389 --> 00:34:13.320 So that one's 0 for this problem. 00:34:13.320 --> 00:34:16.500 And this term, Coriolis. 00:34:16.500 --> 00:34:20.521 0 because our dot is 0. 00:34:20.521 --> 00:34:22.604 We're really treating this like a statics problem. 00:34:22.604 --> 00:34:24.110 It hasn't started to move yet. 00:34:24.110 --> 00:34:26.800 So we're only left with one term here. 00:34:26.800 --> 00:34:31.040 So the sum of the forces in the normal direction, which is 00:34:31.040 --> 00:34:36.300 the theta hat direction are 0. 00:34:36.300 --> 00:34:39.764 And that allows us to solve for the normal force. 00:34:43.530 --> 00:34:48.739 So you need Newton's law to find the normal force to start with. 00:34:48.739 --> 00:34:51.480 And you need the normal force to find the friction force. 00:34:51.480 --> 00:34:57.610 So now let's do the sum of the forces in the that direction. 00:34:57.610 --> 00:35:01.200 And maybe, let's do the acceleration first. 00:35:01.200 --> 00:35:04.930 It's the mass times the acceleration 00:35:04.930 --> 00:35:08.090 in the that direction. 00:35:08.090 --> 00:35:09.500 And that's the mass. 00:35:09.500 --> 00:35:13.210 And now, what's the acceleration? 00:35:13.210 --> 00:35:15.480 These are the r directed terms. 00:35:15.480 --> 00:35:19.430 We have one term, right. 00:35:19.430 --> 00:35:21.150 Minus r theta dot squared. 00:35:26.430 --> 00:35:27.520 No other terms. 00:35:27.520 --> 00:35:30.270 And those are going to be equal to the external forces 00:35:30.270 --> 00:35:31.270 in the radial direction. 00:35:31.270 --> 00:35:33.520 And what are they? 00:35:33.520 --> 00:35:36.786 From the [INAUDIBLE] diagram. 00:35:36.786 --> 00:35:39.719 AUDIENCE: [INAUDIBLE]. 00:35:39.719 --> 00:35:40.760 PROFESSOR: Plus or minus? 00:35:40.760 --> 00:35:41.752 AUDIENCE: Plus. 00:35:41.752 --> 00:35:44.360 PROFESSOR: Plus mu n. 00:35:44.360 --> 00:35:49.370 But we know n is mg cosine theta. 00:35:49.370 --> 00:35:52.480 OK, what else? 00:35:52.480 --> 00:35:58.250 Minus mg sine theta. 00:35:58.250 --> 00:35:58.956 All right. 00:36:07.720 --> 00:36:11.060 You know everything in this expression. 00:36:11.060 --> 00:36:12.330 You know given theta dot. 00:36:12.330 --> 00:36:13.585 You're given r. 00:36:13.585 --> 00:36:15.580 You know mg. 00:36:15.580 --> 00:36:19.130 You know [INAUDIBLE] given theta. 00:36:19.130 --> 00:36:20.950 You could solve this expression for mu. 00:36:25.860 --> 00:36:27.150 All right? 00:36:27.150 --> 00:36:31.970 And the mg goes to the other side. 00:36:31.970 --> 00:36:35.710 And you have a minus r theta squared. 00:36:35.710 --> 00:36:37.810 Notice the M's cancel all the way through, right. 00:36:42.860 --> 00:36:47.450 So mu would looks like it equals, to me, 00:36:47.450 --> 00:37:04.860 g sine theta minus r theta dot squared all divided by what? 00:37:04.860 --> 00:37:08.320 g cosine theta. 00:37:13.770 --> 00:37:15.720 Break these two apart, this gives me 00:37:15.720 --> 00:37:29.180 a tan theta minus r theta dot squared over g QoS theta. 00:37:29.180 --> 00:37:30.870 That has units of acceleration. 00:37:30.870 --> 00:37:32.540 g has units of acceleration. 00:37:32.540 --> 00:37:33.592 This is dimensionless. 00:37:33.592 --> 00:37:36.430 The answer has got to be dimensionless. 00:37:36.430 --> 00:37:39.780 So there's your answer. 00:37:39.780 --> 00:37:42.410 But you really had to use the equations in motion. 00:37:42.410 --> 00:37:44.020 OK now there's another thing I want 00:37:44.020 --> 00:37:47.130 to dress because I heard it pop up two or three times. 00:37:47.130 --> 00:37:55.860 Do not confuse accelerations with forces. 00:37:55.860 --> 00:37:59.050 Newtons second law makes it really clear 00:37:59.050 --> 00:38:00.870 where each one goes. 00:38:00.870 --> 00:38:04.310 The sum of the external forces go one side of the equal sign. 00:38:04.310 --> 00:38:08.610 The mass times the acceleration goes on the other. 00:38:08.610 --> 00:38:11.290 Don't get the two mixed up. 00:38:11.290 --> 00:38:15.100 Solve for the accelerations, if you can. 00:38:15.100 --> 00:38:15.750 Plug them in. 00:38:15.750 --> 00:38:16.880 And multiply them by mass. 00:38:16.880 --> 00:38:19.620 And now you have that one side of the equation. 00:38:19.620 --> 00:38:20.970 The free body diagram. 00:38:20.970 --> 00:38:27.760 The only vectors that should show up on a free body diagram 00:38:27.760 --> 00:38:30.330 are what? 00:38:30.330 --> 00:38:31.180 AUDIENCE: Forces. 00:38:31.180 --> 00:38:32.580 PROFESSOR: Forces. 00:38:32.580 --> 00:38:35.030 The real forces in the problem. 00:38:35.030 --> 00:38:37.540 So in that problem that free body diagram, 00:38:37.540 --> 00:38:42.340 there is no r theta dot squared term. 00:38:42.340 --> 00:38:44.270 It doesn't belong there. 00:38:44.270 --> 00:38:46.430 And it'll keep you from making these sign errors 00:38:46.430 --> 00:38:47.370 and things like that. 00:38:47.370 --> 00:38:50.020 So the business about this thing comes up minus 00:38:50.020 --> 00:38:54.574 because it's minus right out of the acceleration equation. 00:38:54.574 --> 00:39:05.825 All right, what would happen if I had not made the-- we're 00:39:05.825 --> 00:39:06.450 a go? 00:39:10.760 --> 00:39:11.615 This term 0. 00:39:11.615 --> 00:39:15.190 I allow this to have some angular acceleration to it. 00:39:15.190 --> 00:39:18.720 That's what's got to happen to. 00:39:18.720 --> 00:39:20.990 Well you don't have to do it. 00:39:20.990 --> 00:39:22.350 I can't do it constant rate. 00:39:22.350 --> 00:39:25.910 But if you can make this constant rate fast enough, 00:39:25.910 --> 00:39:28.750 it goes up. 00:39:28.750 --> 00:39:31.100 So even if it's constant rate fast enough, 00:39:31.100 --> 00:39:33.470 this thing will slide up the thing, immediately, 00:39:33.470 --> 00:39:35.052 when you start it. 00:39:35.052 --> 00:39:35.760 Almost immediate. 00:39:35.760 --> 00:39:36.885 You start down here, the g. 00:39:36.885 --> 00:39:38.360 The friction force is pretty high. 00:39:38.360 --> 00:39:39.190 And the friction force, of course, 00:39:39.190 --> 00:39:40.564 diminishes as you get further up. 00:39:40.564 --> 00:39:44.210 And eventually, it takes off, right. 00:39:44.210 --> 00:39:46.750 That's because that minus r theta 00:39:46.750 --> 00:39:50.681 dot squared, that's a centripetal acceleration 00:39:50.681 --> 00:39:51.180 inwards. 00:39:51.180 --> 00:39:53.080 And if you don't provide the force that 00:39:53.080 --> 00:39:55.720 causes that centripetal acceleration to happen, 00:39:55.720 --> 00:39:59.600 it says, it says I want to leave town. 00:39:59.600 --> 00:40:03.570 Now how about, though, if I had angular acceleration, as well? 00:40:03.570 --> 00:40:05.900 If I allow angular acceleration to this problem, 00:40:05.900 --> 00:40:09.310 how does it change the two equations that we wrote? 00:40:11.910 --> 00:40:13.020 What does it change? 00:40:13.020 --> 00:40:14.490 Does it change the forces? 00:40:14.490 --> 00:40:16.360 Does it change the free body diagram? 00:40:16.360 --> 00:40:16.910 Not a bit. 00:40:16.910 --> 00:40:20.530 It changes the acceleration side of the equation and what term 00:40:20.530 --> 00:40:26.340 it now turns up that you didn't have before. 00:40:26.340 --> 00:40:26.960 This one. 00:40:26.960 --> 00:40:31.100 So now you have a dynamic term in the theta hat equation. 00:40:31.100 --> 00:40:33.620 And it comes into this expression 00:40:33.620 --> 00:40:36.070 for some of the forces in this direction. 00:40:36.070 --> 00:40:42.110 And you would end up equals to m r theta double dot. 00:40:42.110 --> 00:40:45.680 Now it's a entirely different problem. 00:40:45.680 --> 00:40:48.650 It gets a little harder to solve. 00:40:48.650 --> 00:40:54.450 OK, but really good fundamental practices. 00:40:54.450 --> 00:40:55.830 f equals ma. 00:40:55.830 --> 00:40:57.780 And write out both sides carefully. 00:40:57.780 --> 00:41:00.810 And then, you won't make sign mistakes and so forth. 00:41:00.810 --> 00:41:01.440 Thanks. 00:41:01.440 --> 00:41:03.430 Good to see you.