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 --> 00:00:12.680 To make a donation or to view additional materials 00:00:12.680 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.305 at ocw.mit.edu. 00:00:25.820 --> 00:00:27.470 PROFESSOR: There's a reading. 00:00:27.470 --> 00:00:31.570 The new reading is on Stellar. 00:00:31.570 --> 00:00:35.180 It's an excerpt from a textbook on dynamics by Professor Jim 00:00:35.180 --> 00:00:39.140 Williams who was in the mechanical engineering 00:00:39.140 --> 00:00:43.030 department here at MIT and just recently retired. 00:00:43.030 --> 00:00:47.610 This reading is essentially a review of everything 00:00:47.610 --> 00:00:49.700 we've done in kinematics. 00:00:49.700 --> 00:00:51.700 So it gives you a different point of view. 00:00:51.700 --> 00:00:54.570 It's extremely well written. 00:00:54.570 --> 00:00:56.990 So it goes through things like derivatives 00:00:56.990 --> 00:00:58.610 of rotating vectors. 00:00:58.610 --> 00:01:02.440 It has three or four really terrific examples of problems 00:01:02.440 --> 00:01:06.430 that are solved using the techniques that we've used up 00:01:06.430 --> 00:01:07.710 to this point. 00:01:07.710 --> 00:01:11.590 So it's a good way to review for the quiz, which 00:01:11.590 --> 00:01:15.890 is coming up a week from Thursday, on October 14. 00:01:15.890 --> 00:01:19.980 That quiz will be here, closed book, one sheet 00:01:19.980 --> 00:01:24.070 of notes, piece of paper both sides as your reference 00:01:24.070 --> 00:01:25.740 material. 00:01:25.740 --> 00:01:27.480 OK, survey results. 00:01:27.480 --> 00:01:29.420 So you have p set four. 00:01:29.420 --> 00:01:31.690 It has I think five questions on that. 00:01:31.690 --> 00:01:34.910 We'll quickly take a look at the problems 00:01:34.910 --> 00:01:36.780 that you're working on right now. 00:01:36.780 --> 00:01:42.800 So here's this yo-yo like thing, a spool. 00:01:42.800 --> 00:01:44.550 There were some questions on nb. 00:01:44.550 --> 00:01:47.155 The inner piece doesn't rotate relative to the out piece. 00:01:47.155 --> 00:01:53.230 It's all one solid spool, but it has two different radii that 00:01:53.230 --> 00:01:55.230 are functioning in the problem. 00:01:55.230 --> 00:01:58.760 And the question is, will the spool roll to the left 00:01:58.760 --> 00:02:01.324 or to the right when the string is pulled to the right? 00:02:01.324 --> 00:02:02.740 So we're going to come back to it. 00:02:02.740 --> 00:02:05.073 I'm just going to run through all the questions quickly. 00:02:05.073 --> 00:02:07.830 The second one was this somewhat familiar problem 00:02:07.830 --> 00:02:11.850 you've done now. 00:02:11.850 --> 00:02:16.090 The question asks you about how fast 00:02:16.090 --> 00:02:20.450 it has to go before the thing starts sliding up. 00:02:20.450 --> 00:02:25.210 And the survey question, if the starting position of the spool 00:02:25.210 --> 00:02:29.460 is moved down the rod, will the angular rate 00:02:29.460 --> 00:02:32.650 be higher when spool begins to slide up the rod? 00:02:32.650 --> 00:02:34.820 Will you have to go faster to get it to go? 00:02:34.820 --> 00:02:37.400 OK, next. 00:02:37.400 --> 00:02:41.452 OK, this is finding the equation of motion for a pendulum. 00:02:41.452 --> 00:02:43.160 And everybody knows the natural frequency 00:02:43.160 --> 00:02:44.826 of a pendulum's square root of g over l. 00:02:44.826 --> 00:02:47.635 But this pendulum has a torsional spring added to it. 00:02:47.635 --> 00:02:50.570 And the question is, the square root of g over l 00:02:50.570 --> 00:02:53.080 doesn't involve the mass. 00:02:53.080 --> 00:02:54.660 But will the natural frequency, now 00:02:54.660 --> 00:02:58.160 that you've put the torsional spring on it, will it now 00:02:58.160 --> 00:02:59.489 involve the mass? 00:02:59.489 --> 00:03:00.780 No reason you should know this. 00:03:00.780 --> 00:03:02.950 You've never done this problem before. 00:03:02.950 --> 00:03:04.821 And we'll come back to that. 00:03:04.821 --> 00:03:05.320 Next. 00:03:07.809 --> 00:03:08.600 The fourth problem. 00:03:08.600 --> 00:03:10.460 Two degree of freedom system. 00:03:10.460 --> 00:03:14.670 And wants you to come up with the equation of motion. 00:03:14.670 --> 00:03:18.520 But the survey question, if k1 didn't exist, 00:03:18.520 --> 00:03:22.150 this is just two masses coupled by spring, 00:03:22.150 --> 00:03:25.060 this thing is no longer constrained in the x direction. 00:03:25.060 --> 00:03:29.450 There's no x constraints on the system. 00:03:29.450 --> 00:03:33.950 So if that's the case, this system still 00:03:33.950 --> 00:03:36.510 has two natural frequencies. 00:03:36.510 --> 00:03:40.100 A system that can vibrant essentially 00:03:40.100 --> 00:03:43.860 has as many natural frequencies as it does equations of motion. 00:03:43.860 --> 00:03:45.760 This one requires two equations of motion, 00:03:45.760 --> 00:03:50.050 but it will have one 0 natural frequency and the other one 00:03:50.050 --> 00:03:51.740 greater than 0. 00:03:51.740 --> 00:03:55.920 In the second case when it's greater than 0, 00:03:55.920 --> 00:03:57.750 this thing is oscillating somehow. 00:03:57.750 --> 00:03:58.990 It has no extra strength. 00:03:58.990 --> 00:04:03.460 What can you say about the position of the center of mass? 00:04:03.460 --> 00:04:04.490 So click at the results. 00:04:04.490 --> 00:04:07.010 So this is the first one is the spool problem. 00:04:13.420 --> 00:04:16.180 There's the spool. 00:04:16.180 --> 00:04:17.940 Here's table. 00:04:17.940 --> 00:04:19.410 Here's the inner radius. 00:04:19.410 --> 00:04:21.760 It's got a string wrapped on it. 00:04:21.760 --> 00:04:27.200 You're pulling the string this way at v in the i direction. 00:04:27.200 --> 00:04:35.120 This'll be o xy. 00:04:35.120 --> 00:04:37.600 And now the question is, when you pull on this, 00:04:37.600 --> 00:04:40.330 which way will the spool go? 00:04:40.330 --> 00:04:43.230 And most of you said it will go to the left. 00:04:43.230 --> 00:04:44.866 So I made this up. 00:04:44.866 --> 00:04:46.240 We're going to do the experiment. 00:04:53.560 --> 00:04:59.640 So I've wrapped Teflon tape on this spool. 00:04:59.640 --> 00:05:02.210 It's got a C battery in the center 00:05:02.210 --> 00:05:04.730 to give it a little bit of weight. 00:05:04.730 --> 00:05:09.750 And here's the experiment. 00:05:09.750 --> 00:05:13.390 And the guys in the booth I hope have in a minute 00:05:13.390 --> 00:05:15.535 to zoom in on it. 00:05:15.535 --> 00:05:17.660 I'm going to let go and I'm going to start pulling. 00:05:17.660 --> 00:05:18.710 Oops. 00:05:18.710 --> 00:05:21.690 I need to put a little more out here. 00:05:21.690 --> 00:05:22.500 Pull on the tape. 00:05:26.140 --> 00:05:29.630 You believe it? 00:05:29.630 --> 00:05:31.640 So I've done the reverse of what's in there. 00:05:31.640 --> 00:05:32.550 I'm pulling that way. 00:05:32.550 --> 00:05:35.680 But it goes in the same direction as you pull it. 00:05:35.680 --> 00:05:41.760 So the answer to this question is, it would go to the right. 00:05:41.760 --> 00:05:44.848 Kind of unbelievable almost, right? 00:05:44.848 --> 00:05:47.276 How does it do that? 00:05:47.276 --> 00:05:48.025 Let's do it again. 00:05:57.164 --> 00:05:58.650 All right. 00:05:58.650 --> 00:06:01.240 Now, show you something else. 00:06:01.240 --> 00:06:05.580 If that hasn't boggled your mind. 00:06:05.580 --> 00:06:11.110 Then what's it going to do if I pull straight up? 00:06:11.110 --> 00:06:14.310 How many think it's going to go that way? 00:06:14.310 --> 00:06:16.130 Raise your hand. 00:06:16.130 --> 00:06:18.390 How many think it's going to go that way? 00:06:18.390 --> 00:06:19.890 Raise your hand. 00:06:19.890 --> 00:06:22.330 All right. 00:06:22.330 --> 00:06:23.360 Sure enough. 00:06:23.360 --> 00:06:25.800 That means it must be some angle between in which 00:06:25.800 --> 00:06:27.630 it will do neither. 00:06:27.630 --> 00:06:28.910 It'll start to slip. 00:06:35.152 --> 00:06:36.144 Right about there. 00:06:42.697 --> 00:06:43.530 Neat little problem. 00:06:47.660 --> 00:06:54.270 And now doing this problem is all about computing velocities, 00:06:54.270 --> 00:06:56.365 using concepts like. 00:06:59.140 --> 00:07:05.660 So if we call this a, b, c. 00:07:05.660 --> 00:07:08.240 I think it's the other way around in the problem. 00:07:08.240 --> 00:07:09.610 Excuse me. 00:07:09.610 --> 00:07:14.090 A, b, c, and d. 00:07:21.710 --> 00:07:24.810 The key to this problem is being able to figure out 00:07:24.810 --> 00:07:29.990 what the velocity of this point c is. 00:07:29.990 --> 00:07:31.660 And the first thing to understand, 00:07:31.660 --> 00:07:40.120 this is the velocity of the end of the tape in a fixed 00:07:40.120 --> 00:07:41.622 reference frame. 00:07:41.622 --> 00:07:45.070 So every point on the tape moves at the same speed. 00:07:45.070 --> 00:07:47.010 That means where it touches right here 00:07:47.010 --> 00:07:49.810 is also moving at that speed. 00:07:49.810 --> 00:07:58.455 So the velocity of c with respect to o is to vi. 00:07:58.455 --> 00:08:00.925 And that's kind of the key to the problem. 00:08:05.140 --> 00:08:10.810 That's also equal to the velocity of d with respect 00:08:10.810 --> 00:08:16.538 to o plus the velocity of c with respect to d. 00:08:16.538 --> 00:08:21.870 And the velocity of d is this point 00:08:21.870 --> 00:08:23.700 of contact with the ground. 00:08:23.700 --> 00:08:26.090 What's that velocity? 00:08:26.090 --> 00:08:26.680 0. 00:08:26.680 --> 00:08:28.900 This is back to those instantaneous center 00:08:28.900 --> 00:08:30.430 of rotation things. 00:08:30.430 --> 00:08:33.820 0 there is what makes this equation easy. 00:08:33.820 --> 00:08:39.430 And then you know how to compute the velocity of a rotating 00:08:39.430 --> 00:08:39.929 vector. 00:08:39.929 --> 00:08:42.309 This one's not changing length, but it is surely 00:08:42.309 --> 00:08:44.290 rotating about this point, and so 00:08:44.290 --> 00:08:47.234 there's going to be an omega cross r here. 00:08:47.234 --> 00:08:48.900 And that'll allow you to solve for omega 00:08:48.900 --> 00:08:50.810 and then once you know for omega then you 00:08:50.810 --> 00:08:55.510 can solve for a, which is the real question that was asked. 00:08:55.510 --> 00:08:57.410 OK. 00:08:57.410 --> 00:08:58.010 Next one. 00:09:04.830 --> 00:09:07.235 So the starting position, this is a spool. 00:09:11.200 --> 00:09:14.140 There's the arm. 00:09:14.140 --> 00:09:19.830 And here, the question is at 0.25 meters up here. 00:09:19.830 --> 00:09:26.840 And if you move it down to 0.15 will 00:09:26.840 --> 00:09:30.730 the system be able to go faster before it breaks loose or not? 00:09:30.730 --> 00:09:33.190 And your answer was most people said yes. 00:09:33.190 --> 00:09:36.120 And basically that's true. 00:09:36.120 --> 00:09:39.260 What is it that's driving that thing, 00:09:39.260 --> 00:09:43.660 providing the force that is trying 00:09:43.660 --> 00:09:44.950 to push it up the sleeve? 00:09:48.534 --> 00:09:49.450 AUDIENCE: [INAUDIBLE]. 00:09:52.450 --> 00:09:55.110 PROFESSOR: So the rod as it's going around 00:09:55.110 --> 00:09:57.280 is trying to drive that thing in a circle. 00:09:57.280 --> 00:10:02.930 And therefore, it is being accelerated 00:10:02.930 --> 00:10:04.020 to the central points. 00:10:04.020 --> 00:10:06.430 So it has centripetal acceleration. 00:10:06.430 --> 00:10:11.750 And that requires forces and they come through the rod. 00:10:11.750 --> 00:10:15.440 And that will develop the normal forces and friction. 00:10:15.440 --> 00:10:22.160 So when you slide down the rod some to that new position, 00:10:22.160 --> 00:10:24.870 is has that centripetal acceleration gone up or down. 00:10:28.270 --> 00:10:29.690 r's gotten smaller. 00:10:29.690 --> 00:10:33.080 Centripetal acceleration is? 00:10:33.080 --> 00:10:35.520 r omega squared, right? 00:10:35.520 --> 00:10:37.510 So you might be speaking algebraically. 00:10:37.510 --> 00:10:41.900 So the magnitude of that centripetal acceleration 00:10:41.900 --> 00:10:46.700 has gone up or down if you bring it closer to the axis. 00:10:46.700 --> 00:10:47.490 Down. 00:10:47.490 --> 00:10:51.450 So you can go faster to get up to the same level 00:10:51.450 --> 00:10:53.341 of acceleration or it'll break free. 00:10:53.341 --> 00:10:53.840 Next. 00:10:57.130 --> 00:11:00.820 This is, oh, the natural frequency of the pendulum. 00:11:00.820 --> 00:11:03.479 Some of you said yes, some of you said no, some not sure. 00:11:03.479 --> 00:11:04.520 You're going to find out. 00:11:04.520 --> 00:11:06.561 You're going to figure out the equation of motion 00:11:06.561 --> 00:11:07.240 of the system. 00:11:07.240 --> 00:11:10.565 And it includes that torsional spring and mass is definitely 00:11:10.565 --> 00:11:11.620 going to be in it. 00:11:11.620 --> 00:11:15.242 So look for an answer that includes-- 00:11:15.242 --> 00:11:19.100 that you'll see that mass is going to show up in a way 00:11:19.100 --> 00:11:23.260 that if you solve for the natural frequency, 00:11:23.260 --> 00:11:25.421 it won't leave the final equation. 00:11:25.421 --> 00:11:25.920 OK. 00:11:37.110 --> 00:11:40.240 So this is the one about the two masses. 00:11:40.240 --> 00:11:44.780 And if the spring were 0, the second natural frequency 00:11:44.780 --> 00:11:46.030 of the system. 00:11:46.030 --> 00:11:47.410 The first natural frequency, when 00:11:47.410 --> 00:11:48.784 you have a system that has what's 00:11:48.784 --> 00:11:52.130 called a rigid body degree of freedom. 00:11:52.130 --> 00:11:55.200 When this thing went, there's no k1 spring. 00:11:55.200 --> 00:12:01.070 This system looks like two masses on wheels connected 00:12:01.070 --> 00:12:02.770 by a spring and a dashpot. 00:12:07.510 --> 00:12:10.715 And in the x direction, there are no constraints. 00:12:13.280 --> 00:12:16.232 So if I set this system to vibrating, 00:12:16.232 --> 00:12:18.565 what can you say about the motion of its center of mass? 00:12:27.090 --> 00:12:29.896 So could you find the center of mass of that system? 00:12:29.896 --> 00:12:31.270 Got to give you the masses of it. 00:12:31.270 --> 00:12:36.810 But you have some m1 and an m2. 00:12:36.810 --> 00:12:45.150 And we'll call this the x in that direction. 00:12:45.150 --> 00:12:47.910 Could you determine the center of mass of the system? 00:12:47.910 --> 00:12:49.700 You know how to do that, right? 00:12:49.700 --> 00:12:52.050 If they were equal in mass, it'd be in the center. 00:12:52.050 --> 00:12:55.950 If this was bigger, it'd be over here or something like that. 00:12:55.950 --> 00:13:00.380 So we know that for a system of particles, 00:13:00.380 --> 00:13:02.910 the sum of the external forces on the system 00:13:02.910 --> 00:13:06.210 is equal to mass of the system times the acceleration 00:13:06.210 --> 00:13:10.020 of the what? 00:13:10.020 --> 00:13:12.256 Center of mass. 00:13:12.256 --> 00:13:14.630 What are the external forces in the horizontal direction? 00:13:17.400 --> 00:13:19.110 How big are they? 00:13:19.110 --> 00:13:19.920 0. 00:13:19.920 --> 00:13:21.580 Therefore, what can be the acceleration 00:13:21.580 --> 00:13:23.140 of the center of mass? 00:13:23.140 --> 00:13:23.830 0. 00:13:23.830 --> 00:13:26.877 This thing will actually in this second natural frequency, 00:13:26.877 --> 00:13:28.960 it'll sit there and just oscillate back and forth, 00:13:28.960 --> 00:13:31.970 one mask on one directional, one on the other. 00:13:31.970 --> 00:13:34.610 But its center of mass won't move. 00:13:34.610 --> 00:13:38.000 The other degree of freedom and the other natural frequency, 00:13:38.000 --> 00:13:40.060 the one that has 0 natural frequency, 00:13:40.060 --> 00:13:41.340 is a rigid body motion. 00:13:41.340 --> 00:13:42.770 No relative motion. 00:13:42.770 --> 00:13:44.980 Give this thing a little push, some initial momentum, 00:13:44.980 --> 00:13:46.472 it'll just sit their role forever. 00:13:46.472 --> 00:13:48.430 The mass is not moving relative to one another. 00:13:48.430 --> 00:13:50.475 It just rolls and rolls and rolls and rolls 00:13:50.475 --> 00:13:51.730 and never comes back. 00:13:51.730 --> 00:13:55.049 Infinite period 0 natural frequency. 00:13:55.049 --> 00:13:56.840 OK, we were talking about fictitious forces 00:13:56.840 --> 00:13:57.760 the other day. 00:13:57.760 --> 00:14:04.020 And there are times when they're useful. 00:14:04.020 --> 00:14:09.930 But they're sort of like giving a little kid a loaded gun. 00:14:09.930 --> 00:14:13.100 You really get yourself in trouble quickly. 00:14:13.100 --> 00:14:16.530 So I'm not advocating a lot of use of it. 00:14:16.530 --> 00:14:18.520 But it's good to have a little insight. 00:14:18.520 --> 00:14:21.590 One of the students after the lecture last time, 00:14:21.590 --> 00:14:23.950 a student mentioned to me, he said, 00:14:23.950 --> 00:14:29.150 the way I was taught about fictitious forces 00:14:29.150 --> 00:14:33.030 is that we use them to patch up a problem when 00:14:33.030 --> 00:14:36.560 we're trying to work in a non inertial frame. 00:14:36.560 --> 00:14:40.490 How can you make Newton's laws apply if you 00:14:40.490 --> 00:14:43.650 are in a non inertial frame? 00:14:43.650 --> 00:14:46.230 Otherwise an accelerating reference frame. 00:14:46.230 --> 00:14:48.720 So I just thought I would revisit a little bit of what 00:14:48.720 --> 00:14:54.110 we said the other day with explaining things that way 00:14:54.110 --> 00:14:55.130 a little bit. 00:14:55.130 --> 00:14:58.610 So it might just turn out to be familiar for you. 00:14:58.610 --> 00:14:59.940 And let's just look. 00:15:07.960 --> 00:15:12.760 Let's just go back quickly and revisit that elevator problem. 00:15:12.760 --> 00:15:15.570 So you remember you're here. 00:15:15.570 --> 00:15:17.575 You've got some mass m. 00:15:17.575 --> 00:15:20.260 You're standing on some scales. 00:15:20.260 --> 00:15:22.649 And I want to know what the scales are going to read. 00:15:22.649 --> 00:15:24.440 Now, we did this one quickly the other day. 00:15:24.440 --> 00:15:25.648 I'm just going to revisit it. 00:15:28.260 --> 00:15:32.710 And this accelerating, this is point A. 00:15:32.710 --> 00:15:35.235 And here down here is o. 00:15:35.235 --> 00:15:36.325 My inertial frame. 00:15:38.830 --> 00:15:43.040 And the acceleration of A with respect to o 00:15:43.040 --> 00:15:50.546 is plus a quarter of g in the j hat direction. 00:15:50.546 --> 00:15:52.980 OK? 00:15:52.980 --> 00:15:56.420 Now, when you think about this, if you're 00:15:56.420 --> 00:15:59.200 trying to do physics in the elevator, 00:15:59.200 --> 00:16:01.880 but it's an accelerating frame, so Newton's laws don't apply. 00:16:04.510 --> 00:16:05.780 So you're in the elevator. 00:16:05.780 --> 00:16:08.600 You can't see you're accelerating. 00:16:08.600 --> 00:16:11.700 But you could measure your weight, for example. 00:16:11.700 --> 00:16:18.730 So how do you fix up the equations 00:16:18.730 --> 00:16:22.750 so that you can predict the right answer? 00:16:22.750 --> 00:16:28.570 And to do that, basically you I have to add in a force, 00:16:28.570 --> 00:16:31.720 a fictitious force, that accounts 00:16:31.720 --> 00:16:35.870 for the additional forces you will 00:16:35.870 --> 00:16:40.046 feel in this system due to the acceleration. 00:16:42.750 --> 00:16:47.190 So remember the way we did this problem. 00:16:47.190 --> 00:16:49.470 You do this problem just in a straightforward 00:16:49.470 --> 00:16:56.670 way, a summation of the external forces in the y direction 00:16:56.670 --> 00:17:02.800 is the mass times acceleration, a with respect to o. 00:17:02.800 --> 00:17:05.200 And we need a free body diagram. 00:17:05.200 --> 00:17:12.119 In the free body diagrams, this mass here, 00:17:12.119 --> 00:17:18.720 you have an mg downwards and an n upwards. 00:17:18.720 --> 00:17:27.470 And so you'd say this is equal to n minus mg. 00:17:27.470 --> 00:17:33.740 Now, if you're doing this experiment inside, 00:17:33.740 --> 00:17:37.080 you can measure this n that's going to be your weight. 00:17:37.080 --> 00:17:40.010 And you would like to do this by fictitious forces, 00:17:40.010 --> 00:17:52.560 you say that the summation of the external real forces 00:17:52.560 --> 00:17:58.150 minus this mass times the acceleration of a with respect 00:17:58.150 --> 00:18:04.470 to o has got to be equal to 0. 00:18:04.470 --> 00:18:13.940 And that's then n minus mg minus mg over 4. 00:18:13.940 --> 00:18:15.265 Here's your fictitious force. 00:18:19.270 --> 00:18:22.610 And that's as if then you've come over here 00:18:22.610 --> 00:18:29.800 and you've added an additional force on this free body diagram 00:18:29.800 --> 00:18:34.270 that looks like mg over 4. 00:18:34.270 --> 00:18:38.990 And now you say the sum of these forces, this is the correction. 00:18:38.990 --> 00:18:44.000 It's allowing you now to work in this accelerating frame. 00:18:44.000 --> 00:18:48.260 If do this free body diagram and sum these up, 00:18:48.260 --> 00:18:53.220 then you will find this expression, 00:18:53.220 --> 00:18:55.500 which you can solve for n. 00:18:55.500 --> 00:18:59.730 And you'll get m5 over 4g. 00:19:02.167 --> 00:19:04.625 In fact, if you stood on the scales, that's what you'd get. 00:19:15.060 --> 00:19:17.630 All right, so this is example one. 00:19:17.630 --> 00:19:19.730 I want to do sort of example 1.5. 00:19:19.730 --> 00:19:23.735 The cable breaks on this elevator. 00:19:44.038 --> 00:19:49.070 Your free body diagram still has an mg on it. 00:19:49.070 --> 00:19:51.110 You're still inside. 00:19:51.110 --> 00:19:53.720 You're still measuring forces. 00:19:53.720 --> 00:19:56.110 And you want to use this idea of a fictitious force 00:19:56.110 --> 00:19:58.970 to help you sort out what forces will you 00:19:58.970 --> 00:20:00.570 feel inside of that elevator. 00:20:08.730 --> 00:20:12.330 You know how to do the answer in this straightforward way. 00:20:12.330 --> 00:20:19.390 The sum of external forces is minus mg. 00:20:19.390 --> 00:20:22.630 And that's equal to the mass times the acceleration 00:20:22.630 --> 00:20:24.680 of a with respect to o. 00:20:24.680 --> 00:20:29.545 And therefore acceleration of a with respect to o 00:20:29.545 --> 00:20:31.410 is just minus g. 00:20:37.670 --> 00:20:39.080 And you could have told me that. 00:20:39.080 --> 00:20:39.704 That's trivial. 00:20:39.704 --> 00:20:42.550 If you drop the thing, it accelerates the acceleration 00:20:42.550 --> 00:20:43.460 of gravity. 00:20:43.460 --> 00:20:46.820 So I want to know what's the-- you're 00:20:46.820 --> 00:20:48.220 doing the experiment inside. 00:20:48.220 --> 00:20:49.920 You don't know that you're accelerating. 00:20:49.920 --> 00:20:51.720 All you're doing is measurements. 00:20:51.720 --> 00:20:54.510 But you've been told how to correct it. 00:20:54.510 --> 00:20:58.270 So let's put it in the fictitious force, which 00:20:58.270 --> 00:21:01.310 is minus ma 0. 00:21:01.310 --> 00:21:11.820 And that means the mass times acceleration is downwards. 00:21:11.820 --> 00:21:13.890 So minus that. 00:21:13.890 --> 00:21:18.180 The fictitious force is an mg force upwards. 00:21:18.180 --> 00:21:19.090 And what do you feel? 00:21:22.280 --> 00:21:24.670 What do the scales read? 00:21:24.670 --> 00:21:25.550 Nothing. 00:21:25.550 --> 00:21:26.985 You feel absolutely nothing. 00:21:31.040 --> 00:21:32.440 OK. 00:21:32.440 --> 00:21:38.340 So just an aside kind of question, because I know this 00:21:38.340 --> 00:21:39.800 is confused. 00:21:39.800 --> 00:21:41.830 Occasionally we all get confused around this. 00:21:45.360 --> 00:21:49.823 In these problems that we do, is gravity an acceleration? 00:21:56.080 --> 00:21:58.670 We say all the time, I use the phrase 00:21:58.670 --> 00:22:00.006 the acceleration of gravity. 00:22:00.006 --> 00:22:01.630 How many times have you ever said that? 00:22:01.630 --> 00:22:03.254 Have you ever said that? 00:22:03.254 --> 00:22:04.420 The acceleration of gravity. 00:22:04.420 --> 00:22:05.350 What do we mean? 00:22:05.350 --> 00:22:06.332 What we mean is this. 00:22:06.332 --> 00:22:08.040 When you let something drop in free fall, 00:22:08.040 --> 00:22:09.950 it'll accelerate at g. 00:22:09.950 --> 00:22:13.630 But the way we apply when we're writing out equations 00:22:13.630 --> 00:22:16.940 of motion, we say the sum of the external forces 00:22:16.940 --> 00:22:18.480 is equal to mass times acceleration. 00:22:18.480 --> 00:22:20.311 Where does gravity go? 00:22:20.311 --> 00:22:23.250 It's a force. 00:22:23.250 --> 00:22:25.730 So this is another thing not to get 00:22:25.730 --> 00:22:30.800 confused by is to think about gravity as being 00:22:30.800 --> 00:22:31.570 an acceleration. 00:22:31.570 --> 00:22:33.410 We spend a lot of time figuring out ways 00:22:33.410 --> 00:22:36.680 to write accelerations in rotating, 00:22:36.680 --> 00:22:38.200 translating, reference frames. 00:22:38.200 --> 00:22:39.780 Gravity never pops up in there. 00:22:39.780 --> 00:22:43.760 It always comes up on the force side of the equation. 00:22:43.760 --> 00:22:44.260 OK. 00:22:48.620 --> 00:22:56.130 All right, so one last example from last time. 00:22:56.130 --> 00:23:00.610 But it's relevant to a problem on the homework. 00:23:00.610 --> 00:23:04.530 So we have this shaft spinning. 00:23:04.530 --> 00:23:08.830 This is the z-axis spinning at constant rate omega. 00:23:12.160 --> 00:23:20.740 And we'll use this is r hat and this is z k hat. 00:23:20.740 --> 00:23:25.340 These are the lengths r hat direction and the height z. 00:23:25.340 --> 00:23:27.580 We're using cylindrical coordinates. 00:23:27.580 --> 00:23:32.600 And now let's just use this notion of fictitious forces 00:23:32.600 --> 00:23:36.910 to figure out first what's the what's 00:23:36.910 --> 00:23:38.610 the fictitious force in this system. 00:23:42.819 --> 00:23:44.610 Remember, this is now this notion of you're 00:23:44.610 --> 00:23:46.920 out there where this mass is. 00:23:46.920 --> 00:23:48.620 You're riding with this mass. 00:23:48.620 --> 00:23:52.407 And you're trying to say, what force am I going to feel? 00:23:52.407 --> 00:23:54.240 You want to be able to predict it correctly. 00:23:54.240 --> 00:23:57.370 So you have to find a fictitious force 00:23:57.370 --> 00:24:01.380 that you can bring in that accounts for what 00:24:01.380 --> 00:24:06.000 acceleration are we feeling out there. 00:24:06.000 --> 00:24:07.020 Centripetal, right? 00:24:07.020 --> 00:24:10.910 And the centripetal acceleration, 00:24:10.910 --> 00:24:14.750 we want the sum of the external forces. 00:24:14.750 --> 00:24:17.951 And this is going to be in the r hat 00:24:17.951 --> 00:24:27.530 direction minus m times the-- and now we 00:24:27.530 --> 00:24:29.090 need reference frames here. 00:24:29.090 --> 00:24:35.240 So you have a fixed frame o xy. 00:24:35.240 --> 00:24:40.470 You have a rotating frame here, a x prime, y prime, z prime. 00:24:40.470 --> 00:24:44.470 And this is your point b up here. 00:24:44.470 --> 00:24:51.030 So some of the external forces minus the mass 00:24:51.030 --> 00:24:55.410 times the acceleration of b with respect to 0 00:24:55.410 --> 00:25:00.990 has got to be equal to 0. 00:25:00.990 --> 00:25:04.160 That's how this fictitious force thing works. 00:25:04.160 --> 00:25:08.480 So on the free body diagram of this thing, 00:25:08.480 --> 00:25:10.220 what are the actual forces on it? 00:25:10.220 --> 00:25:17.090 Well, there's certainly a weight downwards. 00:25:17.090 --> 00:25:20.770 There must be some supporting force upwards. 00:25:20.770 --> 00:25:24.130 And we'll put that in the z direction here. 00:25:24.130 --> 00:25:26.670 Because this thing certainly doesn't accelerate up and down. 00:25:26.670 --> 00:25:29.110 So the rods lifting it up somehow. 00:25:29.110 --> 00:25:35.820 The rod has potentially some radial component. 00:25:35.820 --> 00:25:38.270 And I'm just drawing it in the positive direction. 00:25:38.270 --> 00:25:41.470 And those are my forces. 00:25:41.470 --> 00:25:46.510 And in this problem, omega dot theta double dot 00:25:46.510 --> 00:25:51.660 equals 0 and r dot and r double dot were 0. 00:25:51.660 --> 00:25:53.960 So it's constant rotation rate. 00:25:53.960 --> 00:25:55.790 No geometry changes here. 00:25:58.750 --> 00:26:01.910 So we did, we cranked through this problem before. 00:26:01.910 --> 00:26:04.270 And now I wanted to make this correction. 00:26:04.270 --> 00:26:07.060 I want to add this fictitious force in. 00:26:07.060 --> 00:26:11.430 The acceleration is minus r omega 00:26:11.430 --> 00:26:13.330 squared in the r hat direction. 00:26:13.330 --> 00:26:17.290 So minus m times that is a plus. 00:26:17.290 --> 00:26:24.120 I get an m r omega squared. 00:26:28.730 --> 00:26:32.650 And pointing outwards. 00:26:32.650 --> 00:26:41.210 And so the sum of the forces in the radial direction 00:26:41.210 --> 00:26:47.470 minus-- some of the forces in the radial direction now 00:26:47.470 --> 00:26:52.270 minus this mab with respect to o. 00:26:52.270 --> 00:26:54.270 Well, what are some of the real forces? 00:26:54.270 --> 00:26:56.970 Just nr. 00:26:56.970 --> 00:27:01.414 And this minus mass times the acceleration, that's your mr 00:27:01.414 --> 00:27:01.913 squared. 00:27:04.730 --> 00:27:06.840 mr omega squared. 00:27:06.840 --> 00:27:09.360 And that's got to be 0. 00:27:09.360 --> 00:27:12.035 So you find out that there is a-- 00:27:23.600 --> 00:27:24.490 This all is 0. 00:27:24.490 --> 00:27:35.370 Therefore solve for nr is minus mr omega squared. 00:27:35.370 --> 00:27:41.280 So that rod out here is constantly pulling it in 00:27:41.280 --> 00:27:43.510 to make it go around the circle. 00:27:43.510 --> 00:27:45.660 But we've just done this now by this notion 00:27:45.660 --> 00:27:48.650 of a fictitious force. 00:27:48.650 --> 00:27:52.640 So let's go back to this problem now. 00:27:52.640 --> 00:27:55.910 Well actually, I'll just take one final step. 00:27:55.910 --> 00:28:01.630 Here's your free body diagram touched up 00:28:01.630 --> 00:28:03.720 with this fictitious force so that we 00:28:03.720 --> 00:28:06.760 can work as if we're here. 00:28:06.760 --> 00:28:09.937 And by the way, if you could measure, 00:28:09.937 --> 00:28:11.520 if you had a strain gauge or something 00:28:11.520 --> 00:28:17.810 so you can measure the force that this rod 00:28:17.810 --> 00:28:19.800 is putting on that mass. 00:28:19.800 --> 00:28:21.347 If you can measure it out there, this 00:28:21.347 --> 00:28:22.680 would be what you would measure. 00:28:27.820 --> 00:28:35.990 Now what about the torque that force creates around here? 00:28:40.900 --> 00:28:43.330 So you have outward force mr omega 00:28:43.330 --> 00:28:48.830 squared at centrifugal force, this fictitious force. 00:28:48.830 --> 00:28:52.110 Going pulling it out up here. 00:28:52.110 --> 00:28:53.880 What moment does that create down here? 00:28:57.870 --> 00:28:59.135 What's the magnitude first? 00:29:02.550 --> 00:29:11.380 And that'll be a torque around in the theta direction. 00:29:16.810 --> 00:29:18.164 So what's the moment arm? 00:29:18.164 --> 00:29:20.060 AUDIENCE: [INAUDIBLE]. 00:29:20.060 --> 00:29:23.200 PROFESSOR: The moment arm? 00:29:23.200 --> 00:29:24.870 The moment arm. 00:29:24.870 --> 00:29:25.580 z. 00:29:25.580 --> 00:29:26.360 OK. 00:29:26.360 --> 00:29:30.110 So then the torque is going to be 4r cross f 00:29:30.110 --> 00:29:38.440 and it's going to be the mrz omega squared. 00:29:38.440 --> 00:29:44.360 And it's at centripetal forces trying to bend it out. 00:29:44.360 --> 00:29:46.660 Therefore the torque down here required 00:29:46.660 --> 00:29:50.000 to keep it from doing so is actually 00:29:50.000 --> 00:29:52.230 in the other direction. 00:29:52.230 --> 00:29:59.640 So the torque that this system must put on this bar is minus. 00:29:59.640 --> 00:30:01.320 OK. 00:30:01.320 --> 00:30:03.630 So let's go to this problem. 00:30:03.630 --> 00:30:07.260 If this is the point about which we're computing the moments, 00:30:07.260 --> 00:30:11.980 then there's centripetal acceleration in this problem, 00:30:11.980 --> 00:30:13.361 but in what direction are they? 00:30:16.105 --> 00:30:17.230 Either one of those masses. 00:30:22.500 --> 00:30:24.980 Pardon? 00:30:24.980 --> 00:30:28.150 So each little mass, like the upper mass m over 2. 00:30:28.150 --> 00:30:31.190 It's spinning and it's shown as it's coming around the wheel. 00:30:31.190 --> 00:30:32.850 It's up at the top. 00:30:32.850 --> 00:30:36.505 What direction is the fictitious force 00:30:36.505 --> 00:30:39.350 that it's putting on that system? 00:30:39.350 --> 00:30:41.340 Radial outwards, right? 00:30:41.340 --> 00:30:43.600 And does it have any moment arm that 00:30:43.600 --> 00:30:46.800 force going up with respect to that point in the center? 00:30:46.800 --> 00:30:47.300 No. 00:30:47.300 --> 00:30:48.540 So it creates no moment. 00:30:48.540 --> 00:30:51.550 The bottom one by symmetry doesn't either. 00:30:51.550 --> 00:30:58.100 This one over here, it also has a fictitious force 00:30:58.100 --> 00:30:58.820 you can think of. 00:30:58.820 --> 00:31:01.200 It's just outward mr omega squared. 00:31:01.200 --> 00:31:05.951 What's its moment arm with respect to our origin here? 00:31:05.951 --> 00:31:06.450 0. 00:31:06.450 --> 00:31:08.010 Again, it produces no moment. 00:31:08.010 --> 00:31:10.810 It produces an unbalanced force though, right? 00:31:10.810 --> 00:31:13.520 It's going around producing an unbalanced force. 00:31:13.520 --> 00:31:17.340 This problem, the upper ones, the fictitious force is up. 00:31:17.340 --> 00:31:19.621 The lower one, which direction is it? 00:31:19.621 --> 00:31:20.120 Down. 00:31:20.120 --> 00:31:22.170 It's perfectly balanced. 00:31:22.170 --> 00:31:25.896 They're just equal and opposite, just going around. 00:31:25.896 --> 00:31:27.270 You'd feel nothing uncomfortable. 00:31:27.270 --> 00:31:29.603 This one you'd feel, this motorcycle trying to hop down. 00:31:32.135 --> 00:31:34.000 It really happens, right? 00:31:34.000 --> 00:31:37.180 But this one, what happens? 00:31:37.180 --> 00:31:39.980 Does this mass m over 2, what direction is 00:31:39.980 --> 00:31:42.720 the centripetal acceleration? 00:31:42.720 --> 00:31:44.450 I've changed the wording a little bit. 00:31:44.450 --> 00:31:47.450 The upper mass in case b, which direction 00:31:47.450 --> 00:31:50.920 is the centripetal acceleration? 00:31:50.920 --> 00:31:52.290 Straight down, right? 00:31:52.290 --> 00:31:54.290 And the bottom mass, what direction? 00:31:54.290 --> 00:31:55.330 Straight up. 00:31:55.330 --> 00:31:59.930 And the fictitious forces that reflect 00:31:59.930 --> 00:32:03.555 those two accelerations are the top one is up 00:32:03.555 --> 00:32:05.620 and the bottom one is down. 00:32:05.620 --> 00:32:08.660 But they produce a net. 00:32:08.660 --> 00:32:11.360 Do they have a moment arm that they act on with respect 00:32:11.360 --> 00:32:12.647 to that center? 00:32:12.647 --> 00:32:13.980 Yeah, this little distance here. 00:32:13.980 --> 00:32:16.120 I think it's called b in the problem. 00:32:16.120 --> 00:32:20.240 So this thing, m over 2 r omega squared times b, 00:32:20.240 --> 00:32:22.780 is a torque about the x-axis. 00:32:22.780 --> 00:32:25.100 And this one down here is another m 00:32:25.100 --> 00:32:28.150 over 2 r omega squared and it's another torque 00:32:28.150 --> 00:32:29.110 in the same direction. 00:32:29.110 --> 00:32:33.410 So this thing produces quite a strong moment. 00:32:33.410 --> 00:32:43.100 But as the wheel turns, that moment starts off, 00:32:43.100 --> 00:32:45.630 it's about the x-axis here. 00:32:45.630 --> 00:32:47.680 But when that weight gets to 90 degrees, 00:32:47.680 --> 00:32:49.000 what direction is the moment? 00:32:51.730 --> 00:32:57.600 So we just do all the way around the y-axis. 00:32:57.600 --> 00:32:58.810 So it's going to be a moment. 00:32:58.810 --> 00:33:00.860 The answer to this problem is in time 00:33:00.860 --> 00:33:05.650 it's going to look something like mr omega squared z 00:33:05.650 --> 00:33:08.270 sine or cosine omega t. 00:33:08.270 --> 00:33:10.700 So the moment in this around the x-axis 00:33:10.700 --> 00:33:13.900 is going to be trying to-- the bike's going to try 00:33:13.900 --> 00:33:17.300 to go back and forth like this. 00:33:17.300 --> 00:33:18.801 With frequency omega. 00:33:18.801 --> 00:33:19.300 OK. 00:33:26.880 --> 00:33:29.209 Oh, I was going to say that's it for fictitious forces, 00:33:29.209 --> 00:33:29.750 but it's not. 00:33:29.750 --> 00:33:31.270 I've got a great demo for you. 00:33:40.620 --> 00:33:41.130 All right. 00:33:49.986 --> 00:33:50.820 We have a slope. 00:33:55.820 --> 00:34:00.310 I've got a cart on wheels there. 00:34:00.310 --> 00:34:01.110 And I actually do. 00:34:05.500 --> 00:34:06.000 OK. 00:34:10.737 --> 00:34:14.999 And I'm going to put a box here. 00:34:18.730 --> 00:34:20.230 I got the wrong color here, but I'll 00:34:20.230 --> 00:34:22.239 try colored chalk for the first time. 00:34:22.239 --> 00:34:23.520 In the box I've got a fluid. 00:34:28.110 --> 00:34:30.580 And the box has a lid on it so I don't make a mess. 00:34:33.880 --> 00:34:38.360 Now, I'm actually going to do this experiment. 00:34:38.360 --> 00:34:39.639 I brought my ramp. 00:34:39.639 --> 00:34:41.810 We're going to set up this ramp. 00:34:41.810 --> 00:34:44.929 This box is going to be rolling down this hill. 00:34:44.929 --> 00:34:48.110 And basically no losses. 00:34:48.110 --> 00:34:49.898 No friction holding it back. 00:34:49.898 --> 00:34:52.314 It's just going to be able to take off and accelerate down 00:34:52.314 --> 00:34:52.920 that hill. 00:34:56.219 --> 00:35:03.980 I want to know a, b, c. 00:35:03.980 --> 00:35:14.050 When it's rolling down that hill, will the fluid in the box 00:35:14.050 --> 00:35:30.370 look like that, look like that, or like that? 00:35:37.970 --> 00:35:41.010 So we're going to take our poll here. 00:35:41.010 --> 00:35:47.690 So how many believe that when it's rolling down the hill, 00:35:47.690 --> 00:35:50.730 the fluid will stay level, parallel to the ground? 00:35:50.730 --> 00:35:52.120 Let's raise your hands. 00:35:52.120 --> 00:35:54.007 I want everybody to make a guess here. 00:35:54.007 --> 00:35:54.840 Everybody get in it. 00:35:54.840 --> 00:35:58.750 Raise them high if you think it's going to stay level. 00:35:58.750 --> 00:36:00.520 OK, maybe a quarter of you. 00:36:00.520 --> 00:36:02.912 How many think it's going to do this? 00:36:02.912 --> 00:36:05.290 OK, a few more. 00:36:05.290 --> 00:36:08.100 And how many think it's going to do this? 00:36:08.100 --> 00:36:12.500 All right, so I want you get in natural little groups there 00:36:12.500 --> 00:36:15.610 of two or three people and talk about this for a couple minutes 00:36:15.610 --> 00:36:18.750 and see if you can convince your neighbor that you're right. 00:36:18.750 --> 00:36:19.950 Figure this out. 00:36:19.950 --> 00:36:20.960 Stop and talk about it. 00:36:24.590 --> 00:36:26.900 And let's set up to do the experiment. 00:36:33.023 --> 00:36:36.914 Actually, I think if we do it up here. 00:36:36.914 --> 00:36:38.320 Yeah. 00:36:38.320 --> 00:36:39.630 So like that. 00:36:42.510 --> 00:36:44.805 So you hold the ramp. 00:36:44.805 --> 00:36:46.186 OK, you come here. 00:36:49.850 --> 00:36:53.030 We'll just practice our set up here for a second. 00:36:53.030 --> 00:36:56.730 You're going to have to hold it quite a bit higher. 00:36:56.730 --> 00:36:59.044 And actually when we get ready to go-- 00:36:59.044 --> 00:37:00.460 AUDIENCE: How well sealed is that? 00:37:00.460 --> 00:37:03.240 PROFESSOR: It's very well sealed. 00:37:03.240 --> 00:37:05.231 You hold it and you release it. 00:37:05.231 --> 00:37:07.730 And that's about the right-- because I got to catch it so it 00:37:07.730 --> 00:37:10.241 doesn't-- now, we're going to stop. 00:37:10.241 --> 00:37:11.740 We're not going to do it right away. 00:37:11.740 --> 00:37:15.340 I'm going to get them to answer the question a second time. 00:37:20.930 --> 00:37:21.845 Good idea. 00:37:26.750 --> 00:37:28.015 All right. 00:37:28.015 --> 00:37:30.610 And I think that'll work just great. 00:37:30.610 --> 00:37:33.230 And we're not going to give it away. 00:37:33.230 --> 00:37:36.730 OK, so set it down and we'll do the thing. 00:37:36.730 --> 00:37:37.870 Just drop it. 00:37:54.660 --> 00:37:59.010 OK, let's do the quiz again. 00:37:59.010 --> 00:38:01.430 I want to see if any of you convinced your neighbor 00:38:01.430 --> 00:38:03.305 to change their vote. 00:38:03.305 --> 00:38:07.970 So how many believe that it's going to be A? 00:38:07.970 --> 00:38:10.190 Hm, one or two. 00:38:10.190 --> 00:38:12.010 Interesting, that kind of went down. 00:38:12.010 --> 00:38:14.366 How many believe it's going to be B? 00:38:14.366 --> 00:38:15.610 A lot of people went up. 00:38:15.610 --> 00:38:18.010 How many think it's going to be C? 00:38:18.010 --> 00:38:19.990 OK, still a couple hold outs. 00:38:19.990 --> 00:38:21.005 Let's do the experiment. 00:38:26.000 --> 00:38:28.330 So guys in the booth, can you get a good shot of this, 00:38:28.330 --> 00:38:29.200 I hope? 00:38:29.200 --> 00:38:33.000 I want this so it [INAUDIBLE]. 00:38:33.000 --> 00:38:35.510 All right. 00:38:35.510 --> 00:38:36.700 Move the chalk. 00:38:36.700 --> 00:38:38.410 Yep. 00:38:38.410 --> 00:38:42.980 And I'm going to have to catch it. 00:38:42.980 --> 00:38:44.410 All right. 00:38:44.410 --> 00:38:47.630 Let her rip. 00:38:47.630 --> 00:38:48.680 Which way? 00:38:48.680 --> 00:38:49.180 AUDIENCE: B. 00:38:49.180 --> 00:38:49.846 PROFESSOR: Yeah. 00:38:49.846 --> 00:38:50.636 Let's do it again. 00:38:56.480 --> 00:38:59.820 Just make sure the guys in the booth got a good shot of this. 00:38:59.820 --> 00:39:00.980 OK, once more. 00:39:03.740 --> 00:39:04.690 Pretty neat, huh? 00:39:10.800 --> 00:39:16.640 If you're skiing downhill, you're going down the hill, 00:39:16.640 --> 00:39:22.545 accelerating, which way do you feel the forces on your body? 00:39:27.580 --> 00:39:30.560 You feel the forces pulling you down straight? 00:39:30.560 --> 00:39:33.630 Do you feel the forces pulling you down the hill? 00:39:33.630 --> 00:39:36.075 Do you feel the forces pulling you into the hill? 00:39:40.870 --> 00:39:43.520 So this is straightforward application of this. 00:39:46.295 --> 00:39:49.210 The slope of that liquid was parallel to the slope. 00:40:00.685 --> 00:40:02.490 Get my notes back out here. 00:40:25.630 --> 00:40:29.550 So let's quickly do this problem rigorously, 00:40:29.550 --> 00:40:34.320 the way we'd do it straightforward 00:40:34.320 --> 00:40:37.440 with accelerations and free body diagrams and so forth. 00:40:37.440 --> 00:40:43.009 We're going to have some normal force pushing up. 00:40:43.009 --> 00:40:43.800 I said no friction. 00:40:43.800 --> 00:40:47.420 This thing's a slippery surface, just as if it's sliding down. 00:40:47.420 --> 00:40:50.590 Got to do rollers in the classroom. 00:40:50.590 --> 00:40:51.320 It's got an mg. 00:40:56.800 --> 00:41:00.420 And there are no other external forces. 00:41:00.420 --> 00:41:05.750 And we know then that the summation 00:41:05.750 --> 00:41:09.380 of the external forces, that's got 00:41:09.380 --> 00:41:14.470 to be the mass times the acceleration. 00:41:14.470 --> 00:41:17.950 And I'm going to make this here. 00:41:17.950 --> 00:41:21.310 I'll call this point A in the center. 00:41:21.310 --> 00:41:24.290 And I'm going to make my inertial reference 00:41:24.290 --> 00:41:28.445 frame x down the hill, y perpendicular to the hill. 00:41:31.240 --> 00:41:35.240 So the mass times the acceleration of a with respect 00:41:35.240 --> 00:41:38.300 to o is then equal. 00:41:38.300 --> 00:41:40.840 And this is my then external forces. 00:41:40.840 --> 00:41:43.980 So this is my x direction. 00:41:43.980 --> 00:41:50.620 So I'm going to break this and mg force into two components. 00:41:50.620 --> 00:41:52.090 One down the hill. 00:41:52.090 --> 00:41:53.450 Guess I need an angle. 00:42:02.020 --> 00:42:02.590 Let's see. 00:42:06.520 --> 00:42:17.820 So if this is theta, then this is theta here. 00:42:22.380 --> 00:42:26.540 And the force down the hill is that component 00:42:26.540 --> 00:42:34.640 is mg sine theta. 00:42:34.640 --> 00:42:38.975 And the force this direction is then going to be. 00:42:44.720 --> 00:42:47.710 And I'll just give us some unit vectors here 00:42:47.710 --> 00:42:49.230 to keep things straight. 00:42:49.230 --> 00:43:00.850 This force is minus mg cosine theta in the j hat direction. 00:43:00.850 --> 00:43:06.790 So the sum of the forces here, all of them, 00:43:06.790 --> 00:43:22.190 are mg sine theta i hat minus mg cosine theta j hat 00:43:22.190 --> 00:43:25.680 plus n in the j hat direction. 00:43:25.680 --> 00:43:28.670 So that's the total vector sum. 00:43:28.670 --> 00:43:35.590 And that's got to be equal to my mass times the acceleration 00:43:35.590 --> 00:43:38.580 of a with respect to o. 00:43:38.580 --> 00:43:40.380 And we know that's straight down the hill. 00:43:40.380 --> 00:43:45.770 So this one's only going to be in the i hat direction. 00:43:45.770 --> 00:43:47.865 We know the acceleration perpendicular 00:43:47.865 --> 00:43:50.080 hill has got to be 0. 00:43:50.080 --> 00:43:56.110 OK, so if we just add up the components of this thing. 00:43:56.110 --> 00:44:00.360 In the i hat direction, just the x direction, 00:44:00.360 --> 00:44:06.300 then I have only this term and that term. 00:44:06.300 --> 00:44:15.630 And I find that the mg sine theta equals. 00:44:25.630 --> 00:44:31.300 And so the acceleration of a with respect to o 00:44:31.300 --> 00:44:34.630 is just g sine theta. 00:44:34.630 --> 00:44:37.370 OK, well that's pretty trivial. 00:44:37.370 --> 00:44:47.910 And it doesn't give us much insight about-- that 00:44:47.910 --> 00:44:50.300 doesn't give me much insight about why that's the answer. 00:44:53.460 --> 00:44:55.630 So actually this is a time when maybe thinking 00:44:55.630 --> 00:44:59.180 about if you're in that free, you're going down the hill, 00:44:59.180 --> 00:45:02.760 you're skiing, and you're now in an accelerating reference 00:45:02.760 --> 00:45:05.220 frame, and you want to say I want 00:45:05.220 --> 00:45:09.300 to work from this frame, what fictitious force do 00:45:09.300 --> 00:45:11.570 I have to add to that free body diagram 00:45:11.570 --> 00:45:14.315 so I can account for my acceleration? 00:45:33.500 --> 00:45:35.720 So remember the fictitious force we 00:45:35.720 --> 00:45:39.190 want to deal with has to do with this guy. 00:45:39.190 --> 00:45:42.470 We're in the moving frame. 00:45:42.470 --> 00:45:43.880 We can't judge this. 00:45:43.880 --> 00:45:46.540 We're just told that it exists and we need to correct for it. 00:45:46.540 --> 00:45:48.420 So we need a fictitious force that's 00:45:48.420 --> 00:45:52.000 minus ma with respect to o. 00:45:52.000 --> 00:45:56.050 So that would be minus mg sine theta. 00:45:56.050 --> 00:45:57.943 And that would be in this direction. 00:45:57.943 --> 00:46:03.870 So dashed lines here is my mg sine theta fictitious force. 00:46:08.740 --> 00:46:12.840 In the other direction was the real force, the component 00:46:12.840 --> 00:46:16.140 of gravity mg sine theta. 00:46:16.140 --> 00:46:19.980 In this direction was my normal force. 00:46:19.980 --> 00:46:24.630 And in this direction is my component of gravity 00:46:24.630 --> 00:46:28.445 that is my mg cosine theta. 00:46:33.350 --> 00:46:34.360 It's a real force. 00:46:34.360 --> 00:46:37.810 So yeah, these real forces and this fictitious force. 00:46:37.810 --> 00:46:42.860 But now we can say that all these forces have to sum to 0. 00:46:45.710 --> 00:46:48.960 But look what happens here. 00:46:48.960 --> 00:46:52.940 What force do you feel? 00:46:52.940 --> 00:46:56.320 In the reference frame, you're the skier going down the hill. 00:46:56.320 --> 00:47:01.030 What net forces do you feel on you that are in the x direction 00:47:01.030 --> 00:47:01.680 down the hill? 00:47:06.960 --> 00:47:09.370 0. 00:47:09.370 --> 00:47:10.270 It's exactly. 00:47:10.270 --> 00:47:13.300 The reason I did that silly free fall problem a minute 00:47:13.300 --> 00:47:16.385 ago when you drop something, when you're weightless, 00:47:16.385 --> 00:47:18.760 the scales are reading-- the cable broke in the elevator, 00:47:18.760 --> 00:47:20.190 scales read 0. 00:47:20.190 --> 00:47:23.030 The fictitious force was equal and opposite 00:47:23.030 --> 00:47:25.150 to the gravitational force you feel 00:47:25.150 --> 00:47:30.290 and the scales would show no force in the free fall 00:47:30.290 --> 00:47:30.810 direction. 00:47:30.810 --> 00:47:32.790 In this case, you're not free fall, 00:47:32.790 --> 00:47:35.010 but you're free fall down a slope. 00:47:35.010 --> 00:47:37.270 In the component down the slope, you 00:47:37.270 --> 00:47:38.880 have no net force in that direction. 00:47:42.866 --> 00:47:46.410 In the reference frame that moving cart, 00:47:46.410 --> 00:47:50.880 the only actual force you feel is in. 00:47:50.880 --> 00:47:58.610 You will feel a force on your feet equal to mg cosine theta. 00:47:58.610 --> 00:48:03.030 So if theta goes to 90 degrees, you're in free fall, 00:48:03.030 --> 00:48:05.380 and it goes to 0. 00:48:05.380 --> 00:48:08.740 Say it is 0 degrees, it's max, and you would feel mg. 00:48:12.830 --> 00:48:15.042 This is [INAUDIBLE] me where this notion 00:48:15.042 --> 00:48:16.500 of a fictitious force kind of helps 00:48:16.500 --> 00:48:18.310 me figure out what's going on. 00:48:22.450 --> 00:48:24.550 Pretty amazing. 00:48:24.550 --> 00:48:25.280 OK. 00:48:25.280 --> 00:48:29.380 So we're going to move on from this kind of discussion 00:48:29.380 --> 00:48:31.770 of fictitious forces of things. 00:48:31.770 --> 00:48:35.150 So it's been an hour. 00:48:35.150 --> 00:48:37.526 We'll take a little break for a second. 00:48:37.526 --> 00:48:38.900 Also, if you've got any questions 00:48:38.900 --> 00:48:41.439 about what we've covered up to this point, think about them 00:48:41.439 --> 00:48:42.480 and ask them in a minute. 00:48:42.480 --> 00:48:44.490 Let's take a short break. 00:48:44.490 --> 00:48:47.510 So while you've been taking a break, a couple of people 00:48:47.510 --> 00:48:48.870 came up and asked questions. 00:48:48.870 --> 00:48:53.060 One of them was, are fictitious forces real? 00:48:53.060 --> 00:48:55.915 And the other is, why does the water move? 00:49:00.950 --> 00:49:04.830 So the answer to the first question. 00:49:04.830 --> 00:49:11.520 If you want to be careful and do the problem rigorously, 00:49:11.520 --> 00:49:15.650 don't mess with fictitious forces. 00:49:15.650 --> 00:49:19.910 The only real forces in the problem 00:49:19.910 --> 00:49:24.530 are the ones that you would put on a free body diagram, such 00:49:24.530 --> 00:49:26.690 that the sum of the external forces 00:49:26.690 --> 00:49:29.230 equals the mass times acceleration. 00:49:29.230 --> 00:49:31.470 These forces are always real ones. 00:49:31.470 --> 00:49:33.100 They are gravity. 00:49:33.100 --> 00:49:34.360 They are normal forces. 00:49:34.360 --> 00:49:35.510 They're friction. 00:49:35.510 --> 00:49:36.480 They're spring forces. 00:49:36.480 --> 00:49:37.780 They're dashpot forces. 00:49:37.780 --> 00:49:39.320 They're all these mechanical things, 00:49:39.320 --> 00:49:41.560 but they're real forces. 00:49:41.560 --> 00:49:44.530 And the rod, that thing spinning around, 00:49:44.530 --> 00:49:48.000 that rod, since it's a rigid thing, it can support bending. 00:49:48.000 --> 00:49:52.200 So it has shear force, axial forces. 00:49:52.200 --> 00:49:55.780 Those are real forces that it can put on the mass. 00:49:55.780 --> 00:49:59.050 Fictitious forces are not forces. 00:49:59.050 --> 00:50:05.885 They are minus a mass times a real what? 00:50:05.885 --> 00:50:06.426 Acceleration. 00:50:09.290 --> 00:50:11.820 So real forces, real accelerations. 00:50:11.820 --> 00:50:14.585 We use this notion of a fictitious force 00:50:14.585 --> 00:50:16.370 as a convenience. 00:50:16.370 --> 00:50:20.547 But it is a highly dangerous tool. 00:50:20.547 --> 00:50:22.630 But actually once you get used to it a little bit, 00:50:22.630 --> 00:50:24.540 it can give you some really quick insight. 00:50:24.540 --> 00:50:26.950 But I would always, if I have doubts, I'd always go back 00:50:26.950 --> 00:50:29.380 and just do the problem the rigorous way and just 00:50:29.380 --> 00:50:32.400 check that your insight was right. 00:50:32.400 --> 00:50:35.040 But it can give you a great insight to starting a problem. 00:50:35.040 --> 00:50:39.085 You know to expect a centrifugal term in your results. 00:50:39.085 --> 00:50:42.495 You know to expect a Coriolis term. 00:50:42.495 --> 00:50:51.390 You know that they're going to be there because sometimes you 00:50:51.390 --> 00:50:52.632 feel them. 00:50:52.632 --> 00:50:55.400 The reason we like to use them is because you actually 00:50:55.400 --> 00:50:58.310 can feel them. 00:50:58.310 --> 00:51:01.120 They actually affect what you feel. 00:51:01.120 --> 00:51:17.040 So in that first problem, the elevator problem, the answer-- 00:51:17.040 --> 00:51:19.680 I didn't actually write it. 00:51:19.680 --> 00:51:21.180 Here it is. 00:51:21.180 --> 00:51:24.490 The normal force that you feel, that the scale weighs 00:51:24.490 --> 00:51:26.570 is 5 force g. 00:51:26.570 --> 00:51:30.073 And it's the sum of gravity and that fictitious force. 00:51:30.073 --> 00:51:34.160 It's mg plus m times a quarter of a g. 00:51:38.214 --> 00:51:40.630 It's in this moving references, this accelerator reference 00:51:40.630 --> 00:51:43.080 frame inside an elevator where you can't directly 00:51:43.080 --> 00:51:44.220 measure the acceleration. 00:51:44.220 --> 00:51:47.050 All you can measure is the force. 00:51:47.050 --> 00:51:49.150 But you know what the acceleration is. 00:51:49.150 --> 00:51:49.900 It's given. 00:51:49.900 --> 00:51:53.180 So you stick it in as a fix and you find out 00:51:53.180 --> 00:51:56.250 that the real normal force that you measure actually has 00:51:56.250 --> 00:52:00.130 in it that piece comes from the fictitious force. 00:52:00.130 --> 00:52:01.100 OK. 00:52:01.100 --> 00:52:03.700 Why does the water move? 00:52:03.700 --> 00:52:07.190 This is really the same discussion. 00:52:07.190 --> 00:52:11.630 The fictitious force up the hill and the real force 00:52:11.630 --> 00:52:14.490 down the hill exactly cancel. 00:52:14.490 --> 00:52:17.340 If you're standing on scales, it actually 00:52:17.340 --> 00:52:21.980 can measure side force as well as normal force. 00:52:21.980 --> 00:52:26.520 Standing on the scales, you would feel no side forces 00:52:26.520 --> 00:52:27.680 up and down the hill. 00:52:27.680 --> 00:52:29.470 None. 00:52:29.470 --> 00:52:32.950 You measure a normal force and the normal force you measure 00:52:32.950 --> 00:52:37.110 will be mg cosine theta. 00:52:37.110 --> 00:52:43.840 So the force on any object in that accelerating reference 00:52:43.840 --> 00:52:46.890 frame, it essentially thinks that all forces are 00:52:46.890 --> 00:52:50.040 perpendicular, that you feel in that reference frame, 00:52:50.040 --> 00:52:52.850 are perpendicular to the hill. 00:52:52.850 --> 00:52:57.900 So why does water seek level? 00:52:57.900 --> 00:53:00.630 Because the force gravity is down 00:53:00.630 --> 00:53:02.930 and the surface of the water goes perpendicular 00:53:02.930 --> 00:53:04.220 to the gravitational force. 00:53:04.220 --> 00:53:06.980 That's just the nature of fluids. 00:53:06.980 --> 00:53:11.490 So if the force on all those fluid particles is this angle, 00:53:11.490 --> 00:53:13.580 what's the fluid level going to do? 00:53:13.580 --> 00:53:17.130 So that the pressure on the surface is equal everywhere. 00:53:17.130 --> 00:53:20.090 And if you go a foot down, you get equal pressure 00:53:20.090 --> 00:53:21.440 everywhere in the fluid. 00:53:29.460 --> 00:53:32.300 So the water moves so that its surface 00:53:32.300 --> 00:53:35.820 will be perpendicular to what it thinks. 00:53:35.820 --> 00:53:38.285 It thinks that this is kind of being a local gravity. 00:53:38.285 --> 00:53:40.660 The local gravity is pointed in that direction, 00:53:40.660 --> 00:53:43.880 so the surface level seeks that level. 00:53:43.880 --> 00:53:46.440 If somebody else can say it better than that. 00:53:46.440 --> 00:53:48.420 Tom, Dave, Matt, anybody? 00:53:54.694 --> 00:53:55.662 Yeah. 00:53:55.662 --> 00:53:57.224 AUDIENCE: Can I ask you a question? 00:53:57.224 --> 00:53:57.890 PROFESSOR: Yeah. 00:54:01.422 --> 00:54:02.338 AUDIENCE: [INAUDIBLE]. 00:54:19.160 --> 00:54:21.490 PROFESSOR: Yeah, as if you're on the body 00:54:21.490 --> 00:54:23.800 and you're trying to sum up all the forces you feel 00:54:23.800 --> 00:54:27.480 and accounting also for the fact that this is accelerating. 00:54:27.480 --> 00:54:31.090 And the fix it force is this fictitious force that's 00:54:31.090 --> 00:54:33.750 due to the body's acceleration. 00:54:33.750 --> 00:54:35.550 You have to know that you're accelerating 00:54:35.550 --> 00:54:38.870 to be able to put this term in. 00:54:38.870 --> 00:54:42.530 And it comes from this equation, thinking 00:54:42.530 --> 00:54:44.820 of this as a generic equation by taking this term 00:54:44.820 --> 00:54:46.944 and just moving it to this side and setting 00:54:46.944 --> 00:54:47.860 everything equal to 0. 00:54:54.600 --> 00:54:56.668 Lots of puzzlement around this. 00:54:56.668 --> 00:54:59.536 AUDIENCE: What happens if you make the angle of the ramp very 00:54:59.536 --> 00:55:00.754 steep? 00:55:00.754 --> 00:55:01.670 PROFESSOR: Very steep. 00:55:01.670 --> 00:55:05.380 So if it goes to 90 degrees, you're in free fall. 00:55:09.380 --> 00:55:13.350 This cosine theta goes to 0, n goes to 0. 00:55:13.350 --> 00:55:14.580 This term stays the same. 00:55:14.580 --> 00:55:16.360 This goes to mg that way. 00:55:16.360 --> 00:55:17.540 This goes mg that way. 00:55:17.540 --> 00:55:18.915 And it's an equilibrium in both-- 00:55:18.915 --> 00:55:19.831 AUDIENCE: [INAUDIBLE]. 00:55:21.600 --> 00:55:23.100 PROFESSOR: Then you're going to have 00:55:23.100 --> 00:55:25.140 very, very small, normal force. 00:55:27.378 --> 00:55:28.294 AUDIENCE: [INAUDIBLE]. 00:55:31.452 --> 00:55:33.285 PROFESSOR: I couldn't hear the last comment. 00:55:33.285 --> 00:55:35.710 AUDIENCE: The water wouldn't go all the way to the side. 00:55:35.710 --> 00:55:37.650 It'd be vertical length [INAUDIBLE]. 00:55:40.570 --> 00:55:43.740 PROFESSOR: Well, water, unfortunately, 00:55:43.740 --> 00:55:48.460 with the tiniest little bit of normal force here 00:55:48.460 --> 00:55:49.950 the water will seek that level. 00:55:49.950 --> 00:55:57.210 But when it gets to 0 the water is going to be in free fall. 00:55:57.210 --> 00:55:58.940 Surface tension takes over. 00:55:58.940 --> 00:55:59.910 Yeah? 00:55:59.910 --> 00:56:03.305 AUDIENCE: Is this like the argument that [INAUDIBLE] 00:56:03.305 --> 00:56:06.215 is the same as if it was just on a flat surface accelerating? 00:56:09.872 --> 00:56:11.080 PROFESSOR: Oh, good question. 00:56:11.080 --> 00:56:14.810 If you were on a flat surface accelerating. 00:56:14.810 --> 00:56:19.110 If you were accelerating in this cart. 00:56:19.110 --> 00:56:21.540 Accelerates with the fluid on it and it's 00:56:21.540 --> 00:56:24.590 accelerating in that direction, what 00:56:24.590 --> 00:56:26.160 would the surface of the liquid do? 00:56:29.050 --> 00:56:32.610 Stay level or change? 00:56:32.610 --> 00:56:35.620 OK, does it slope back or does it slope forward? 00:56:35.620 --> 00:56:38.120 I'm accelerating in that direction. 00:56:38.120 --> 00:56:45.020 There's a fictitious force pushing back on it, 00:56:45.020 --> 00:56:48.590 makes it pile up at the back. 00:56:48.590 --> 00:56:50.800 OK one more. 00:56:50.800 --> 00:56:52.566 You're in the elevator. 00:56:52.566 --> 00:56:54.415 You're accelerating a quarter of a g. 00:56:58.070 --> 00:57:03.440 The scales tells you you weigh 25% more than you used to. 00:57:03.440 --> 00:57:04.490 And you have a pendulum. 00:57:04.490 --> 00:57:06.770 And before the elevator starts, it's 00:57:06.770 --> 00:57:10.510 got a one second natural period. 00:57:10.510 --> 00:57:11.860 And now the elevator gets going. 00:57:11.860 --> 00:57:14.820 Does the period change? 00:57:14.820 --> 00:57:15.800 Higher or lower? 00:57:15.800 --> 00:57:16.716 AUDIENCE: [INAUDIBLE]. 00:57:21.045 --> 00:57:22.670 PROFESSOR: OK, so the natural frequency 00:57:22.670 --> 00:57:25.044 of the pendulum, simple pendulum square root of g over l. 00:57:25.044 --> 00:57:27.630 What do you guess that the natural frequency is if you're 00:57:27.630 --> 00:57:29.950 going up at a quarter of a g? 00:57:34.855 --> 00:57:35.530 Say again? 00:57:35.530 --> 00:57:36.446 AUDIENCE: [INAUDIBLE]. 00:57:44.890 --> 00:57:49.590 PROFESSOR: It's as if you have a higher or lower acceleration 00:57:49.590 --> 00:57:52.880 of gravity around. 00:57:52.880 --> 00:57:54.610 But you have a higher gravitational force 00:57:54.610 --> 00:57:57.070 is what you think you're feeling. 00:57:57.070 --> 00:58:00.050 Is that going to make the tension in this string higher? 00:58:00.050 --> 00:58:02.690 Is that going to make the weight of the pendulum feel greater? 00:58:02.690 --> 00:58:04.900 Yeah, by 5 force. 00:58:04.900 --> 00:58:06.630 And you'll find out, you'll work out. 00:58:06.630 --> 00:58:09.915 It gets kind of messy to prove this rigorously, 00:58:09.915 --> 00:58:12.415 but it's going to look something like the square root of 5/4 00:58:12.415 --> 00:58:12.965 g over l. 00:58:15.610 --> 00:58:16.110 OK. 00:58:31.682 --> 00:58:32.890 We've got a little time left. 00:58:32.890 --> 00:58:35.300 I might not be able to completely finish this. 00:58:35.300 --> 00:58:36.900 But I do want to show you something. 00:58:59.321 --> 00:59:00.570 Hold it up so they can see it. 00:59:06.157 --> 00:59:07.490 Gets pretty violent, doesn't it? 00:59:07.490 --> 00:59:08.280 Pass it around. 00:59:16.120 --> 00:59:17.110 OK. 00:59:17.110 --> 00:59:18.975 This thing is shaking like crazy. 00:59:23.515 --> 00:59:24.015 All right. 00:59:30.300 --> 00:59:30.800 All right. 00:59:30.800 --> 00:59:33.022 In the absence of external forces, 00:59:33.022 --> 00:59:34.980 what can you say about the motion of its center 00:59:34.980 --> 00:59:36.490 of gravity, do you think? 00:59:36.490 --> 00:59:39.760 In at least this direction. 00:59:39.760 --> 00:59:41.160 Can't move, right? 00:59:41.160 --> 00:59:44.310 Up and down direction the center of gravity 00:59:44.310 --> 00:59:48.120 does move, because I'm putting a force on it through the cord. 00:59:48.120 --> 00:59:52.360 This has a relation to a problem that you 00:59:52.360 --> 00:59:56.289 had on problem set two. 00:59:56.289 --> 00:59:58.330 I want to go back and talk about it a little bit. 01:00:06.920 --> 01:00:09.900 And this is that thing with the track in it. 01:00:09.900 --> 01:00:11.550 And a roller going around. 01:00:11.550 --> 01:00:13.415 And it's going to around at constant speed. 01:00:13.415 --> 01:00:15.310 And this has mass m. 01:00:15.310 --> 01:00:16.990 This is mass. 01:00:16.990 --> 01:00:20.720 The outside thing is just the mass of this block. 01:00:20.720 --> 01:00:22.991 And there's an internal donut in here, [INAUDIBLE], 01:00:22.991 --> 01:00:24.865 and this mass is made to go around and around 01:00:24.865 --> 01:00:26.750 and around and around inside. 01:00:26.750 --> 01:00:33.500 And you were asked to figure out the acceleration of that guy. 01:00:33.500 --> 01:00:35.058 But this thing's on wheels. 01:00:46.780 --> 01:00:52.380 Now, this problem, and I apologize 01:00:52.380 --> 01:00:55.730 for posing this problem the way I did. 01:00:55.730 --> 01:00:59.510 I should have posed in the following way. 01:00:59.510 --> 01:01:04.910 These two problems are really equal to one another. 01:01:04.910 --> 01:01:06.720 I have a rod. 01:01:06.720 --> 01:01:08.850 I have a mass on the end. 01:01:08.850 --> 01:01:12.680 This radius here we said was e. 01:01:12.680 --> 01:01:14.370 This radius is e. 01:01:14.370 --> 01:01:15.670 The mass is the same. 01:01:15.670 --> 01:01:16.980 It's m. 01:01:16.980 --> 01:01:20.530 This was going around at omega. 01:01:20.530 --> 01:01:24.580 This is going around at omega. 01:01:24.580 --> 01:01:25.920 This is on wheels. 01:01:25.920 --> 01:01:30.970 Exactly these two problems are identical. 01:01:30.970 --> 01:01:33.294 Because if the geometry's the same, 01:01:33.294 --> 01:01:34.960 the rotation rate's the same, the masses 01:01:34.960 --> 01:01:37.770 are the same, the acceleration of this bead here, 01:01:37.770 --> 01:01:39.790 this roller is the same in this problem 01:01:39.790 --> 01:01:41.880 as it is in that problem. 01:01:41.880 --> 01:01:43.600 Just in this problem it's much easier 01:01:43.600 --> 01:01:46.360 to figure out the forces. 01:01:46.360 --> 01:01:48.100 This turned out to be kind of nasty. 01:01:48.100 --> 01:01:49.840 And I said last time that the solution 01:01:49.840 --> 01:01:53.430 to b part, which is the normal force that this 01:01:53.430 --> 01:01:56.450 is put on this roller, I said that answer was wrong. 01:01:56.450 --> 01:01:59.790 I actually am wrong about that. 01:01:59.790 --> 01:02:01.480 I was thinking of a different answer. 01:02:01.480 --> 01:02:04.780 I was thinking of this problem. 01:02:04.780 --> 01:02:08.590 The answer that was given, the normal force that 01:02:08.590 --> 01:02:11.770 is by the track places on the roller or vice 01:02:11.770 --> 01:02:15.390 versa is actually what was given is correct. 01:02:15.390 --> 01:02:17.070 It wasn't what I actually was seeking 01:02:17.070 --> 01:02:18.111 when I wrote the problem. 01:02:18.111 --> 01:02:20.290 Sometimes when you're making up problems, 01:02:20.290 --> 01:02:23.040 you have an end result in mind and then you wrote it wrong. 01:02:23.040 --> 01:02:24.560 I wrote it wrong. 01:02:24.560 --> 01:02:25.950 So let's do this problem. 01:02:29.470 --> 01:02:34.550 This is partly to review a couple of things for you. 01:02:34.550 --> 01:02:40.700 What I really want is I want to find equation of motion 01:02:40.700 --> 01:02:44.549 in-- [INAUDIBLE]. 01:02:50.760 --> 01:02:54.980 So here's my xyz inertial reference frame. 01:02:54.980 --> 01:02:57.284 I just lined it up here with the center 01:02:57.284 --> 01:02:58.450 of this thing to start with. 01:02:58.450 --> 01:02:59.780 Doesn't really matter. 01:02:59.780 --> 01:03:02.570 So I want to find an equation of motion in the x direction. 01:03:06.049 --> 01:03:08.090 We don't have time to do this whole problem here, 01:03:08.090 --> 01:03:12.850 so I'm going to do the first piece of it. 01:03:12.850 --> 01:03:15.910 Just to give you some review, an exercise in figuring out 01:03:15.910 --> 01:03:18.910 degrees of freedom. 01:03:18.910 --> 01:03:21.075 I'm implying that I can write one equation. 01:03:21.075 --> 01:03:23.140 It may take more than that. 01:03:23.140 --> 01:03:27.230 So the number of degrees of freedom we said 01:03:27.230 --> 01:03:35.122 is equal to the number of required-- 01:03:35.122 --> 01:03:46.550 I can't write today-- independent coordinates. 01:03:46.550 --> 01:03:50.540 So the number of coordinates required to completely describe 01:03:50.540 --> 01:03:51.710 the motion. 01:03:51.710 --> 01:03:54.120 And we came up a little expression for that. 01:03:54.120 --> 01:03:56.890 It's 6 times the number rigid bodies 01:03:56.890 --> 01:04:00.700 plus 3 times the number of particles. 01:04:00.700 --> 01:04:04.052 So this is rigid bodies. 01:04:04.052 --> 01:04:05.195 This is particles. 01:04:11.650 --> 01:04:13.360 Minus c. 01:04:13.360 --> 01:04:14.610 And these are the constraints. 01:04:19.430 --> 01:04:21.450 So let's figure that out for this problem. 01:04:37.002 --> 01:04:38.390 So how many constraints? 01:04:38.390 --> 01:04:41.650 It's just a quick practice. 01:04:41.650 --> 01:04:43.880 So constraints and I'm going to start off 01:04:43.880 --> 01:04:50.370 with on the main block. 01:04:50.370 --> 01:04:56.710 So on the main block, there's no y motion. 01:04:56.710 --> 01:04:57.930 There's no z motion. 01:04:57.930 --> 01:05:00.450 These rollers don't let it come in or out. 01:05:00.450 --> 01:05:06.385 So no y or z motion. 01:05:09.557 --> 01:05:10.640 So that's two constraints. 01:05:13.600 --> 01:05:15.750 What about rotations on the main block? 01:05:15.750 --> 01:05:19.670 How many constraints are we assuming here? 01:05:19.670 --> 01:05:20.540 Three. 01:05:20.540 --> 01:05:23.480 We're not allowing it the role in x, the y, or the z. 01:05:33.238 --> 01:05:36.400 So there's three more constraints. 01:05:36.400 --> 01:05:38.215 And now on the little mass. 01:05:42.090 --> 01:05:44.590 On the little mass, there's no z motion. 01:05:44.590 --> 01:05:45.590 That's out of the board. 01:05:54.200 --> 01:05:56.110 That's one more constraint. 01:05:56.110 --> 01:06:16.030 And then omega t is the angle that this thing. 01:06:16.030 --> 01:06:17.980 It's theta. 01:06:17.980 --> 01:06:18.760 This is specified. 01:06:25.704 --> 01:06:26.870 It's a given in the problem. 01:06:26.870 --> 01:06:27.589 It's not unknown. 01:06:27.589 --> 01:06:29.130 It's just that's what's going, that's 01:06:29.130 --> 01:06:30.720 what makes this thing do its thing. 01:06:30.720 --> 01:06:32.570 It's going round and round. 01:06:32.570 --> 01:06:35.130 And so this is not an unknown. 01:06:35.130 --> 01:06:47.410 And therefore you can say that the y motion of this particle 01:06:47.410 --> 01:06:52.030 is e sine omega t. 01:06:55.180 --> 01:06:57.320 And that's a constraint. 01:06:57.320 --> 01:07:07.820 And the x motion is the motion of the main block 01:07:07.820 --> 01:07:17.260 to which it's attached plus an e cosine omega t. 01:07:17.260 --> 01:07:19.187 And these are knowns. 01:07:19.187 --> 01:07:20.520 You've already counted for this. 01:07:20.520 --> 01:07:23.150 That's the main mass's motion. 01:07:23.150 --> 01:07:25.190 So this is something known. 01:07:25.190 --> 01:07:27.840 So these mount up, each one of these. 01:07:27.840 --> 01:07:32.946 This gives you a constraint and this gives you a constraint. 01:07:32.946 --> 01:07:34.690 And so let's add them up. 01:07:34.690 --> 01:07:36.970 So 2 and 3 are 5. 01:07:36.970 --> 01:07:41.390 6, 7, 8. 01:07:41.390 --> 01:07:44.150 So you can actually completely describe 01:07:44.150 --> 01:07:48.185 the motion of this thing with one equation of motion. 01:08:00.480 --> 01:08:01.540 So I'm seeking. 01:08:01.540 --> 01:08:10.593 So I want to find the sum of the external forces 01:08:10.593 --> 01:08:19.310 on the main block in the x direction. 01:08:19.310 --> 01:08:21.050 And that's going to be. 01:08:21.050 --> 01:08:25.245 We know that that's the mass of that block times x double dot. 01:08:25.245 --> 01:08:28.270 So x is going to be the coordinate that describes 01:08:28.270 --> 01:08:30.620 the motion that mass mb. 01:08:37.640 --> 01:08:41.010 So we need a free body diagram of mb. 01:08:55.150 --> 01:08:56.370 So here's the block. 01:08:59.850 --> 01:09:03.030 There'll be a normal force up. 01:09:03.030 --> 01:09:08.775 There's going to be mbg downwards. 01:09:08.775 --> 01:09:12.640 And the only other forces acting on this block 01:09:12.640 --> 01:09:22.090 are a force that comes through this rod that's holding 01:09:22.090 --> 01:09:23.659 that mass, the little mass. 01:09:34.100 --> 01:09:36.660 I'm just going to draw it at some arbitrary angle, 01:09:36.660 --> 01:09:39.130 not necessarily theta. 01:09:39.130 --> 01:09:41.460 It's going to be a force. 01:09:41.460 --> 01:09:48.229 I'm going to call that force on this main mass. 01:09:48.229 --> 01:09:49.630 And it's a vector for the moment. 01:09:49.630 --> 01:09:51.955 It'll have two components in the x and y directions. 01:10:02.200 --> 01:10:05.280 So in order to write an equation-- in the y direction 01:10:05.280 --> 01:10:10.520 it's trivial. n is equal to mbg. 01:10:10.520 --> 01:10:14.240 In this direction, we have to have to break this 01:10:14.240 --> 01:10:15.320 into two components. 01:10:15.320 --> 01:10:19.240 We need to know the component of this in the x direction 01:10:19.240 --> 01:10:23.800 and put it in here and we'll be done. 01:10:23.800 --> 01:10:32.570 So now I'm going to use third law. 01:10:32.570 --> 01:10:39.420 The force that rod places on this body 01:10:39.420 --> 01:10:43.400 is equal and opposite to the force at the rod 01:10:43.400 --> 01:10:46.900 places on the mass going round and round. 01:10:46.900 --> 01:10:47.650 You agree with me? 01:10:51.010 --> 01:10:55.060 So what we need to do next is here's 01:10:55.060 --> 01:10:58.500 the mass up here going around and round. 01:10:58.500 --> 01:11:07.060 And the force is on it, because it has an mg downward. 01:11:07.060 --> 01:11:12.350 And it has some force on it that is 01:11:12.350 --> 01:11:15.700 the force on the little mass. 01:11:15.700 --> 01:11:19.170 And you're going to find that f on the little mass 01:11:19.170 --> 01:11:27.060 equals minus f on the big mass due to third law consideration. 01:11:27.060 --> 01:11:30.050 So all I have to do is then figure out 01:11:30.050 --> 01:11:34.350 the x component of this force, put a minus sign in front 01:11:34.350 --> 01:11:37.900 of it, and I'm basically done. 01:11:37.900 --> 01:11:39.870 And to do that I say f equals ma. 01:11:39.870 --> 01:11:41.210 I don't have to finish it today. 01:11:41.210 --> 01:11:42.870 I'll show you next time. 01:11:42.870 --> 01:11:47.596 And we'll figure out the equation of motion 01:11:47.596 --> 01:11:48.220 for this thing. 01:11:48.220 --> 01:11:51.590 Now, in dragging all this stuff here today, 01:11:51.590 --> 01:11:54.532 I forgot to bring my muddy cards. 01:11:54.532 --> 01:11:55.490 So you've got a minute. 01:11:55.490 --> 01:11:57.160 And if any of you have just a brilliant question 01:11:57.160 --> 01:11:58.909 you want to answer or something you really 01:11:58.909 --> 01:12:02.130 liked, if you don't mind, just take a piece of paper, 01:12:02.130 --> 01:12:05.310 make a note on it, and drop it off on the way out. 01:12:05.310 --> 01:12:05.960 Thanks a lot. 01:12:05.960 --> 01:12:08.320 See you on Thursday.