1 00:00:00,070 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,810 Commons license. 3 00:00:03,810 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,680 To make a donation or to view additional materials 6 00:00:12,680 --> 00:00:16,600 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,600 --> 00:00:17,305 at ocw.mit.edu. 8 00:00:25,820 --> 00:00:27,470 PROFESSOR: There's a reading. 9 00:00:27,470 --> 00:00:31,570 The new reading is on Stellar. 10 00:00:31,570 --> 00:00:35,180 It's an excerpt from a textbook on dynamics by Professor Jim 11 00:00:35,180 --> 00:00:39,140 Williams who was in the mechanical engineering 12 00:00:39,140 --> 00:00:43,030 department here at MIT and just recently retired. 13 00:00:43,030 --> 00:00:47,610 This reading is essentially a review of everything 14 00:00:47,610 --> 00:00:49,700 we've done in kinematics. 15 00:00:49,700 --> 00:00:51,700 So it gives you a different point of view. 16 00:00:51,700 --> 00:00:54,570 It's extremely well written. 17 00:00:54,570 --> 00:00:56,990 So it goes through things like derivatives 18 00:00:56,990 --> 00:00:58,610 of rotating vectors. 19 00:00:58,610 --> 00:01:02,440 It has three or four really terrific examples of problems 20 00:01:02,440 --> 00:01:06,430 that are solved using the techniques that we've used up 21 00:01:06,430 --> 00:01:07,710 to this point. 22 00:01:07,710 --> 00:01:11,590 So it's a good way to review for the quiz, which 23 00:01:11,590 --> 00:01:15,890 is coming up a week from Thursday, on October 14. 24 00:01:15,890 --> 00:01:19,980 That quiz will be here, closed book, one sheet 25 00:01:19,980 --> 00:01:24,070 of notes, piece of paper both sides as your reference 26 00:01:24,070 --> 00:01:25,740 material. 27 00:01:25,740 --> 00:01:27,480 OK, survey results. 28 00:01:27,480 --> 00:01:29,420 So you have p set four. 29 00:01:29,420 --> 00:01:31,690 It has I think five questions on that. 30 00:01:31,690 --> 00:01:34,910 We'll quickly take a look at the problems 31 00:01:34,910 --> 00:01:36,780 that you're working on right now. 32 00:01:36,780 --> 00:01:42,800 So here's this yo-yo like thing, a spool. 33 00:01:42,800 --> 00:01:44,550 There were some questions on nb. 34 00:01:44,550 --> 00:01:47,155 The inner piece doesn't rotate relative to the out piece. 35 00:01:47,155 --> 00:01:53,230 It's all one solid spool, but it has two different radii that 36 00:01:53,230 --> 00:01:55,230 are functioning in the problem. 37 00:01:55,230 --> 00:01:58,760 And the question is, will the spool roll to the left 38 00:01:58,760 --> 00:02:01,324 or to the right when the string is pulled to the right? 39 00:02:01,324 --> 00:02:02,740 So we're going to come back to it. 40 00:02:02,740 --> 00:02:05,073 I'm just going to run through all the questions quickly. 41 00:02:05,073 --> 00:02:07,830 The second one was this somewhat familiar problem 42 00:02:07,830 --> 00:02:11,850 you've done now. 43 00:02:11,850 --> 00:02:16,090 The question asks you about how fast 44 00:02:16,090 --> 00:02:20,450 it has to go before the thing starts sliding up. 45 00:02:20,450 --> 00:02:25,210 And the survey question, if the starting position of the spool 46 00:02:25,210 --> 00:02:29,460 is moved down the rod, will the angular rate 47 00:02:29,460 --> 00:02:32,650 be higher when spool begins to slide up the rod? 48 00:02:32,650 --> 00:02:34,820 Will you have to go faster to get it to go? 49 00:02:34,820 --> 00:02:37,400 OK, next. 50 00:02:37,400 --> 00:02:41,452 OK, this is finding the equation of motion for a pendulum. 51 00:02:41,452 --> 00:02:43,160 And everybody knows the natural frequency 52 00:02:43,160 --> 00:02:44,826 of a pendulum's square root of g over l. 53 00:02:44,826 --> 00:02:47,635 But this pendulum has a torsional spring added to it. 54 00:02:47,635 --> 00:02:50,570 And the question is, the square root of g over l 55 00:02:50,570 --> 00:02:53,080 doesn't involve the mass. 56 00:02:53,080 --> 00:02:54,660 But will the natural frequency, now 57 00:02:54,660 --> 00:02:58,160 that you've put the torsional spring on it, will it now 58 00:02:58,160 --> 00:02:59,489 involve the mass? 59 00:02:59,489 --> 00:03:00,780 No reason you should know this. 60 00:03:00,780 --> 00:03:02,950 You've never done this problem before. 61 00:03:02,950 --> 00:03:04,821 And we'll come back to that. 62 00:03:04,821 --> 00:03:05,320 Next. 63 00:03:07,809 --> 00:03:08,600 The fourth problem. 64 00:03:08,600 --> 00:03:10,460 Two degree of freedom system. 65 00:03:10,460 --> 00:03:14,670 And wants you to come up with the equation of motion. 66 00:03:14,670 --> 00:03:18,520 But the survey question, if k1 didn't exist, 67 00:03:18,520 --> 00:03:22,150 this is just two masses coupled by spring, 68 00:03:22,150 --> 00:03:25,060 this thing is no longer constrained in the x direction. 69 00:03:25,060 --> 00:03:29,450 There's no x constraints on the system. 70 00:03:29,450 --> 00:03:33,950 So if that's the case, this system still 71 00:03:33,950 --> 00:03:36,510 has two natural frequencies. 72 00:03:36,510 --> 00:03:40,100 A system that can vibrant essentially 73 00:03:40,100 --> 00:03:43,860 has as many natural frequencies as it does equations of motion. 74 00:03:43,860 --> 00:03:45,760 This one requires two equations of motion, 75 00:03:45,760 --> 00:03:50,050 but it will have one 0 natural frequency and the other one 76 00:03:50,050 --> 00:03:51,740 greater than 0. 77 00:03:51,740 --> 00:03:55,920 In the second case when it's greater than 0, 78 00:03:55,920 --> 00:03:57,750 this thing is oscillating somehow. 79 00:03:57,750 --> 00:03:58,990 It has no extra strength. 80 00:03:58,990 --> 00:04:03,460 What can you say about the position of the center of mass? 81 00:04:03,460 --> 00:04:04,490 So click at the results. 82 00:04:04,490 --> 00:04:07,010 So this is the first one is the spool problem. 83 00:04:13,420 --> 00:04:16,180 There's the spool. 84 00:04:16,180 --> 00:04:17,940 Here's table. 85 00:04:17,940 --> 00:04:19,410 Here's the inner radius. 86 00:04:19,410 --> 00:04:21,760 It's got a string wrapped on it. 87 00:04:21,760 --> 00:04:27,200 You're pulling the string this way at v in the i direction. 88 00:04:27,200 --> 00:04:35,120 This'll be o xy. 89 00:04:35,120 --> 00:04:37,600 And now the question is, when you pull on this, 90 00:04:37,600 --> 00:04:40,330 which way will the spool go? 91 00:04:40,330 --> 00:04:43,230 And most of you said it will go to the left. 92 00:04:43,230 --> 00:04:44,866 So I made this up. 93 00:04:44,866 --> 00:04:46,240 We're going to do the experiment. 94 00:04:53,560 --> 00:04:59,640 So I've wrapped Teflon tape on this spool. 95 00:04:59,640 --> 00:05:02,210 It's got a C battery in the center 96 00:05:02,210 --> 00:05:04,730 to give it a little bit of weight. 97 00:05:04,730 --> 00:05:09,750 And here's the experiment. 98 00:05:09,750 --> 00:05:13,390 And the guys in the booth I hope have in a minute 99 00:05:13,390 --> 00:05:15,535 to zoom in on it. 100 00:05:15,535 --> 00:05:17,660 I'm going to let go and I'm going to start pulling. 101 00:05:17,660 --> 00:05:18,710 Oops. 102 00:05:18,710 --> 00:05:21,690 I need to put a little more out here. 103 00:05:21,690 --> 00:05:22,500 Pull on the tape. 104 00:05:26,140 --> 00:05:29,630 You believe it? 105 00:05:29,630 --> 00:05:31,640 So I've done the reverse of what's in there. 106 00:05:31,640 --> 00:05:32,550 I'm pulling that way. 107 00:05:32,550 --> 00:05:35,680 But it goes in the same direction as you pull it. 108 00:05:35,680 --> 00:05:41,760 So the answer to this question is, it would go to the right. 109 00:05:41,760 --> 00:05:44,848 Kind of unbelievable almost, right? 110 00:05:44,848 --> 00:05:47,276 How does it do that? 111 00:05:47,276 --> 00:05:48,025 Let's do it again. 112 00:05:57,164 --> 00:05:58,650 All right. 113 00:05:58,650 --> 00:06:01,240 Now, show you something else. 114 00:06:01,240 --> 00:06:05,580 If that hasn't boggled your mind. 115 00:06:05,580 --> 00:06:11,110 Then what's it going to do if I pull straight up? 116 00:06:11,110 --> 00:06:14,310 How many think it's going to go that way? 117 00:06:14,310 --> 00:06:16,130 Raise your hand. 118 00:06:16,130 --> 00:06:18,390 How many think it's going to go that way? 119 00:06:18,390 --> 00:06:19,890 Raise your hand. 120 00:06:19,890 --> 00:06:22,330 All right. 121 00:06:22,330 --> 00:06:23,360 Sure enough. 122 00:06:23,360 --> 00:06:25,800 That means it must be some angle between in which 123 00:06:25,800 --> 00:06:27,630 it will do neither. 124 00:06:27,630 --> 00:06:28,910 It'll start to slip. 125 00:06:35,152 --> 00:06:36,144 Right about there. 126 00:06:42,697 --> 00:06:43,530 Neat little problem. 127 00:06:47,660 --> 00:06:54,270 And now doing this problem is all about computing velocities, 128 00:06:54,270 --> 00:06:56,365 using concepts like. 129 00:06:59,140 --> 00:07:05,660 So if we call this a, b, c. 130 00:07:05,660 --> 00:07:08,240 I think it's the other way around in the problem. 131 00:07:08,240 --> 00:07:09,610 Excuse me. 132 00:07:09,610 --> 00:07:14,090 A, b, c, and d. 133 00:07:21,710 --> 00:07:24,810 The key to this problem is being able to figure out 134 00:07:24,810 --> 00:07:29,990 what the velocity of this point c is. 135 00:07:29,990 --> 00:07:31,660 And the first thing to understand, 136 00:07:31,660 --> 00:07:40,120 this is the velocity of the end of the tape in a fixed 137 00:07:40,120 --> 00:07:41,622 reference frame. 138 00:07:41,622 --> 00:07:45,070 So every point on the tape moves at the same speed. 139 00:07:45,070 --> 00:07:47,010 That means where it touches right here 140 00:07:47,010 --> 00:07:49,810 is also moving at that speed. 141 00:07:49,810 --> 00:07:58,455 So the velocity of c with respect to o is to vi. 142 00:07:58,455 --> 00:08:00,925 And that's kind of the key to the problem. 143 00:08:05,140 --> 00:08:10,810 That's also equal to the velocity of d with respect 144 00:08:10,810 --> 00:08:16,538 to o plus the velocity of c with respect to d. 145 00:08:16,538 --> 00:08:21,870 And the velocity of d is this point 146 00:08:21,870 --> 00:08:23,700 of contact with the ground. 147 00:08:23,700 --> 00:08:26,090 What's that velocity? 148 00:08:26,090 --> 00:08:26,680 0. 149 00:08:26,680 --> 00:08:28,900 This is back to those instantaneous center 150 00:08:28,900 --> 00:08:30,430 of rotation things. 151 00:08:30,430 --> 00:08:33,820 0 there is what makes this equation easy. 152 00:08:33,820 --> 00:08:39,430 And then you know how to compute the velocity of a rotating 153 00:08:39,430 --> 00:08:39,929 vector. 154 00:08:39,929 --> 00:08:42,309 This one's not changing length, but it is surely 155 00:08:42,309 --> 00:08:44,290 rotating about this point, and so 156 00:08:44,290 --> 00:08:47,234 there's going to be an omega cross r here. 157 00:08:47,234 --> 00:08:48,900 And that'll allow you to solve for omega 158 00:08:48,900 --> 00:08:50,810 and then once you know for omega then you 159 00:08:50,810 --> 00:08:55,510 can solve for a, which is the real question that was asked. 160 00:08:55,510 --> 00:08:57,410 OK. 161 00:08:57,410 --> 00:08:58,010 Next one. 162 00:09:04,830 --> 00:09:07,235 So the starting position, this is a spool. 163 00:09:11,200 --> 00:09:14,140 There's the arm. 164 00:09:14,140 --> 00:09:19,830 And here, the question is at 0.25 meters up here. 165 00:09:19,830 --> 00:09:26,840 And if you move it down to 0.15 will 166 00:09:26,840 --> 00:09:30,730 the system be able to go faster before it breaks loose or not? 167 00:09:30,730 --> 00:09:33,190 And your answer was most people said yes. 168 00:09:33,190 --> 00:09:36,120 And basically that's true. 169 00:09:36,120 --> 00:09:39,260 What is it that's driving that thing, 170 00:09:39,260 --> 00:09:43,660 providing the force that is trying 171 00:09:43,660 --> 00:09:44,950 to push it up the sleeve? 172 00:09:48,534 --> 00:09:49,450 AUDIENCE: [INAUDIBLE]. 173 00:09:52,450 --> 00:09:55,110 PROFESSOR: So the rod as it's going around 174 00:09:55,110 --> 00:09:57,280 is trying to drive that thing in a circle. 175 00:09:57,280 --> 00:10:02,930 And therefore, it is being accelerated 176 00:10:02,930 --> 00:10:04,020 to the central points. 177 00:10:04,020 --> 00:10:06,430 So it has centripetal acceleration. 178 00:10:06,430 --> 00:10:11,750 And that requires forces and they come through the rod. 179 00:10:11,750 --> 00:10:15,440 And that will develop the normal forces and friction. 180 00:10:15,440 --> 00:10:22,160 So when you slide down the rod some to that new position, 181 00:10:22,160 --> 00:10:24,870 is has that centripetal acceleration gone up or down. 182 00:10:28,270 --> 00:10:29,690 r's gotten smaller. 183 00:10:29,690 --> 00:10:33,080 Centripetal acceleration is? 184 00:10:33,080 --> 00:10:35,520 r omega squared, right? 185 00:10:35,520 --> 00:10:37,510 So you might be speaking algebraically. 186 00:10:37,510 --> 00:10:41,900 So the magnitude of that centripetal acceleration 187 00:10:41,900 --> 00:10:46,700 has gone up or down if you bring it closer to the axis. 188 00:10:46,700 --> 00:10:47,490 Down. 189 00:10:47,490 --> 00:10:51,450 So you can go faster to get up to the same level 190 00:10:51,450 --> 00:10:53,341 of acceleration or it'll break free. 191 00:10:53,341 --> 00:10:53,840 Next. 192 00:10:57,130 --> 00:11:00,820 This is, oh, the natural frequency of the pendulum. 193 00:11:00,820 --> 00:11:03,479 Some of you said yes, some of you said no, some not sure. 194 00:11:03,479 --> 00:11:04,520 You're going to find out. 195 00:11:04,520 --> 00:11:06,561 You're going to figure out the equation of motion 196 00:11:06,561 --> 00:11:07,240 of the system. 197 00:11:07,240 --> 00:11:10,565 And it includes that torsional spring and mass is definitely 198 00:11:10,565 --> 00:11:11,620 going to be in it. 199 00:11:11,620 --> 00:11:15,242 So look for an answer that includes-- 200 00:11:15,242 --> 00:11:19,100 that you'll see that mass is going to show up in a way 201 00:11:19,100 --> 00:11:23,260 that if you solve for the natural frequency, 202 00:11:23,260 --> 00:11:25,421 it won't leave the final equation. 203 00:11:25,421 --> 00:11:25,920 OK. 204 00:11:37,110 --> 00:11:40,240 So this is the one about the two masses. 205 00:11:40,240 --> 00:11:44,780 And if the spring were 0, the second natural frequency 206 00:11:44,780 --> 00:11:46,030 of the system. 207 00:11:46,030 --> 00:11:47,410 The first natural frequency, when 208 00:11:47,410 --> 00:11:48,784 you have a system that has what's 209 00:11:48,784 --> 00:11:52,130 called a rigid body degree of freedom. 210 00:11:52,130 --> 00:11:55,200 When this thing went, there's no k1 spring. 211 00:11:55,200 --> 00:12:01,070 This system looks like two masses on wheels connected 212 00:12:01,070 --> 00:12:02,770 by a spring and a dashpot. 213 00:12:07,510 --> 00:12:10,715 And in the x direction, there are no constraints. 214 00:12:13,280 --> 00:12:16,232 So if I set this system to vibrating, 215 00:12:16,232 --> 00:12:18,565 what can you say about the motion of its center of mass? 216 00:12:27,090 --> 00:12:29,896 So could you find the center of mass of that system? 217 00:12:29,896 --> 00:12:31,270 Got to give you the masses of it. 218 00:12:31,270 --> 00:12:36,810 But you have some m1 and an m2. 219 00:12:36,810 --> 00:12:45,150 And we'll call this the x in that direction. 220 00:12:45,150 --> 00:12:47,910 Could you determine the center of mass of the system? 221 00:12:47,910 --> 00:12:49,700 You know how to do that, right? 222 00:12:49,700 --> 00:12:52,050 If they were equal in mass, it'd be in the center. 223 00:12:52,050 --> 00:12:55,950 If this was bigger, it'd be over here or something like that. 224 00:12:55,950 --> 00:13:00,380 So we know that for a system of particles, 225 00:13:00,380 --> 00:13:02,910 the sum of the external forces on the system 226 00:13:02,910 --> 00:13:06,210 is equal to mass of the system times the acceleration 227 00:13:06,210 --> 00:13:10,020 of the what? 228 00:13:10,020 --> 00:13:12,256 Center of mass. 229 00:13:12,256 --> 00:13:14,630 What are the external forces in the horizontal direction? 230 00:13:17,400 --> 00:13:19,110 How big are they? 231 00:13:19,110 --> 00:13:19,920 0. 232 00:13:19,920 --> 00:13:21,580 Therefore, what can be the acceleration 233 00:13:21,580 --> 00:13:23,140 of the center of mass? 234 00:13:23,140 --> 00:13:23,830 0. 235 00:13:23,830 --> 00:13:26,877 This thing will actually in this second natural frequency, 236 00:13:26,877 --> 00:13:28,960 it'll sit there and just oscillate back and forth, 237 00:13:28,960 --> 00:13:31,970 one mask on one directional, one on the other. 238 00:13:31,970 --> 00:13:34,610 But its center of mass won't move. 239 00:13:34,610 --> 00:13:38,000 The other degree of freedom and the other natural frequency, 240 00:13:38,000 --> 00:13:40,060 the one that has 0 natural frequency, 241 00:13:40,060 --> 00:13:41,340 is a rigid body motion. 242 00:13:41,340 --> 00:13:42,770 No relative motion. 243 00:13:42,770 --> 00:13:44,980 Give this thing a little push, some initial momentum, 244 00:13:44,980 --> 00:13:46,472 it'll just sit their role forever. 245 00:13:46,472 --> 00:13:48,430 The mass is not moving relative to one another. 246 00:13:48,430 --> 00:13:50,475 It just rolls and rolls and rolls and rolls 247 00:13:50,475 --> 00:13:51,730 and never comes back. 248 00:13:51,730 --> 00:13:55,049 Infinite period 0 natural frequency. 249 00:13:55,049 --> 00:13:56,840 OK, we were talking about fictitious forces 250 00:13:56,840 --> 00:13:57,760 the other day. 251 00:13:57,760 --> 00:14:04,020 And there are times when they're useful. 252 00:14:04,020 --> 00:14:09,930 But they're sort of like giving a little kid a loaded gun. 253 00:14:09,930 --> 00:14:13,100 You really get yourself in trouble quickly. 254 00:14:13,100 --> 00:14:16,530 So I'm not advocating a lot of use of it. 255 00:14:16,530 --> 00:14:18,520 But it's good to have a little insight. 256 00:14:18,520 --> 00:14:21,590 One of the students after the lecture last time, 257 00:14:21,590 --> 00:14:23,950 a student mentioned to me, he said, 258 00:14:23,950 --> 00:14:29,150 the way I was taught about fictitious forces 259 00:14:29,150 --> 00:14:33,030 is that we use them to patch up a problem when 260 00:14:33,030 --> 00:14:36,560 we're trying to work in a non inertial frame. 261 00:14:36,560 --> 00:14:40,490 How can you make Newton's laws apply if you 262 00:14:40,490 --> 00:14:43,650 are in a non inertial frame? 263 00:14:43,650 --> 00:14:46,230 Otherwise an accelerating reference frame. 264 00:14:46,230 --> 00:14:48,720 So I just thought I would revisit a little bit of what 265 00:14:48,720 --> 00:14:54,110 we said the other day with explaining things that way 266 00:14:54,110 --> 00:14:55,130 a little bit. 267 00:14:55,130 --> 00:14:58,610 So it might just turn out to be familiar for you. 268 00:14:58,610 --> 00:14:59,940 And let's just look. 269 00:15:07,960 --> 00:15:12,760 Let's just go back quickly and revisit that elevator problem. 270 00:15:12,760 --> 00:15:15,570 So you remember you're here. 271 00:15:15,570 --> 00:15:17,575 You've got some mass m. 272 00:15:17,575 --> 00:15:20,260 You're standing on some scales. 273 00:15:20,260 --> 00:15:22,649 And I want to know what the scales are going to read. 274 00:15:22,649 --> 00:15:24,440 Now, we did this one quickly the other day. 275 00:15:24,440 --> 00:15:25,648 I'm just going to revisit it. 276 00:15:28,260 --> 00:15:32,710 And this accelerating, this is point A. 277 00:15:32,710 --> 00:15:35,235 And here down here is o. 278 00:15:35,235 --> 00:15:36,325 My inertial frame. 279 00:15:38,830 --> 00:15:43,040 And the acceleration of A with respect to o 280 00:15:43,040 --> 00:15:50,546 is plus a quarter of g in the j hat direction. 281 00:15:50,546 --> 00:15:52,980 OK? 282 00:15:52,980 --> 00:15:56,420 Now, when you think about this, if you're 283 00:15:56,420 --> 00:15:59,200 trying to do physics in the elevator, 284 00:15:59,200 --> 00:16:01,880 but it's an accelerating frame, so Newton's laws don't apply. 285 00:16:04,510 --> 00:16:05,780 So you're in the elevator. 286 00:16:05,780 --> 00:16:08,600 You can't see you're accelerating. 287 00:16:08,600 --> 00:16:11,700 But you could measure your weight, for example. 288 00:16:11,700 --> 00:16:18,730 So how do you fix up the equations 289 00:16:18,730 --> 00:16:22,750 so that you can predict the right answer? 290 00:16:22,750 --> 00:16:28,570 And to do that, basically you I have to add in a force, 291 00:16:28,570 --> 00:16:31,720 a fictitious force, that accounts 292 00:16:31,720 --> 00:16:35,870 for the additional forces you will 293 00:16:35,870 --> 00:16:40,046 feel in this system due to the acceleration. 294 00:16:42,750 --> 00:16:47,190 So remember the way we did this problem. 295 00:16:47,190 --> 00:16:49,470 You do this problem just in a straightforward 296 00:16:49,470 --> 00:16:56,670 way, a summation of the external forces in the y direction 297 00:16:56,670 --> 00:17:02,800 is the mass times acceleration, a with respect to o. 298 00:17:02,800 --> 00:17:05,200 And we need a free body diagram. 299 00:17:05,200 --> 00:17:12,119 In the free body diagrams, this mass here, 300 00:17:12,119 --> 00:17:18,720 you have an mg downwards and an n upwards. 301 00:17:18,720 --> 00:17:27,470 And so you'd say this is equal to n minus mg. 302 00:17:27,470 --> 00:17:33,740 Now, if you're doing this experiment inside, 303 00:17:33,740 --> 00:17:37,080 you can measure this n that's going to be your weight. 304 00:17:37,080 --> 00:17:40,010 And you would like to do this by fictitious forces, 305 00:17:40,010 --> 00:17:52,560 you say that the summation of the external real forces 306 00:17:52,560 --> 00:17:58,150 minus this mass times the acceleration of a with respect 307 00:17:58,150 --> 00:18:04,470 to o has got to be equal to 0. 308 00:18:04,470 --> 00:18:13,940 And that's then n minus mg minus mg over 4. 309 00:18:13,940 --> 00:18:15,265 Here's your fictitious force. 310 00:18:19,270 --> 00:18:22,610 And that's as if then you've come over here 311 00:18:22,610 --> 00:18:29,800 and you've added an additional force on this free body diagram 312 00:18:29,800 --> 00:18:34,270 that looks like mg over 4. 313 00:18:34,270 --> 00:18:38,990 And now you say the sum of these forces, this is the correction. 314 00:18:38,990 --> 00:18:44,000 It's allowing you now to work in this accelerating frame. 315 00:18:44,000 --> 00:18:48,260 If do this free body diagram and sum these up, 316 00:18:48,260 --> 00:18:53,220 then you will find this expression, 317 00:18:53,220 --> 00:18:55,500 which you can solve for n. 318 00:18:55,500 --> 00:18:59,730 And you'll get m5 over 4g. 319 00:19:02,167 --> 00:19:04,625 In fact, if you stood on the scales, that's what you'd get. 320 00:19:15,060 --> 00:19:17,630 All right, so this is example one. 321 00:19:17,630 --> 00:19:19,730 I want to do sort of example 1.5. 322 00:19:19,730 --> 00:19:23,735 The cable breaks on this elevator. 323 00:19:44,038 --> 00:19:49,070 Your free body diagram still has an mg on it. 324 00:19:49,070 --> 00:19:51,110 You're still inside. 325 00:19:51,110 --> 00:19:53,720 You're still measuring forces. 326 00:19:53,720 --> 00:19:56,110 And you want to use this idea of a fictitious force 327 00:19:56,110 --> 00:19:58,970 to help you sort out what forces will you 328 00:19:58,970 --> 00:20:00,570 feel inside of that elevator. 329 00:20:08,730 --> 00:20:12,330 You know how to do the answer in this straightforward way. 330 00:20:12,330 --> 00:20:19,390 The sum of external forces is minus mg. 331 00:20:19,390 --> 00:20:22,630 And that's equal to the mass times the acceleration 332 00:20:22,630 --> 00:20:24,680 of a with respect to o. 333 00:20:24,680 --> 00:20:29,545 And therefore acceleration of a with respect to o 334 00:20:29,545 --> 00:20:31,410 is just minus g. 335 00:20:37,670 --> 00:20:39,080 And you could have told me that. 336 00:20:39,080 --> 00:20:39,704 That's trivial. 337 00:20:39,704 --> 00:20:42,550 If you drop the thing, it accelerates the acceleration 338 00:20:42,550 --> 00:20:43,460 of gravity. 339 00:20:43,460 --> 00:20:46,820 So I want to know what's the-- you're 340 00:20:46,820 --> 00:20:48,220 doing the experiment inside. 341 00:20:48,220 --> 00:20:49,920 You don't know that you're accelerating. 342 00:20:49,920 --> 00:20:51,720 All you're doing is measurements. 343 00:20:51,720 --> 00:20:54,510 But you've been told how to correct it. 344 00:20:54,510 --> 00:20:58,270 So let's put it in the fictitious force, which 345 00:20:58,270 --> 00:21:01,310 is minus ma 0. 346 00:21:01,310 --> 00:21:11,820 And that means the mass times acceleration is downwards. 347 00:21:11,820 --> 00:21:13,890 So minus that. 348 00:21:13,890 --> 00:21:18,180 The fictitious force is an mg force upwards. 349 00:21:18,180 --> 00:21:19,090 And what do you feel? 350 00:21:22,280 --> 00:21:24,670 What do the scales read? 351 00:21:24,670 --> 00:21:25,550 Nothing. 352 00:21:25,550 --> 00:21:26,985 You feel absolutely nothing. 353 00:21:31,040 --> 00:21:32,440 OK. 354 00:21:32,440 --> 00:21:38,340 So just an aside kind of question, because I know this 355 00:21:38,340 --> 00:21:39,800 is confused. 356 00:21:39,800 --> 00:21:41,830 Occasionally we all get confused around this. 357 00:21:45,360 --> 00:21:49,823 In these problems that we do, is gravity an acceleration? 358 00:21:56,080 --> 00:21:58,670 We say all the time, I use the phrase 359 00:21:58,670 --> 00:22:00,006 the acceleration of gravity. 360 00:22:00,006 --> 00:22:01,630 How many times have you ever said that? 361 00:22:01,630 --> 00:22:03,254 Have you ever said that? 362 00:22:03,254 --> 00:22:04,420 The acceleration of gravity. 363 00:22:04,420 --> 00:22:05,350 What do we mean? 364 00:22:05,350 --> 00:22:06,332 What we mean is this. 365 00:22:06,332 --> 00:22:08,040 When you let something drop in free fall, 366 00:22:08,040 --> 00:22:09,950 it'll accelerate at g. 367 00:22:09,950 --> 00:22:13,630 But the way we apply when we're writing out equations 368 00:22:13,630 --> 00:22:16,940 of motion, we say the sum of the external forces 369 00:22:16,940 --> 00:22:18,480 is equal to mass times acceleration. 370 00:22:18,480 --> 00:22:20,311 Where does gravity go? 371 00:22:20,311 --> 00:22:23,250 It's a force. 372 00:22:23,250 --> 00:22:25,730 So this is another thing not to get 373 00:22:25,730 --> 00:22:30,800 confused by is to think about gravity as being 374 00:22:30,800 --> 00:22:31,570 an acceleration. 375 00:22:31,570 --> 00:22:33,410 We spend a lot of time figuring out ways 376 00:22:33,410 --> 00:22:36,680 to write accelerations in rotating, 377 00:22:36,680 --> 00:22:38,200 translating, reference frames. 378 00:22:38,200 --> 00:22:39,780 Gravity never pops up in there. 379 00:22:39,780 --> 00:22:43,760 It always comes up on the force side of the equation. 380 00:22:43,760 --> 00:22:44,260 OK. 381 00:22:48,620 --> 00:22:56,130 All right, so one last example from last time. 382 00:22:56,130 --> 00:23:00,610 But it's relevant to a problem on the homework. 383 00:23:00,610 --> 00:23:04,530 So we have this shaft spinning. 384 00:23:04,530 --> 00:23:08,830 This is the z-axis spinning at constant rate omega. 385 00:23:12,160 --> 00:23:20,740 And we'll use this is r hat and this is z k hat. 386 00:23:20,740 --> 00:23:25,340 These are the lengths r hat direction and the height z. 387 00:23:25,340 --> 00:23:27,580 We're using cylindrical coordinates. 388 00:23:27,580 --> 00:23:32,600 And now let's just use this notion of fictitious forces 389 00:23:32,600 --> 00:23:36,910 to figure out first what's the what's 390 00:23:36,910 --> 00:23:38,610 the fictitious force in this system. 391 00:23:42,819 --> 00:23:44,610 Remember, this is now this notion of you're 392 00:23:44,610 --> 00:23:46,920 out there where this mass is. 393 00:23:46,920 --> 00:23:48,620 You're riding with this mass. 394 00:23:48,620 --> 00:23:52,407 And you're trying to say, what force am I going to feel? 395 00:23:52,407 --> 00:23:54,240 You want to be able to predict it correctly. 396 00:23:54,240 --> 00:23:57,370 So you have to find a fictitious force 397 00:23:57,370 --> 00:24:01,380 that you can bring in that accounts for what 398 00:24:01,380 --> 00:24:06,000 acceleration are we feeling out there. 399 00:24:06,000 --> 00:24:07,020 Centripetal, right? 400 00:24:07,020 --> 00:24:10,910 And the centripetal acceleration, 401 00:24:10,910 --> 00:24:14,750 we want the sum of the external forces. 402 00:24:14,750 --> 00:24:17,951 And this is going to be in the r hat 403 00:24:17,951 --> 00:24:27,530 direction minus m times the-- and now we 404 00:24:27,530 --> 00:24:29,090 need reference frames here. 405 00:24:29,090 --> 00:24:35,240 So you have a fixed frame o xy. 406 00:24:35,240 --> 00:24:40,470 You have a rotating frame here, a x prime, y prime, z prime. 407 00:24:40,470 --> 00:24:44,470 And this is your point b up here. 408 00:24:44,470 --> 00:24:51,030 So some of the external forces minus the mass 409 00:24:51,030 --> 00:24:55,410 times the acceleration of b with respect to 0 410 00:24:55,410 --> 00:25:00,990 has got to be equal to 0. 411 00:25:00,990 --> 00:25:04,160 That's how this fictitious force thing works. 412 00:25:04,160 --> 00:25:08,480 So on the free body diagram of this thing, 413 00:25:08,480 --> 00:25:10,220 what are the actual forces on it? 414 00:25:10,220 --> 00:25:17,090 Well, there's certainly a weight downwards. 415 00:25:17,090 --> 00:25:20,770 There must be some supporting force upwards. 416 00:25:20,770 --> 00:25:24,130 And we'll put that in the z direction here. 417 00:25:24,130 --> 00:25:26,670 Because this thing certainly doesn't accelerate up and down. 418 00:25:26,670 --> 00:25:29,110 So the rods lifting it up somehow. 419 00:25:29,110 --> 00:25:35,820 The rod has potentially some radial component. 420 00:25:35,820 --> 00:25:38,270 And I'm just drawing it in the positive direction. 421 00:25:38,270 --> 00:25:41,470 And those are my forces. 422 00:25:41,470 --> 00:25:46,510 And in this problem, omega dot theta double dot 423 00:25:46,510 --> 00:25:51,660 equals 0 and r dot and r double dot were 0. 424 00:25:51,660 --> 00:25:53,960 So it's constant rotation rate. 425 00:25:53,960 --> 00:25:55,790 No geometry changes here. 426 00:25:58,750 --> 00:26:01,910 So we did, we cranked through this problem before. 427 00:26:01,910 --> 00:26:04,270 And now I wanted to make this correction. 428 00:26:04,270 --> 00:26:07,060 I want to add this fictitious force in. 429 00:26:07,060 --> 00:26:11,430 The acceleration is minus r omega 430 00:26:11,430 --> 00:26:13,330 squared in the r hat direction. 431 00:26:13,330 --> 00:26:17,290 So minus m times that is a plus. 432 00:26:17,290 --> 00:26:24,120 I get an m r omega squared. 433 00:26:28,730 --> 00:26:32,650 And pointing outwards. 434 00:26:32,650 --> 00:26:41,210 And so the sum of the forces in the radial direction 435 00:26:41,210 --> 00:26:47,470 minus-- some of the forces in the radial direction now 436 00:26:47,470 --> 00:26:52,270 minus this mab with respect to o. 437 00:26:52,270 --> 00:26:54,270 Well, what are some of the real forces? 438 00:26:54,270 --> 00:26:56,970 Just nr. 439 00:26:56,970 --> 00:27:01,414 And this minus mass times the acceleration, that's your mr 440 00:27:01,414 --> 00:27:01,913 squared. 441 00:27:04,730 --> 00:27:06,840 mr omega squared. 442 00:27:06,840 --> 00:27:09,360 And that's got to be 0. 443 00:27:09,360 --> 00:27:12,035 So you find out that there is a-- 444 00:27:23,600 --> 00:27:24,490 This all is 0. 445 00:27:24,490 --> 00:27:35,370 Therefore solve for nr is minus mr omega squared. 446 00:27:35,370 --> 00:27:41,280 So that rod out here is constantly pulling it in 447 00:27:41,280 --> 00:27:43,510 to make it go around the circle. 448 00:27:43,510 --> 00:27:45,660 But we've just done this now by this notion 449 00:27:45,660 --> 00:27:48,650 of a fictitious force. 450 00:27:48,650 --> 00:27:52,640 So let's go back to this problem now. 451 00:27:52,640 --> 00:27:55,910 Well actually, I'll just take one final step. 452 00:27:55,910 --> 00:28:01,630 Here's your free body diagram touched up 453 00:28:01,630 --> 00:28:03,720 with this fictitious force so that we 454 00:28:03,720 --> 00:28:06,760 can work as if we're here. 455 00:28:06,760 --> 00:28:09,937 And by the way, if you could measure, 456 00:28:09,937 --> 00:28:11,520 if you had a strain gauge or something 457 00:28:11,520 --> 00:28:17,810 so you can measure the force that this rod 458 00:28:17,810 --> 00:28:19,800 is putting on that mass. 459 00:28:19,800 --> 00:28:21,347 If you can measure it out there, this 460 00:28:21,347 --> 00:28:22,680 would be what you would measure. 461 00:28:27,820 --> 00:28:35,990 Now what about the torque that force creates around here? 462 00:28:40,900 --> 00:28:43,330 So you have outward force mr omega 463 00:28:43,330 --> 00:28:48,830 squared at centrifugal force, this fictitious force. 464 00:28:48,830 --> 00:28:52,110 Going pulling it out up here. 465 00:28:52,110 --> 00:28:53,880 What moment does that create down here? 466 00:28:57,870 --> 00:28:59,135 What's the magnitude first? 467 00:29:02,550 --> 00:29:11,380 And that'll be a torque around in the theta direction. 468 00:29:16,810 --> 00:29:18,164 So what's the moment arm? 469 00:29:18,164 --> 00:29:20,060 AUDIENCE: [INAUDIBLE]. 470 00:29:20,060 --> 00:29:23,200 PROFESSOR: The moment arm? 471 00:29:23,200 --> 00:29:24,870 The moment arm. 472 00:29:24,870 --> 00:29:25,580 z. 473 00:29:25,580 --> 00:29:26,360 OK. 474 00:29:26,360 --> 00:29:30,110 So then the torque is going to be 4r cross f 475 00:29:30,110 --> 00:29:38,440 and it's going to be the mrz omega squared. 476 00:29:38,440 --> 00:29:44,360 And it's at centripetal forces trying to bend it out. 477 00:29:44,360 --> 00:29:46,660 Therefore the torque down here required 478 00:29:46,660 --> 00:29:50,000 to keep it from doing so is actually 479 00:29:50,000 --> 00:29:52,230 in the other direction. 480 00:29:52,230 --> 00:29:59,640 So the torque that this system must put on this bar is minus. 481 00:29:59,640 --> 00:30:01,320 OK. 482 00:30:01,320 --> 00:30:03,630 So let's go to this problem. 483 00:30:03,630 --> 00:30:07,260 If this is the point about which we're computing the moments, 484 00:30:07,260 --> 00:30:11,980 then there's centripetal acceleration in this problem, 485 00:30:11,980 --> 00:30:13,361 but in what direction are they? 486 00:30:16,105 --> 00:30:17,230 Either one of those masses. 487 00:30:22,500 --> 00:30:24,980 Pardon? 488 00:30:24,980 --> 00:30:28,150 So each little mass, like the upper mass m over 2. 489 00:30:28,150 --> 00:30:31,190 It's spinning and it's shown as it's coming around the wheel. 490 00:30:31,190 --> 00:30:32,850 It's up at the top. 491 00:30:32,850 --> 00:30:36,505 What direction is the fictitious force 492 00:30:36,505 --> 00:30:39,350 that it's putting on that system? 493 00:30:39,350 --> 00:30:41,340 Radial outwards, right? 494 00:30:41,340 --> 00:30:43,600 And does it have any moment arm that 495 00:30:43,600 --> 00:30:46,800 force going up with respect to that point in the center? 496 00:30:46,800 --> 00:30:47,300 No. 497 00:30:47,300 --> 00:30:48,540 So it creates no moment. 498 00:30:48,540 --> 00:30:51,550 The bottom one by symmetry doesn't either. 499 00:30:51,550 --> 00:30:58,100 This one over here, it also has a fictitious force 500 00:30:58,100 --> 00:30:58,820 you can think of. 501 00:30:58,820 --> 00:31:01,200 It's just outward mr omega squared. 502 00:31:01,200 --> 00:31:05,951 What's its moment arm with respect to our origin here? 503 00:31:05,951 --> 00:31:06,450 0. 504 00:31:06,450 --> 00:31:08,010 Again, it produces no moment. 505 00:31:08,010 --> 00:31:10,810 It produces an unbalanced force though, right? 506 00:31:10,810 --> 00:31:13,520 It's going around producing an unbalanced force. 507 00:31:13,520 --> 00:31:17,340 This problem, the upper ones, the fictitious force is up. 508 00:31:17,340 --> 00:31:19,621 The lower one, which direction is it? 509 00:31:19,621 --> 00:31:20,120 Down. 510 00:31:20,120 --> 00:31:22,170 It's perfectly balanced. 511 00:31:22,170 --> 00:31:25,896 They're just equal and opposite, just going around. 512 00:31:25,896 --> 00:31:27,270 You'd feel nothing uncomfortable. 513 00:31:27,270 --> 00:31:29,603 This one you'd feel, this motorcycle trying to hop down. 514 00:31:32,135 --> 00:31:34,000 It really happens, right? 515 00:31:34,000 --> 00:31:37,180 But this one, what happens? 516 00:31:37,180 --> 00:31:39,980 Does this mass m over 2, what direction is 517 00:31:39,980 --> 00:31:42,720 the centripetal acceleration? 518 00:31:42,720 --> 00:31:44,450 I've changed the wording a little bit. 519 00:31:44,450 --> 00:31:47,450 The upper mass in case b, which direction 520 00:31:47,450 --> 00:31:50,920 is the centripetal acceleration? 521 00:31:50,920 --> 00:31:52,290 Straight down, right? 522 00:31:52,290 --> 00:31:54,290 And the bottom mass, what direction? 523 00:31:54,290 --> 00:31:55,330 Straight up. 524 00:31:55,330 --> 00:31:59,930 And the fictitious forces that reflect 525 00:31:59,930 --> 00:32:03,555 those two accelerations are the top one is up 526 00:32:03,555 --> 00:32:05,620 and the bottom one is down. 527 00:32:05,620 --> 00:32:08,660 But they produce a net. 528 00:32:08,660 --> 00:32:11,360 Do they have a moment arm that they act on with respect 529 00:32:11,360 --> 00:32:12,647 to that center? 530 00:32:12,647 --> 00:32:13,980 Yeah, this little distance here. 531 00:32:13,980 --> 00:32:16,120 I think it's called b in the problem. 532 00:32:16,120 --> 00:32:20,240 So this thing, m over 2 r omega squared times b, 533 00:32:20,240 --> 00:32:22,780 is a torque about the x-axis. 534 00:32:22,780 --> 00:32:25,100 And this one down here is another m 535 00:32:25,100 --> 00:32:28,150 over 2 r omega squared and it's another torque 536 00:32:28,150 --> 00:32:29,110 in the same direction. 537 00:32:29,110 --> 00:32:33,410 So this thing produces quite a strong moment. 538 00:32:33,410 --> 00:32:43,100 But as the wheel turns, that moment starts off, 539 00:32:43,100 --> 00:32:45,630 it's about the x-axis here. 540 00:32:45,630 --> 00:32:47,680 But when that weight gets to 90 degrees, 541 00:32:47,680 --> 00:32:49,000 what direction is the moment? 542 00:32:51,730 --> 00:32:57,600 So we just do all the way around the y-axis. 543 00:32:57,600 --> 00:32:58,810 So it's going to be a moment. 544 00:32:58,810 --> 00:33:00,860 The answer to this problem is in time 545 00:33:00,860 --> 00:33:05,650 it's going to look something like mr omega squared z 546 00:33:05,650 --> 00:33:08,270 sine or cosine omega t. 547 00:33:08,270 --> 00:33:10,700 So the moment in this around the x-axis 548 00:33:10,700 --> 00:33:13,900 is going to be trying to-- the bike's going to try 549 00:33:13,900 --> 00:33:17,300 to go back and forth like this. 550 00:33:17,300 --> 00:33:18,801 With frequency omega. 551 00:33:18,801 --> 00:33:19,300 OK. 552 00:33:26,880 --> 00:33:29,209 Oh, I was going to say that's it for fictitious forces, 553 00:33:29,209 --> 00:33:29,750 but it's not. 554 00:33:29,750 --> 00:33:31,270 I've got a great demo for you. 555 00:33:40,620 --> 00:33:41,130 All right. 556 00:33:49,986 --> 00:33:50,820 We have a slope. 557 00:33:55,820 --> 00:34:00,310 I've got a cart on wheels there. 558 00:34:00,310 --> 00:34:01,110 And I actually do. 559 00:34:05,500 --> 00:34:06,000 OK. 560 00:34:10,737 --> 00:34:14,999 And I'm going to put a box here. 561 00:34:18,730 --> 00:34:20,230 I got the wrong color here, but I'll 562 00:34:20,230 --> 00:34:22,239 try colored chalk for the first time. 563 00:34:22,239 --> 00:34:23,520 In the box I've got a fluid. 564 00:34:28,110 --> 00:34:30,580 And the box has a lid on it so I don't make a mess. 565 00:34:33,880 --> 00:34:38,360 Now, I'm actually going to do this experiment. 566 00:34:38,360 --> 00:34:39,639 I brought my ramp. 567 00:34:39,639 --> 00:34:41,810 We're going to set up this ramp. 568 00:34:41,810 --> 00:34:44,929 This box is going to be rolling down this hill. 569 00:34:44,929 --> 00:34:48,110 And basically no losses. 570 00:34:48,110 --> 00:34:49,898 No friction holding it back. 571 00:34:49,898 --> 00:34:52,314 It's just going to be able to take off and accelerate down 572 00:34:52,314 --> 00:34:52,920 that hill. 573 00:34:56,219 --> 00:35:03,980 I want to know a, b, c. 574 00:35:03,980 --> 00:35:14,050 When it's rolling down that hill, will the fluid in the box 575 00:35:14,050 --> 00:35:30,370 look like that, look like that, or like that? 576 00:35:37,970 --> 00:35:41,010 So we're going to take our poll here. 577 00:35:41,010 --> 00:35:47,690 So how many believe that when it's rolling down the hill, 578 00:35:47,690 --> 00:35:50,730 the fluid will stay level, parallel to the ground? 579 00:35:50,730 --> 00:35:52,120 Let's raise your hands. 580 00:35:52,120 --> 00:35:54,007 I want everybody to make a guess here. 581 00:35:54,007 --> 00:35:54,840 Everybody get in it. 582 00:35:54,840 --> 00:35:58,750 Raise them high if you think it's going to stay level. 583 00:35:58,750 --> 00:36:00,520 OK, maybe a quarter of you. 584 00:36:00,520 --> 00:36:02,912 How many think it's going to do this? 585 00:36:02,912 --> 00:36:05,290 OK, a few more. 586 00:36:05,290 --> 00:36:08,100 And how many think it's going to do this? 587 00:36:08,100 --> 00:36:12,500 All right, so I want you get in natural little groups there 588 00:36:12,500 --> 00:36:15,610 of two or three people and talk about this for a couple minutes 589 00:36:15,610 --> 00:36:18,750 and see if you can convince your neighbor that you're right. 590 00:36:18,750 --> 00:36:19,950 Figure this out. 591 00:36:19,950 --> 00:36:20,960 Stop and talk about it. 592 00:36:24,590 --> 00:36:26,900 And let's set up to do the experiment. 593 00:36:33,023 --> 00:36:36,914 Actually, I think if we do it up here. 594 00:36:36,914 --> 00:36:38,320 Yeah. 595 00:36:38,320 --> 00:36:39,630 So like that. 596 00:36:42,510 --> 00:36:44,805 So you hold the ramp. 597 00:36:44,805 --> 00:36:46,186 OK, you come here. 598 00:36:49,850 --> 00:36:53,030 We'll just practice our set up here for a second. 599 00:36:53,030 --> 00:36:56,730 You're going to have to hold it quite a bit higher. 600 00:36:56,730 --> 00:36:59,044 And actually when we get ready to go-- 601 00:36:59,044 --> 00:37:00,460 AUDIENCE: How well sealed is that? 602 00:37:00,460 --> 00:37:03,240 PROFESSOR: It's very well sealed. 603 00:37:03,240 --> 00:37:05,231 You hold it and you release it. 604 00:37:05,231 --> 00:37:07,730 And that's about the right-- because I got to catch it so it 605 00:37:07,730 --> 00:37:10,241 doesn't-- now, we're going to stop. 606 00:37:10,241 --> 00:37:11,740 We're not going to do it right away. 607 00:37:11,740 --> 00:37:15,340 I'm going to get them to answer the question a second time. 608 00:37:20,930 --> 00:37:21,845 Good idea. 609 00:37:26,750 --> 00:37:28,015 All right. 610 00:37:28,015 --> 00:37:30,610 And I think that'll work just great. 611 00:37:30,610 --> 00:37:33,230 And we're not going to give it away. 612 00:37:33,230 --> 00:37:36,730 OK, so set it down and we'll do the thing. 613 00:37:36,730 --> 00:37:37,870 Just drop it. 614 00:37:54,660 --> 00:37:59,010 OK, let's do the quiz again. 615 00:37:59,010 --> 00:38:01,430 I want to see if any of you convinced your neighbor 616 00:38:01,430 --> 00:38:03,305 to change their vote. 617 00:38:03,305 --> 00:38:07,970 So how many believe that it's going to be A? 618 00:38:07,970 --> 00:38:10,190 Hm, one or two. 619 00:38:10,190 --> 00:38:12,010 Interesting, that kind of went down. 620 00:38:12,010 --> 00:38:14,366 How many believe it's going to be B? 621 00:38:14,366 --> 00:38:15,610 A lot of people went up. 622 00:38:15,610 --> 00:38:18,010 How many think it's going to be C? 623 00:38:18,010 --> 00:38:19,990 OK, still a couple hold outs. 624 00:38:19,990 --> 00:38:21,005 Let's do the experiment. 625 00:38:26,000 --> 00:38:28,330 So guys in the booth, can you get a good shot of this, 626 00:38:28,330 --> 00:38:29,200 I hope? 627 00:38:29,200 --> 00:38:33,000 I want this so it [INAUDIBLE]. 628 00:38:33,000 --> 00:38:35,510 All right. 629 00:38:35,510 --> 00:38:36,700 Move the chalk. 630 00:38:36,700 --> 00:38:38,410 Yep. 631 00:38:38,410 --> 00:38:42,980 And I'm going to have to catch it. 632 00:38:42,980 --> 00:38:44,410 All right. 633 00:38:44,410 --> 00:38:47,630 Let her rip. 634 00:38:47,630 --> 00:38:48,680 Which way? 635 00:38:48,680 --> 00:38:49,180 AUDIENCE: B. 636 00:38:49,180 --> 00:38:49,846 PROFESSOR: Yeah. 637 00:38:49,846 --> 00:38:50,636 Let's do it again. 638 00:38:56,480 --> 00:38:59,820 Just make sure the guys in the booth got a good shot of this. 639 00:38:59,820 --> 00:39:00,980 OK, once more. 640 00:39:03,740 --> 00:39:04,690 Pretty neat, huh? 641 00:39:10,800 --> 00:39:16,640 If you're skiing downhill, you're going down the hill, 642 00:39:16,640 --> 00:39:22,545 accelerating, which way do you feel the forces on your body? 643 00:39:27,580 --> 00:39:30,560 You feel the forces pulling you down straight? 644 00:39:30,560 --> 00:39:33,630 Do you feel the forces pulling you down the hill? 645 00:39:33,630 --> 00:39:36,075 Do you feel the forces pulling you into the hill? 646 00:39:40,870 --> 00:39:43,520 So this is straightforward application of this. 647 00:39:46,295 --> 00:39:49,210 The slope of that liquid was parallel to the slope. 648 00:40:00,685 --> 00:40:02,490 Get my notes back out here. 649 00:40:25,630 --> 00:40:29,550 So let's quickly do this problem rigorously, 650 00:40:29,550 --> 00:40:34,320 the way we'd do it straightforward 651 00:40:34,320 --> 00:40:37,440 with accelerations and free body diagrams and so forth. 652 00:40:37,440 --> 00:40:43,009 We're going to have some normal force pushing up. 653 00:40:43,009 --> 00:40:43,800 I said no friction. 654 00:40:43,800 --> 00:40:47,420 This thing's a slippery surface, just as if it's sliding down. 655 00:40:47,420 --> 00:40:50,590 Got to do rollers in the classroom. 656 00:40:50,590 --> 00:40:51,320 It's got an mg. 657 00:40:56,800 --> 00:41:00,420 And there are no other external forces. 658 00:41:00,420 --> 00:41:05,750 And we know then that the summation 659 00:41:05,750 --> 00:41:09,380 of the external forces, that's got 660 00:41:09,380 --> 00:41:14,470 to be the mass times the acceleration. 661 00:41:14,470 --> 00:41:17,950 And I'm going to make this here. 662 00:41:17,950 --> 00:41:21,310 I'll call this point A in the center. 663 00:41:21,310 --> 00:41:24,290 And I'm going to make my inertial reference 664 00:41:24,290 --> 00:41:28,445 frame x down the hill, y perpendicular to the hill. 665 00:41:31,240 --> 00:41:35,240 So the mass times the acceleration of a with respect 666 00:41:35,240 --> 00:41:38,300 to o is then equal. 667 00:41:38,300 --> 00:41:40,840 And this is my then external forces. 668 00:41:40,840 --> 00:41:43,980 So this is my x direction. 669 00:41:43,980 --> 00:41:50,620 So I'm going to break this and mg force into two components. 670 00:41:50,620 --> 00:41:52,090 One down the hill. 671 00:41:52,090 --> 00:41:53,450 Guess I need an angle. 672 00:42:02,020 --> 00:42:02,590 Let's see. 673 00:42:06,520 --> 00:42:17,820 So if this is theta, then this is theta here. 674 00:42:22,380 --> 00:42:26,540 And the force down the hill is that component 675 00:42:26,540 --> 00:42:34,640 is mg sine theta. 676 00:42:34,640 --> 00:42:38,975 And the force this direction is then going to be. 677 00:42:44,720 --> 00:42:47,710 And I'll just give us some unit vectors here 678 00:42:47,710 --> 00:42:49,230 to keep things straight. 679 00:42:49,230 --> 00:43:00,850 This force is minus mg cosine theta in the j hat direction. 680 00:43:00,850 --> 00:43:06,790 So the sum of the forces here, all of them, 681 00:43:06,790 --> 00:43:22,190 are mg sine theta i hat minus mg cosine theta j hat 682 00:43:22,190 --> 00:43:25,680 plus n in the j hat direction. 683 00:43:25,680 --> 00:43:28,670 So that's the total vector sum. 684 00:43:28,670 --> 00:43:35,590 And that's got to be equal to my mass times the acceleration 685 00:43:35,590 --> 00:43:38,580 of a with respect to o. 686 00:43:38,580 --> 00:43:40,380 And we know that's straight down the hill. 687 00:43:40,380 --> 00:43:45,770 So this one's only going to be in the i hat direction. 688 00:43:45,770 --> 00:43:47,865 We know the acceleration perpendicular 689 00:43:47,865 --> 00:43:50,080 hill has got to be 0. 690 00:43:50,080 --> 00:43:56,110 OK, so if we just add up the components of this thing. 691 00:43:56,110 --> 00:44:00,360 In the i hat direction, just the x direction, 692 00:44:00,360 --> 00:44:06,300 then I have only this term and that term. 693 00:44:06,300 --> 00:44:15,630 And I find that the mg sine theta equals. 694 00:44:25,630 --> 00:44:31,300 And so the acceleration of a with respect to o 695 00:44:31,300 --> 00:44:34,630 is just g sine theta. 696 00:44:34,630 --> 00:44:37,370 OK, well that's pretty trivial. 697 00:44:37,370 --> 00:44:47,910 And it doesn't give us much insight about-- that 698 00:44:47,910 --> 00:44:50,300 doesn't give me much insight about why that's the answer. 699 00:44:53,460 --> 00:44:55,630 So actually this is a time when maybe thinking 700 00:44:55,630 --> 00:44:59,180 about if you're in that free, you're going down the hill, 701 00:44:59,180 --> 00:45:02,760 you're skiing, and you're now in an accelerating reference 702 00:45:02,760 --> 00:45:05,220 frame, and you want to say I want 703 00:45:05,220 --> 00:45:09,300 to work from this frame, what fictitious force do 704 00:45:09,300 --> 00:45:11,570 I have to add to that free body diagram 705 00:45:11,570 --> 00:45:14,315 so I can account for my acceleration? 706 00:45:33,500 --> 00:45:35,720 So remember the fictitious force we 707 00:45:35,720 --> 00:45:39,190 want to deal with has to do with this guy. 708 00:45:39,190 --> 00:45:42,470 We're in the moving frame. 709 00:45:42,470 --> 00:45:43,880 We can't judge this. 710 00:45:43,880 --> 00:45:46,540 We're just told that it exists and we need to correct for it. 711 00:45:46,540 --> 00:45:48,420 So we need a fictitious force that's 712 00:45:48,420 --> 00:45:52,000 minus ma with respect to o. 713 00:45:52,000 --> 00:45:56,050 So that would be minus mg sine theta. 714 00:45:56,050 --> 00:45:57,943 And that would be in this direction. 715 00:45:57,943 --> 00:46:03,870 So dashed lines here is my mg sine theta fictitious force. 716 00:46:08,740 --> 00:46:12,840 In the other direction was the real force, the component 717 00:46:12,840 --> 00:46:16,140 of gravity mg sine theta. 718 00:46:16,140 --> 00:46:19,980 In this direction was my normal force. 719 00:46:19,980 --> 00:46:24,630 And in this direction is my component of gravity 720 00:46:24,630 --> 00:46:28,445 that is my mg cosine theta. 721 00:46:33,350 --> 00:46:34,360 It's a real force. 722 00:46:34,360 --> 00:46:37,810 So yeah, these real forces and this fictitious force. 723 00:46:37,810 --> 00:46:42,860 But now we can say that all these forces have to sum to 0. 724 00:46:45,710 --> 00:46:48,960 But look what happens here. 725 00:46:48,960 --> 00:46:52,940 What force do you feel? 726 00:46:52,940 --> 00:46:56,320 In the reference frame, you're the skier going down the hill. 727 00:46:56,320 --> 00:47:01,030 What net forces do you feel on you that are in the x direction 728 00:47:01,030 --> 00:47:01,680 down the hill? 729 00:47:06,960 --> 00:47:09,370 0. 730 00:47:09,370 --> 00:47:10,270 It's exactly. 731 00:47:10,270 --> 00:47:13,300 The reason I did that silly free fall problem a minute 732 00:47:13,300 --> 00:47:16,385 ago when you drop something, when you're weightless, 733 00:47:16,385 --> 00:47:18,760 the scales are reading-- the cable broke in the elevator, 734 00:47:18,760 --> 00:47:20,190 scales read 0. 735 00:47:20,190 --> 00:47:23,030 The fictitious force was equal and opposite 736 00:47:23,030 --> 00:47:25,150 to the gravitational force you feel 737 00:47:25,150 --> 00:47:30,290 and the scales would show no force in the free fall 738 00:47:30,290 --> 00:47:30,810 direction. 739 00:47:30,810 --> 00:47:32,790 In this case, you're not free fall, 740 00:47:32,790 --> 00:47:35,010 but you're free fall down a slope. 741 00:47:35,010 --> 00:47:37,270 In the component down the slope, you 742 00:47:37,270 --> 00:47:38,880 have no net force in that direction. 743 00:47:42,866 --> 00:47:46,410 In the reference frame that moving cart, 744 00:47:46,410 --> 00:47:50,880 the only actual force you feel is in. 745 00:47:50,880 --> 00:47:58,610 You will feel a force on your feet equal to mg cosine theta. 746 00:47:58,610 --> 00:48:03,030 So if theta goes to 90 degrees, you're in free fall, 747 00:48:03,030 --> 00:48:05,380 and it goes to 0. 748 00:48:05,380 --> 00:48:08,740 Say it is 0 degrees, it's max, and you would feel mg. 749 00:48:12,830 --> 00:48:15,042 This is [INAUDIBLE] me where this notion 750 00:48:15,042 --> 00:48:16,500 of a fictitious force kind of helps 751 00:48:16,500 --> 00:48:18,310 me figure out what's going on. 752 00:48:22,450 --> 00:48:24,550 Pretty amazing. 753 00:48:24,550 --> 00:48:25,280 OK. 754 00:48:25,280 --> 00:48:29,380 So we're going to move on from this kind of discussion 755 00:48:29,380 --> 00:48:31,770 of fictitious forces of things. 756 00:48:31,770 --> 00:48:35,150 So it's been an hour. 757 00:48:35,150 --> 00:48:37,526 We'll take a little break for a second. 758 00:48:37,526 --> 00:48:38,900 Also, if you've got any questions 759 00:48:38,900 --> 00:48:41,439 about what we've covered up to this point, think about them 760 00:48:41,439 --> 00:48:42,480 and ask them in a minute. 761 00:48:42,480 --> 00:48:44,490 Let's take a short break. 762 00:48:44,490 --> 00:48:47,510 So while you've been taking a break, a couple of people 763 00:48:47,510 --> 00:48:48,870 came up and asked questions. 764 00:48:48,870 --> 00:48:53,060 One of them was, are fictitious forces real? 765 00:48:53,060 --> 00:48:55,915 And the other is, why does the water move? 766 00:49:00,950 --> 00:49:04,830 So the answer to the first question. 767 00:49:04,830 --> 00:49:11,520 If you want to be careful and do the problem rigorously, 768 00:49:11,520 --> 00:49:15,650 don't mess with fictitious forces. 769 00:49:15,650 --> 00:49:19,910 The only real forces in the problem 770 00:49:19,910 --> 00:49:24,530 are the ones that you would put on a free body diagram, such 771 00:49:24,530 --> 00:49:26,690 that the sum of the external forces 772 00:49:26,690 --> 00:49:29,230 equals the mass times acceleration. 773 00:49:29,230 --> 00:49:31,470 These forces are always real ones. 774 00:49:31,470 --> 00:49:33,100 They are gravity. 775 00:49:33,100 --> 00:49:34,360 They are normal forces. 776 00:49:34,360 --> 00:49:35,510 They're friction. 777 00:49:35,510 --> 00:49:36,480 They're spring forces. 778 00:49:36,480 --> 00:49:37,780 They're dashpot forces. 779 00:49:37,780 --> 00:49:39,320 They're all these mechanical things, 780 00:49:39,320 --> 00:49:41,560 but they're real forces. 781 00:49:41,560 --> 00:49:44,530 And the rod, that thing spinning around, 782 00:49:44,530 --> 00:49:48,000 that rod, since it's a rigid thing, it can support bending. 783 00:49:48,000 --> 00:49:52,200 So it has shear force, axial forces. 784 00:49:52,200 --> 00:49:55,780 Those are real forces that it can put on the mass. 785 00:49:55,780 --> 00:49:59,050 Fictitious forces are not forces. 786 00:49:59,050 --> 00:50:05,885 They are minus a mass times a real what? 787 00:50:05,885 --> 00:50:06,426 Acceleration. 788 00:50:09,290 --> 00:50:11,820 So real forces, real accelerations. 789 00:50:11,820 --> 00:50:14,585 We use this notion of a fictitious force 790 00:50:14,585 --> 00:50:16,370 as a convenience. 791 00:50:16,370 --> 00:50:20,547 But it is a highly dangerous tool. 792 00:50:20,547 --> 00:50:22,630 But actually once you get used to it a little bit, 793 00:50:22,630 --> 00:50:24,540 it can give you some really quick insight. 794 00:50:24,540 --> 00:50:26,950 But I would always, if I have doubts, I'd always go back 795 00:50:26,950 --> 00:50:29,380 and just do the problem the rigorous way and just 796 00:50:29,380 --> 00:50:32,400 check that your insight was right. 797 00:50:32,400 --> 00:50:35,040 But it can give you a great insight to starting a problem. 798 00:50:35,040 --> 00:50:39,085 You know to expect a centrifugal term in your results. 799 00:50:39,085 --> 00:50:42,495 You know to expect a Coriolis term. 800 00:50:42,495 --> 00:50:51,390 You know that they're going to be there because sometimes you 801 00:50:51,390 --> 00:50:52,632 feel them. 802 00:50:52,632 --> 00:50:55,400 The reason we like to use them is because you actually 803 00:50:55,400 --> 00:50:58,310 can feel them. 804 00:50:58,310 --> 00:51:01,120 They actually affect what you feel. 805 00:51:01,120 --> 00:51:17,040 So in that first problem, the elevator problem, the answer-- 806 00:51:17,040 --> 00:51:19,680 I didn't actually write it. 807 00:51:19,680 --> 00:51:21,180 Here it is. 808 00:51:21,180 --> 00:51:24,490 The normal force that you feel, that the scale weighs 809 00:51:24,490 --> 00:51:26,570 is 5 force g. 810 00:51:26,570 --> 00:51:30,073 And it's the sum of gravity and that fictitious force. 811 00:51:30,073 --> 00:51:34,160 It's mg plus m times a quarter of a g. 812 00:51:38,214 --> 00:51:40,630 It's in this moving references, this accelerator reference 813 00:51:40,630 --> 00:51:43,080 frame inside an elevator where you can't directly 814 00:51:43,080 --> 00:51:44,220 measure the acceleration. 815 00:51:44,220 --> 00:51:47,050 All you can measure is the force. 816 00:51:47,050 --> 00:51:49,150 But you know what the acceleration is. 817 00:51:49,150 --> 00:51:49,900 It's given. 818 00:51:49,900 --> 00:51:53,180 So you stick it in as a fix and you find out 819 00:51:53,180 --> 00:51:56,250 that the real normal force that you measure actually has 820 00:51:56,250 --> 00:52:00,130 in it that piece comes from the fictitious force. 821 00:52:00,130 --> 00:52:01,100 OK. 822 00:52:01,100 --> 00:52:03,700 Why does the water move? 823 00:52:03,700 --> 00:52:07,190 This is really the same discussion. 824 00:52:07,190 --> 00:52:11,630 The fictitious force up the hill and the real force 825 00:52:11,630 --> 00:52:14,490 down the hill exactly cancel. 826 00:52:14,490 --> 00:52:17,340 If you're standing on scales, it actually 827 00:52:17,340 --> 00:52:21,980 can measure side force as well as normal force. 828 00:52:21,980 --> 00:52:26,520 Standing on the scales, you would feel no side forces 829 00:52:26,520 --> 00:52:27,680 up and down the hill. 830 00:52:27,680 --> 00:52:29,470 None. 831 00:52:29,470 --> 00:52:32,950 You measure a normal force and the normal force you measure 832 00:52:32,950 --> 00:52:37,110 will be mg cosine theta. 833 00:52:37,110 --> 00:52:43,840 So the force on any object in that accelerating reference 834 00:52:43,840 --> 00:52:46,890 frame, it essentially thinks that all forces are 835 00:52:46,890 --> 00:52:50,040 perpendicular, that you feel in that reference frame, 836 00:52:50,040 --> 00:52:52,850 are perpendicular to the hill. 837 00:52:52,850 --> 00:52:57,900 So why does water seek level? 838 00:52:57,900 --> 00:53:00,630 Because the force gravity is down 839 00:53:00,630 --> 00:53:02,930 and the surface of the water goes perpendicular 840 00:53:02,930 --> 00:53:04,220 to the gravitational force. 841 00:53:04,220 --> 00:53:06,980 That's just the nature of fluids. 842 00:53:06,980 --> 00:53:11,490 So if the force on all those fluid particles is this angle, 843 00:53:11,490 --> 00:53:13,580 what's the fluid level going to do? 844 00:53:13,580 --> 00:53:17,130 So that the pressure on the surface is equal everywhere. 845 00:53:17,130 --> 00:53:20,090 And if you go a foot down, you get equal pressure 846 00:53:20,090 --> 00:53:21,440 everywhere in the fluid. 847 00:53:29,460 --> 00:53:32,300 So the water moves so that its surface 848 00:53:32,300 --> 00:53:35,820 will be perpendicular to what it thinks. 849 00:53:35,820 --> 00:53:38,285 It thinks that this is kind of being a local gravity. 850 00:53:38,285 --> 00:53:40,660 The local gravity is pointed in that direction, 851 00:53:40,660 --> 00:53:43,880 so the surface level seeks that level. 852 00:53:43,880 --> 00:53:46,440 If somebody else can say it better than that. 853 00:53:46,440 --> 00:53:48,420 Tom, Dave, Matt, anybody? 854 00:53:54,694 --> 00:53:55,662 Yeah. 855 00:53:55,662 --> 00:53:57,224 AUDIENCE: Can I ask you a question? 856 00:53:57,224 --> 00:53:57,890 PROFESSOR: Yeah. 857 00:54:01,422 --> 00:54:02,338 AUDIENCE: [INAUDIBLE]. 858 00:54:19,160 --> 00:54:21,490 PROFESSOR: Yeah, as if you're on the body 859 00:54:21,490 --> 00:54:23,800 and you're trying to sum up all the forces you feel 860 00:54:23,800 --> 00:54:27,480 and accounting also for the fact that this is accelerating. 861 00:54:27,480 --> 00:54:31,090 And the fix it force is this fictitious force that's 862 00:54:31,090 --> 00:54:33,750 due to the body's acceleration. 863 00:54:33,750 --> 00:54:35,550 You have to know that you're accelerating 864 00:54:35,550 --> 00:54:38,870 to be able to put this term in. 865 00:54:38,870 --> 00:54:42,530 And it comes from this equation, thinking 866 00:54:42,530 --> 00:54:44,820 of this as a generic equation by taking this term 867 00:54:44,820 --> 00:54:46,944 and just moving it to this side and setting 868 00:54:46,944 --> 00:54:47,860 everything equal to 0. 869 00:54:54,600 --> 00:54:56,668 Lots of puzzlement around this. 870 00:54:56,668 --> 00:54:59,536 AUDIENCE: What happens if you make the angle of the ramp very 871 00:54:59,536 --> 00:55:00,754 steep? 872 00:55:00,754 --> 00:55:01,670 PROFESSOR: Very steep. 873 00:55:01,670 --> 00:55:05,380 So if it goes to 90 degrees, you're in free fall. 874 00:55:09,380 --> 00:55:13,350 This cosine theta goes to 0, n goes to 0. 875 00:55:13,350 --> 00:55:14,580 This term stays the same. 876 00:55:14,580 --> 00:55:16,360 This goes to mg that way. 877 00:55:16,360 --> 00:55:17,540 This goes mg that way. 878 00:55:17,540 --> 00:55:18,915 And it's an equilibrium in both-- 879 00:55:18,915 --> 00:55:19,831 AUDIENCE: [INAUDIBLE]. 880 00:55:21,600 --> 00:55:23,100 PROFESSOR: Then you're going to have 881 00:55:23,100 --> 00:55:25,140 very, very small, normal force. 882 00:55:27,378 --> 00:55:28,294 AUDIENCE: [INAUDIBLE]. 883 00:55:31,452 --> 00:55:33,285 PROFESSOR: I couldn't hear the last comment. 884 00:55:33,285 --> 00:55:35,710 AUDIENCE: The water wouldn't go all the way to the side. 885 00:55:35,710 --> 00:55:37,650 It'd be vertical length [INAUDIBLE]. 886 00:55:40,570 --> 00:55:43,740 PROFESSOR: Well, water, unfortunately, 887 00:55:43,740 --> 00:55:48,460 with the tiniest little bit of normal force here 888 00:55:48,460 --> 00:55:49,950 the water will seek that level. 889 00:55:49,950 --> 00:55:57,210 But when it gets to 0 the water is going to be in free fall. 890 00:55:57,210 --> 00:55:58,940 Surface tension takes over. 891 00:55:58,940 --> 00:55:59,910 Yeah? 892 00:55:59,910 --> 00:56:03,305 AUDIENCE: Is this like the argument that [INAUDIBLE] 893 00:56:03,305 --> 00:56:06,215 is the same as if it was just on a flat surface accelerating? 894 00:56:09,872 --> 00:56:11,080 PROFESSOR: Oh, good question. 895 00:56:11,080 --> 00:56:14,810 If you were on a flat surface accelerating. 896 00:56:14,810 --> 00:56:19,110 If you were accelerating in this cart. 897 00:56:19,110 --> 00:56:21,540 Accelerates with the fluid on it and it's 898 00:56:21,540 --> 00:56:24,590 accelerating in that direction, what 899 00:56:24,590 --> 00:56:26,160 would the surface of the liquid do? 900 00:56:29,050 --> 00:56:32,610 Stay level or change? 901 00:56:32,610 --> 00:56:35,620 OK, does it slope back or does it slope forward? 902 00:56:35,620 --> 00:56:38,120 I'm accelerating in that direction. 903 00:56:38,120 --> 00:56:45,020 There's a fictitious force pushing back on it, 904 00:56:45,020 --> 00:56:48,590 makes it pile up at the back. 905 00:56:48,590 --> 00:56:50,800 OK one more. 906 00:56:50,800 --> 00:56:52,566 You're in the elevator. 907 00:56:52,566 --> 00:56:54,415 You're accelerating a quarter of a g. 908 00:56:58,070 --> 00:57:03,440 The scales tells you you weigh 25% more than you used to. 909 00:57:03,440 --> 00:57:04,490 And you have a pendulum. 910 00:57:04,490 --> 00:57:06,770 And before the elevator starts, it's 911 00:57:06,770 --> 00:57:10,510 got a one second natural period. 912 00:57:10,510 --> 00:57:11,860 And now the elevator gets going. 913 00:57:11,860 --> 00:57:14,820 Does the period change? 914 00:57:14,820 --> 00:57:15,800 Higher or lower? 915 00:57:15,800 --> 00:57:16,716 AUDIENCE: [INAUDIBLE]. 916 00:57:21,045 --> 00:57:22,670 PROFESSOR: OK, so the natural frequency 917 00:57:22,670 --> 00:57:25,044 of the pendulum, simple pendulum square root of g over l. 918 00:57:25,044 --> 00:57:27,630 What do you guess that the natural frequency is if you're 919 00:57:27,630 --> 00:57:29,950 going up at a quarter of a g? 920 00:57:34,855 --> 00:57:35,530 Say again? 921 00:57:35,530 --> 00:57:36,446 AUDIENCE: [INAUDIBLE]. 922 00:57:44,890 --> 00:57:49,590 PROFESSOR: It's as if you have a higher or lower acceleration 923 00:57:49,590 --> 00:57:52,880 of gravity around. 924 00:57:52,880 --> 00:57:54,610 But you have a higher gravitational force 925 00:57:54,610 --> 00:57:57,070 is what you think you're feeling. 926 00:57:57,070 --> 00:58:00,050 Is that going to make the tension in this string higher? 927 00:58:00,050 --> 00:58:02,690 Is that going to make the weight of the pendulum feel greater? 928 00:58:02,690 --> 00:58:04,900 Yeah, by 5 force. 929 00:58:04,900 --> 00:58:06,630 And you'll find out, you'll work out. 930 00:58:06,630 --> 00:58:09,915 It gets kind of messy to prove this rigorously, 931 00:58:09,915 --> 00:58:12,415 but it's going to look something like the square root of 5/4 932 00:58:12,415 --> 00:58:12,965 g over l. 933 00:58:15,610 --> 00:58:16,110 OK. 934 00:58:31,682 --> 00:58:32,890 We've got a little time left. 935 00:58:32,890 --> 00:58:35,300 I might not be able to completely finish this. 936 00:58:35,300 --> 00:58:36,900 But I do want to show you something. 937 00:58:59,321 --> 00:59:00,570 Hold it up so they can see it. 938 00:59:06,157 --> 00:59:07,490 Gets pretty violent, doesn't it? 939 00:59:07,490 --> 00:59:08,280 Pass it around. 940 00:59:16,120 --> 00:59:17,110 OK. 941 00:59:17,110 --> 00:59:18,975 This thing is shaking like crazy. 942 00:59:23,515 --> 00:59:24,015 All right. 943 00:59:30,300 --> 00:59:30,800 All right. 944 00:59:30,800 --> 00:59:33,022 In the absence of external forces, 945 00:59:33,022 --> 00:59:34,980 what can you say about the motion of its center 946 00:59:34,980 --> 00:59:36,490 of gravity, do you think? 947 00:59:36,490 --> 00:59:39,760 In at least this direction. 948 00:59:39,760 --> 00:59:41,160 Can't move, right? 949 00:59:41,160 --> 00:59:44,310 Up and down direction the center of gravity 950 00:59:44,310 --> 00:59:48,120 does move, because I'm putting a force on it through the cord. 951 00:59:48,120 --> 00:59:52,360 This has a relation to a problem that you 952 00:59:52,360 --> 00:59:56,289 had on problem set two. 953 00:59:56,289 --> 00:59:58,330 I want to go back and talk about it a little bit. 954 01:00:06,920 --> 01:00:09,900 And this is that thing with the track in it. 955 01:00:09,900 --> 01:00:11,550 And a roller going around. 956 01:00:11,550 --> 01:00:13,415 And it's going to around at constant speed. 957 01:00:13,415 --> 01:00:15,310 And this has mass m. 958 01:00:15,310 --> 01:00:16,990 This is mass. 959 01:00:16,990 --> 01:00:20,720 The outside thing is just the mass of this block. 960 01:00:20,720 --> 01:00:22,991 And there's an internal donut in here, [INAUDIBLE], 961 01:00:22,991 --> 01:00:24,865 and this mass is made to go around and around 962 01:00:24,865 --> 01:00:26,750 and around and around inside. 963 01:00:26,750 --> 01:00:33,500 And you were asked to figure out the acceleration of that guy. 964 01:00:33,500 --> 01:00:35,058 But this thing's on wheels. 965 01:00:46,780 --> 01:00:52,380 Now, this problem, and I apologize 966 01:00:52,380 --> 01:00:55,730 for posing this problem the way I did. 967 01:00:55,730 --> 01:00:59,510 I should have posed in the following way. 968 01:00:59,510 --> 01:01:04,910 These two problems are really equal to one another. 969 01:01:04,910 --> 01:01:06,720 I have a rod. 970 01:01:06,720 --> 01:01:08,850 I have a mass on the end. 971 01:01:08,850 --> 01:01:12,680 This radius here we said was e. 972 01:01:12,680 --> 01:01:14,370 This radius is e. 973 01:01:14,370 --> 01:01:15,670 The mass is the same. 974 01:01:15,670 --> 01:01:16,980 It's m. 975 01:01:16,980 --> 01:01:20,530 This was going around at omega. 976 01:01:20,530 --> 01:01:24,580 This is going around at omega. 977 01:01:24,580 --> 01:01:25,920 This is on wheels. 978 01:01:25,920 --> 01:01:30,970 Exactly these two problems are identical. 979 01:01:30,970 --> 01:01:33,294 Because if the geometry's the same, 980 01:01:33,294 --> 01:01:34,960 the rotation rate's the same, the masses 981 01:01:34,960 --> 01:01:37,770 are the same, the acceleration of this bead here, 982 01:01:37,770 --> 01:01:39,790 this roller is the same in this problem 983 01:01:39,790 --> 01:01:41,880 as it is in that problem. 984 01:01:41,880 --> 01:01:43,600 Just in this problem it's much easier 985 01:01:43,600 --> 01:01:46,360 to figure out the forces. 986 01:01:46,360 --> 01:01:48,100 This turned out to be kind of nasty. 987 01:01:48,100 --> 01:01:49,840 And I said last time that the solution 988 01:01:49,840 --> 01:01:53,430 to b part, which is the normal force that this 989 01:01:53,430 --> 01:01:56,450 is put on this roller, I said that answer was wrong. 990 01:01:56,450 --> 01:01:59,790 I actually am wrong about that. 991 01:01:59,790 --> 01:02:01,480 I was thinking of a different answer. 992 01:02:01,480 --> 01:02:04,780 I was thinking of this problem. 993 01:02:04,780 --> 01:02:08,590 The answer that was given, the normal force that 994 01:02:08,590 --> 01:02:11,770 is by the track places on the roller or vice 995 01:02:11,770 --> 01:02:15,390 versa is actually what was given is correct. 996 01:02:15,390 --> 01:02:17,070 It wasn't what I actually was seeking 997 01:02:17,070 --> 01:02:18,111 when I wrote the problem. 998 01:02:18,111 --> 01:02:20,290 Sometimes when you're making up problems, 999 01:02:20,290 --> 01:02:23,040 you have an end result in mind and then you wrote it wrong. 1000 01:02:23,040 --> 01:02:24,560 I wrote it wrong. 1001 01:02:24,560 --> 01:02:25,950 So let's do this problem. 1002 01:02:29,470 --> 01:02:34,550 This is partly to review a couple of things for you. 1003 01:02:34,550 --> 01:02:40,700 What I really want is I want to find equation of motion 1004 01:02:40,700 --> 01:02:44,549 in-- [INAUDIBLE]. 1005 01:02:50,760 --> 01:02:54,980 So here's my xyz inertial reference frame. 1006 01:02:54,980 --> 01:02:57,284 I just lined it up here with the center 1007 01:02:57,284 --> 01:02:58,450 of this thing to start with. 1008 01:02:58,450 --> 01:02:59,780 Doesn't really matter. 1009 01:02:59,780 --> 01:03:02,570 So I want to find an equation of motion in the x direction. 1010 01:03:06,049 --> 01:03:08,090 We don't have time to do this whole problem here, 1011 01:03:08,090 --> 01:03:12,850 so I'm going to do the first piece of it. 1012 01:03:12,850 --> 01:03:15,910 Just to give you some review, an exercise in figuring out 1013 01:03:15,910 --> 01:03:18,910 degrees of freedom. 1014 01:03:18,910 --> 01:03:21,075 I'm implying that I can write one equation. 1015 01:03:21,075 --> 01:03:23,140 It may take more than that. 1016 01:03:23,140 --> 01:03:27,230 So the number of degrees of freedom we said 1017 01:03:27,230 --> 01:03:35,122 is equal to the number of required-- 1018 01:03:35,122 --> 01:03:46,550 I can't write today-- independent coordinates. 1019 01:03:46,550 --> 01:03:50,540 So the number of coordinates required to completely describe 1020 01:03:50,540 --> 01:03:51,710 the motion. 1021 01:03:51,710 --> 01:03:54,120 And we came up a little expression for that. 1022 01:03:54,120 --> 01:03:56,890 It's 6 times the number rigid bodies 1023 01:03:56,890 --> 01:04:00,700 plus 3 times the number of particles. 1024 01:04:00,700 --> 01:04:04,052 So this is rigid bodies. 1025 01:04:04,052 --> 01:04:05,195 This is particles. 1026 01:04:11,650 --> 01:04:13,360 Minus c. 1027 01:04:13,360 --> 01:04:14,610 And these are the constraints. 1028 01:04:19,430 --> 01:04:21,450 So let's figure that out for this problem. 1029 01:04:37,002 --> 01:04:38,390 So how many constraints? 1030 01:04:38,390 --> 01:04:41,650 It's just a quick practice. 1031 01:04:41,650 --> 01:04:43,880 So constraints and I'm going to start off 1032 01:04:43,880 --> 01:04:50,370 with on the main block. 1033 01:04:50,370 --> 01:04:56,710 So on the main block, there's no y motion. 1034 01:04:56,710 --> 01:04:57,930 There's no z motion. 1035 01:04:57,930 --> 01:05:00,450 These rollers don't let it come in or out. 1036 01:05:00,450 --> 01:05:06,385 So no y or z motion. 1037 01:05:09,557 --> 01:05:10,640 So that's two constraints. 1038 01:05:13,600 --> 01:05:15,750 What about rotations on the main block? 1039 01:05:15,750 --> 01:05:19,670 How many constraints are we assuming here? 1040 01:05:19,670 --> 01:05:20,540 Three. 1041 01:05:20,540 --> 01:05:23,480 We're not allowing it the role in x, the y, or the z. 1042 01:05:33,238 --> 01:05:36,400 So there's three more constraints. 1043 01:05:36,400 --> 01:05:38,215 And now on the little mass. 1044 01:05:42,090 --> 01:05:44,590 On the little mass, there's no z motion. 1045 01:05:44,590 --> 01:05:45,590 That's out of the board. 1046 01:05:54,200 --> 01:05:56,110 That's one more constraint. 1047 01:05:56,110 --> 01:06:16,030 And then omega t is the angle that this thing. 1048 01:06:16,030 --> 01:06:17,980 It's theta. 1049 01:06:17,980 --> 01:06:18,760 This is specified. 1050 01:06:25,704 --> 01:06:26,870 It's a given in the problem. 1051 01:06:26,870 --> 01:06:27,589 It's not unknown. 1052 01:06:27,589 --> 01:06:29,130 It's just that's what's going, that's 1053 01:06:29,130 --> 01:06:30,720 what makes this thing do its thing. 1054 01:06:30,720 --> 01:06:32,570 It's going round and round. 1055 01:06:32,570 --> 01:06:35,130 And so this is not an unknown. 1056 01:06:35,130 --> 01:06:47,410 And therefore you can say that the y motion of this particle 1057 01:06:47,410 --> 01:06:52,030 is e sine omega t. 1058 01:06:55,180 --> 01:06:57,320 And that's a constraint. 1059 01:06:57,320 --> 01:07:07,820 And the x motion is the motion of the main block 1060 01:07:07,820 --> 01:07:17,260 to which it's attached plus an e cosine omega t. 1061 01:07:17,260 --> 01:07:19,187 And these are knowns. 1062 01:07:19,187 --> 01:07:20,520 You've already counted for this. 1063 01:07:20,520 --> 01:07:23,150 That's the main mass's motion. 1064 01:07:23,150 --> 01:07:25,190 So this is something known. 1065 01:07:25,190 --> 01:07:27,840 So these mount up, each one of these. 1066 01:07:27,840 --> 01:07:32,946 This gives you a constraint and this gives you a constraint. 1067 01:07:32,946 --> 01:07:34,690 And so let's add them up. 1068 01:07:34,690 --> 01:07:36,970 So 2 and 3 are 5. 1069 01:07:36,970 --> 01:07:41,390 6, 7, 8. 1070 01:07:41,390 --> 01:07:44,150 So you can actually completely describe 1071 01:07:44,150 --> 01:07:48,185 the motion of this thing with one equation of motion. 1072 01:08:00,480 --> 01:08:01,540 So I'm seeking. 1073 01:08:01,540 --> 01:08:10,593 So I want to find the sum of the external forces 1074 01:08:10,593 --> 01:08:19,310 on the main block in the x direction. 1075 01:08:19,310 --> 01:08:21,050 And that's going to be. 1076 01:08:21,050 --> 01:08:25,245 We know that that's the mass of that block times x double dot. 1077 01:08:25,245 --> 01:08:28,270 So x is going to be the coordinate that describes 1078 01:08:28,270 --> 01:08:30,620 the motion that mass mb. 1079 01:08:37,640 --> 01:08:41,010 So we need a free body diagram of mb. 1080 01:08:55,150 --> 01:08:56,370 So here's the block. 1081 01:08:59,850 --> 01:09:03,030 There'll be a normal force up. 1082 01:09:03,030 --> 01:09:08,775 There's going to be mbg downwards. 1083 01:09:08,775 --> 01:09:12,640 And the only other forces acting on this block 1084 01:09:12,640 --> 01:09:22,090 are a force that comes through this rod that's holding 1085 01:09:22,090 --> 01:09:23,659 that mass, the little mass. 1086 01:09:34,100 --> 01:09:36,660 I'm just going to draw it at some arbitrary angle, 1087 01:09:36,660 --> 01:09:39,130 not necessarily theta. 1088 01:09:39,130 --> 01:09:41,460 It's going to be a force. 1089 01:09:41,460 --> 01:09:48,229 I'm going to call that force on this main mass. 1090 01:09:48,229 --> 01:09:49,630 And it's a vector for the moment. 1091 01:09:49,630 --> 01:09:51,955 It'll have two components in the x and y directions. 1092 01:10:02,200 --> 01:10:05,280 So in order to write an equation-- in the y direction 1093 01:10:05,280 --> 01:10:10,520 it's trivial. n is equal to mbg. 1094 01:10:10,520 --> 01:10:14,240 In this direction, we have to have to break this 1095 01:10:14,240 --> 01:10:15,320 into two components. 1096 01:10:15,320 --> 01:10:19,240 We need to know the component of this in the x direction 1097 01:10:19,240 --> 01:10:23,800 and put it in here and we'll be done. 1098 01:10:23,800 --> 01:10:32,570 So now I'm going to use third law. 1099 01:10:32,570 --> 01:10:39,420 The force that rod places on this body 1100 01:10:39,420 --> 01:10:43,400 is equal and opposite to the force at the rod 1101 01:10:43,400 --> 01:10:46,900 places on the mass going round and round. 1102 01:10:46,900 --> 01:10:47,650 You agree with me? 1103 01:10:51,010 --> 01:10:55,060 So what we need to do next is here's 1104 01:10:55,060 --> 01:10:58,500 the mass up here going around and round. 1105 01:10:58,500 --> 01:11:07,060 And the force is on it, because it has an mg downward. 1106 01:11:07,060 --> 01:11:12,350 And it has some force on it that is 1107 01:11:12,350 --> 01:11:15,700 the force on the little mass. 1108 01:11:15,700 --> 01:11:19,170 And you're going to find that f on the little mass 1109 01:11:19,170 --> 01:11:27,060 equals minus f on the big mass due to third law consideration. 1110 01:11:27,060 --> 01:11:30,050 So all I have to do is then figure out 1111 01:11:30,050 --> 01:11:34,350 the x component of this force, put a minus sign in front 1112 01:11:34,350 --> 01:11:37,900 of it, and I'm basically done. 1113 01:11:37,900 --> 01:11:39,870 And to do that I say f equals ma. 1114 01:11:39,870 --> 01:11:41,210 I don't have to finish it today. 1115 01:11:41,210 --> 01:11:42,870 I'll show you next time. 1116 01:11:42,870 --> 01:11:47,596 And we'll figure out the equation of motion 1117 01:11:47,596 --> 01:11:48,220 for this thing. 1118 01:11:48,220 --> 01:11:51,590 Now, in dragging all this stuff here today, 1119 01:11:51,590 --> 01:11:54,532 I forgot to bring my muddy cards. 1120 01:11:54,532 --> 01:11:55,490 So you've got a minute. 1121 01:11:55,490 --> 01:11:57,160 And if any of you have just a brilliant question 1122 01:11:57,160 --> 01:11:58,909 you want to answer or something you really 1123 01:11:58,909 --> 01:12:02,130 liked, if you don't mind, just take a piece of paper, 1124 01:12:02,130 --> 01:12:05,310 make a note on it, and drop it off on the way out. 1125 01:12:05,310 --> 01:12:05,960 Thanks a lot. 1126 01:12:05,960 --> 01:12:08,320 See you on Thursday.