1 00:00:00,100 --> 00:00:02,450 The following content is provided under a Creative 2 00:00:02,450 --> 00:00:03,830 Commons license. 3 00:00:03,830 --> 00:00:06,070 Your support will help MIT OpenCourseWare 4 00:00:06,070 --> 00:00:10,170 continue to offer high quality educational resources for free. 5 00:00:10,170 --> 00:00:12,710 To make a donation or to view additional materials 6 00:00:12,710 --> 00:00:16,620 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,620 --> 00:00:17,325 at ocw.mit.edu. 8 00:00:21,156 --> 00:00:23,670 PROFESSOR: In a minute, we're going 9 00:00:23,670 --> 00:00:26,430 to go through this exercise of key concepts for the week. 10 00:00:26,430 --> 00:00:30,220 But for you early birds, I'll offer, 11 00:00:30,220 --> 00:00:34,070 and we may get to some of this. 12 00:00:34,070 --> 00:00:37,130 Is there just something from the weak or the problem set? 13 00:00:37,130 --> 00:00:39,820 Just some topic that's still bugging you 14 00:00:39,820 --> 00:00:41,420 that you just don't get and you'd 15 00:00:41,420 --> 00:00:44,170 like me to put on a little list and I might get to it? 16 00:00:47,544 --> 00:00:50,436 AUDIENCE: So if we solve for velocity, 17 00:00:50,436 --> 00:00:51,882 is it enough to just take the time 18 00:00:51,882 --> 00:00:53,810 derivative of that velocity. 19 00:00:53,810 --> 00:00:55,410 PROFESSOR: Yes. 20 00:00:55,410 --> 00:01:00,770 So if you have all of the terms, the full velocity 21 00:01:00,770 --> 00:01:05,110 expression in vector notation, and you 22 00:01:05,110 --> 00:01:08,520 want to know the acceleration of that point, 23 00:01:08,520 --> 00:01:12,560 just take the derivatives taking care of all the rotating unit 24 00:01:12,560 --> 00:01:13,929 vectors and all that stuff. 25 00:01:13,929 --> 00:01:14,429 Yeah? 26 00:01:14,429 --> 00:01:16,720 AUDIENCE: When you take the derivative of the velocity, 27 00:01:16,720 --> 00:01:20,057 where does the Coriolis term drop from? 28 00:01:20,057 --> 00:01:21,790 PROFESSOR: Where does it fall out of? 29 00:01:21,790 --> 00:01:24,380 AUDIENCE: Yeah. 30 00:01:24,380 --> 00:01:26,410 PROFESSOR: I can't answer it in words. 31 00:01:26,410 --> 00:01:28,070 I can't remember exactly. 32 00:01:28,070 --> 00:01:32,610 But Coriolis term-- curiously, it 33 00:01:32,610 --> 00:01:36,370 comes from two different places. 34 00:01:36,370 --> 00:01:42,380 It's two omega v rel kind of term. 35 00:01:42,380 --> 00:01:45,720 And actually, it comes from two different places 36 00:01:45,720 --> 00:01:48,916 when you grind out the derivatives of the velocity. 37 00:01:48,916 --> 00:01:49,862 AUDIENCE: [INAUDIBLE]. 38 00:01:54,592 --> 00:02:01,930 PROFESSOR: So in terms of using the general-- so I remember. 39 00:02:01,930 --> 00:02:03,980 Now what I want you to do and, partly, 40 00:02:03,980 --> 00:02:07,820 what this session is going to do is 41 00:02:07,820 --> 00:02:09,930 get you to use the acceleration formulas enough 42 00:02:09,930 --> 00:02:12,380 that you have those just committed to memory. 43 00:02:12,380 --> 00:02:13,620 Now we're going to let you have crib sheets when 44 00:02:13,620 --> 00:02:14,450 you go into quizzes. 45 00:02:14,450 --> 00:02:16,658 And I would always have the velocity and acceleration 46 00:02:16,658 --> 00:02:21,320 formulas in polar coordinates and in full vector form. 47 00:02:21,320 --> 00:02:23,260 If you identify each of the terms correctly, 48 00:02:23,260 --> 00:02:26,880 that formula will work just fine. 49 00:02:26,880 --> 00:02:29,020 Getting the right components of rotation rates 50 00:02:29,020 --> 00:02:30,560 can be a little tricky, as you might 51 00:02:30,560 --> 00:02:34,920 have found in that homework problem from this week. 52 00:02:34,920 --> 00:02:39,440 In that time derivative of a rotating vector, 53 00:02:39,440 --> 00:02:41,360 the omega cross the vector. 54 00:02:41,360 --> 00:02:43,250 What omega is that? 55 00:02:43,250 --> 00:02:44,390 That's tricky. 56 00:02:44,390 --> 00:02:47,170 That you have to be careful with. 57 00:02:47,170 --> 00:02:48,537 But really remember. 58 00:02:48,537 --> 00:02:50,870 If you remember the velocity and acceleration equations, 59 00:02:50,870 --> 00:02:52,050 you can trust them. 60 00:02:52,050 --> 00:02:54,490 They will work. 61 00:02:54,490 --> 00:02:58,270 So OK, we can start now. 62 00:02:58,270 --> 00:03:02,015 And I'll go back to here. 63 00:03:06,330 --> 00:03:10,440 So take that minute, like we did last time. 64 00:03:10,440 --> 00:03:12,700 Write down on a piece of paper two, three, four. 65 00:03:12,700 --> 00:03:15,200 What you think are the key concepts 66 00:03:15,200 --> 00:03:18,030 that were important in the past week. 67 00:03:18,030 --> 00:03:22,290 By important, you need to know them to do a good job 68 00:03:22,290 --> 00:03:24,340 on the quiz coming up. 69 00:03:24,340 --> 00:03:26,750 Important concepts. 70 00:03:26,750 --> 00:03:29,546 OK, let's make a list. 71 00:03:29,546 --> 00:03:31,212 AUDIENCE: Did you say polar coordinates? 72 00:03:31,212 --> 00:03:33,296 PROFESSOR: Polar coordinates. 73 00:03:33,296 --> 00:03:36,340 And I'm going to generalize that to call 74 00:03:36,340 --> 00:03:47,050 it choosing coordinate systems. 75 00:03:50,220 --> 00:03:52,030 There's an art to that. 76 00:03:52,030 --> 00:03:54,639 So when you to use them. 77 00:03:54,639 --> 00:03:55,930 It's [INAUDIBLE] of that, yeah. 78 00:03:55,930 --> 00:03:58,305 AUDIENCE: The equation for some of the parts? 79 00:03:58,305 --> 00:03:59,400 PROFESSOR: Oh, yeah. 80 00:03:59,400 --> 00:04:01,720 The torque equation. 81 00:04:01,720 --> 00:04:08,897 Some of the external torques vectors is dh/dt. 82 00:04:13,470 --> 00:04:17,144 And torques are always with respect to some point. 83 00:04:17,144 --> 00:04:20,690 So dh/dt. 84 00:04:20,690 --> 00:04:24,680 Angular momentum is with respect to some point. 85 00:04:24,680 --> 00:04:27,220 And there's a second term to this, though. 86 00:04:27,220 --> 00:04:33,580 And it is velocity of the point in the inertial frame 87 00:04:33,580 --> 00:04:37,570 cross the linear momentum in the inertial frame. 88 00:04:40,110 --> 00:04:42,020 That one, you haven't had to use much yet. 89 00:04:42,020 --> 00:04:47,561 But that's one of the two really important Newton law kind 90 00:04:47,561 --> 00:04:49,060 of things that we use in the course. 91 00:04:49,060 --> 00:04:51,350 Sum of forces equals mass times acceleration. 92 00:04:51,350 --> 00:04:53,800 Sum of torques equals that. 93 00:04:53,800 --> 00:04:55,574 OK, I have another one. 94 00:04:55,574 --> 00:04:57,562 AUDIENCE: Derivative of rotating unit vectors. 95 00:04:57,562 --> 00:04:59,550 PROFESSOR: Derivative rotating vectors. 96 00:05:15,480 --> 00:05:19,260 And taking derivatives rotating vectors always boils down to, 97 00:05:19,260 --> 00:05:22,040 eventually, it's down to you got to do the unit vector. 98 00:05:22,040 --> 00:05:23,130 So that's part of it. 99 00:05:23,130 --> 00:05:24,046 How about another one? 100 00:05:27,160 --> 00:05:29,080 AUDIENCE: [INAUDIBLE]. 101 00:05:29,080 --> 00:05:34,020 PROFESSOR: OK, so I'll generalize that to, 102 00:05:34,020 --> 00:05:45,740 essentially, being able to find equations of motion. 103 00:05:45,740 --> 00:06:05,410 So from sum of forces and-- all right. 104 00:06:05,410 --> 00:06:08,580 OK, anything else? 105 00:06:15,460 --> 00:06:26,615 I'll tell you one really important one that you've been. 106 00:06:26,615 --> 00:06:29,930 Maybe you think we did it last week. 107 00:06:29,930 --> 00:06:36,620 I'd say being able to find accelerations 108 00:06:36,620 --> 00:06:38,500 and moving coordinate systems. 109 00:06:38,500 --> 00:06:40,410 We did mostly velocities last week. 110 00:06:40,410 --> 00:06:43,650 But finding velocities and accelerations 111 00:06:43,650 --> 00:06:45,555 in rotating and moving frames. 112 00:07:05,240 --> 00:07:07,540 OK, so that's a pretty good list. 113 00:07:07,540 --> 00:07:10,370 [INAUDIBLE] ask me what I thought the important things 114 00:07:10,370 --> 00:07:11,350 from the week were? 115 00:07:15,090 --> 00:07:17,065 That hits the important stuff. 116 00:07:17,065 --> 00:07:17,565 OK. 117 00:07:22,890 --> 00:07:26,140 What we were talking about as people were arriving is 118 00:07:26,140 --> 00:07:28,410 and I might not get this today if we run out of time. 119 00:07:28,410 --> 00:07:30,243 But does anybody just got a burning question 120 00:07:30,243 --> 00:07:33,902 about some concept that just didn't work out for you 121 00:07:33,902 --> 00:07:36,310 or a problem set that came up? 122 00:07:36,310 --> 00:07:36,810 Yeah? 123 00:07:36,810 --> 00:07:39,792 AUDIENCE: The derivatives for [INAUDIBLE] vectors 124 00:07:39,792 --> 00:07:43,271 and [INAUDIBLE]. 125 00:07:43,271 --> 00:07:47,831 PROFESSOR: So that's really the derivative of rotation rate 126 00:07:47,831 --> 00:07:48,330 vector. 127 00:07:48,330 --> 00:07:48,430 AUDIENCE: Yeah. 128 00:07:48,430 --> 00:07:49,840 PROFESSOR: That was, kind of, a nasty problem 129 00:07:49,840 --> 00:07:51,280 the first time you hit that. 130 00:07:51,280 --> 00:07:54,300 That was new, right? 131 00:07:54,300 --> 00:07:55,510 Let me write that one down. 132 00:08:06,849 --> 00:08:08,920 And rotating frames and all that. 133 00:08:08,920 --> 00:08:09,420 I get it. 134 00:08:09,420 --> 00:08:10,747 Does somebody else have another one? 135 00:08:10,747 --> 00:08:12,496 AUDIENCE: [INAUDIBLE] of that. [INAUDIBLE] 136 00:08:12,496 --> 00:08:15,494 just like which omega is it when you're looking [INAUDIBLE]. 137 00:08:15,494 --> 00:08:16,202 PROFESSOR: Right. 138 00:08:16,202 --> 00:08:17,150 Oh, yeah. 139 00:08:17,150 --> 00:08:18,100 Right. 140 00:08:18,100 --> 00:08:22,300 OK, anybody else a burning question? 141 00:08:22,300 --> 00:08:24,160 I think I'm actually going to start there. 142 00:08:24,160 --> 00:08:28,040 We're going to spend just a minute on this one. 143 00:08:28,040 --> 00:08:37,039 And as an example problem yesterday in the lecture, 144 00:08:37,039 --> 00:08:38,450 I basically did the same. 145 00:08:38,450 --> 00:08:40,970 I had a rotor here. 146 00:08:40,970 --> 00:08:43,789 It could have a disk on it just to make it clear. 147 00:08:43,789 --> 00:08:47,360 So this thing is rotating at some omega 2. 148 00:08:47,360 --> 00:08:53,790 And it was on a merry go round going at some omega 1. 149 00:08:53,790 --> 00:08:56,400 And I'll pick at coordinate system here. 150 00:08:56,400 --> 00:09:05,020 I think it was yxz that I wrote in the lecture notes. 151 00:09:05,020 --> 00:09:06,440 And that's the rotating one. 152 00:09:06,440 --> 00:09:09,660 So you have an o and an a here. 153 00:09:09,660 --> 00:09:17,875 And you have a fixed frame that but the y 154 00:09:17,875 --> 00:09:19,980 ones, the little ones, attach to the platform. 155 00:09:19,980 --> 00:09:21,330 It's going around. 156 00:09:21,330 --> 00:09:22,410 OK? 157 00:09:22,410 --> 00:09:29,060 So the rotation rate of the platform here is what? 158 00:09:34,370 --> 00:09:38,670 In magnitude and unit vector. 159 00:09:38,670 --> 00:09:40,470 Pardon? 160 00:09:40,470 --> 00:09:43,750 Mega 1 k hat. 161 00:09:43,750 --> 00:09:47,760 And does it matter which k because they're parallel. 162 00:09:47,760 --> 00:09:49,460 So the little k and the big k. 163 00:09:49,460 --> 00:10:01,190 OK, what's the rotation rate of this shift, 164 00:10:01,190 --> 00:10:03,740 in terms of a magnitude and a unit vector? 165 00:10:03,740 --> 00:10:07,324 AUDIENCE: Omega 2 lower case y dot. 166 00:10:07,324 --> 00:10:08,282 PROFESSOR: OK, omega 2. 167 00:10:08,282 --> 00:10:14,310 And I'll call it at j1 here. 168 00:10:14,310 --> 00:10:16,710 It's hard to distinguish my handwriting on the board 169 00:10:16,710 --> 00:10:19,050 between uppers lowers. 170 00:10:19,050 --> 00:10:21,590 So with z1's, x1's. 171 00:10:21,590 --> 00:10:24,565 OK, so it's in the J1 direction, right? 172 00:10:24,565 --> 00:10:26,440 And I've just intentionally made it positive. 173 00:10:26,440 --> 00:10:28,810 If it were in the other direction, have a minus there. 174 00:10:28,810 --> 00:10:37,290 OK, so now at the end, the problem 175 00:10:37,290 --> 00:10:43,520 asked for the time derivative of the rotation 176 00:10:43,520 --> 00:10:50,320 rate of the spinning. 177 00:10:50,320 --> 00:10:52,510 The problem that in the homework, 178 00:10:52,510 --> 00:10:55,000 this thing was actually inclined, basically. 179 00:10:59,990 --> 00:11:09,090 It's easy to just say the total rotation rate of the shaft. 180 00:11:09,090 --> 00:11:10,590 Let's get that settled first. 181 00:11:10,590 --> 00:11:15,770 In the fixed inertial frame, the total rotation rate of this 182 00:11:15,770 --> 00:11:17,102 is what? 183 00:11:17,102 --> 00:11:19,026 AUDIENCE: [INAUDIBLE]. 184 00:11:19,026 --> 00:11:21,200 PROFESSOR: Yeah, and you can sum rotation rates. 185 00:11:21,200 --> 00:11:30,100 So it's omega 1 k hat plus omega 2 j1. 186 00:11:30,100 --> 00:11:35,540 And I want to take the time derivative of this omega chafed 187 00:11:35,540 --> 00:11:36,040 total. 188 00:11:39,150 --> 00:11:43,430 So it's d by dt of these things. 189 00:11:43,430 --> 00:11:45,210 So I can start to do that. 190 00:11:45,210 --> 00:11:48,990 So d by by dt is, in this case, is k change a direction. 191 00:11:48,990 --> 00:11:50,260 No problem. 192 00:11:50,260 --> 00:11:52,610 Could that change? 193 00:11:52,610 --> 00:11:54,230 Could it? 194 00:11:54,230 --> 00:11:55,284 Yeah, sure. 195 00:11:55,284 --> 00:11:56,950 I mean, it could be accelerating, right. 196 00:11:56,950 --> 00:11:58,990 If it is, then you get a term here 197 00:11:58,990 --> 00:12:04,530 that comes from this one that would be an omega 1 dot k hat 198 00:12:04,530 --> 00:12:05,030 plus. 199 00:12:05,030 --> 00:12:08,036 And now we need to take the derivative of this piece. 200 00:12:08,036 --> 00:12:09,012 All right. 201 00:12:11,940 --> 00:12:14,600 And that's now a rotating vector. 202 00:12:14,600 --> 00:12:18,040 And we've said that the derivative of any rotating 203 00:12:18,040 --> 00:12:28,180 vector dq dt is the derivative of t as if the rotation is 0. 204 00:12:28,180 --> 00:12:32,200 It's its magnitude increasing, not its direction changing. 205 00:12:32,200 --> 00:12:35,670 Plus omega cross q. 206 00:12:35,670 --> 00:12:38,790 And I've intentionally left sub-scripts and server scripts 207 00:12:38,790 --> 00:12:40,540 off here because that's the issue here. 208 00:12:40,540 --> 00:12:44,270 Which ones go where? 209 00:12:44,270 --> 00:12:47,650 All right, so this is what we need to apply 210 00:12:47,650 --> 00:12:50,430 to taking that derivative. 211 00:12:50,430 --> 00:12:54,790 We took this derivative and found this term 212 00:12:54,790 --> 00:12:59,370 was what gave us the omega 1.k. 213 00:12:59,370 --> 00:13:04,687 And when taking derivative of this, what was the answer here? 214 00:13:04,687 --> 00:13:05,312 AUDIENCE: Zero. 215 00:13:05,312 --> 00:13:08,649 PROFESSOR: Zero because the rotation rate is omega 1. 216 00:13:08,649 --> 00:13:09,440 You're crossing it. 217 00:13:09,440 --> 00:13:11,785 Well omega 1k crossed with omega 1k. 218 00:13:11,785 --> 00:13:12,760 That goes with zero. 219 00:13:12,760 --> 00:13:14,520 There's no change in direction. 220 00:13:14,520 --> 00:13:16,680 So we applied this formula once to this. 221 00:13:16,680 --> 00:13:19,140 Now we're going to apply this formula here. 222 00:13:19,140 --> 00:13:22,910 So I'll do it this way. 223 00:13:22,910 --> 00:13:24,170 So this is term one. 224 00:13:29,640 --> 00:13:32,250 Term one gave us this. 225 00:13:32,250 --> 00:13:34,560 Term two is going to give us this. 226 00:13:34,560 --> 00:13:37,820 So take this. 227 00:13:37,820 --> 00:13:42,270 Take the derivative of-- where we go here-- this guy. 228 00:13:44,890 --> 00:13:46,430 Now when I say this is the-- it's 229 00:13:46,430 --> 00:13:49,520 the partial derivative of this thing, 230 00:13:49,520 --> 00:13:53,990 ignoring the contribution from the change of direction. 231 00:13:53,990 --> 00:13:56,610 So it's as if some rotation were zero. 232 00:13:56,610 --> 00:13:57,785 Which rotation? 233 00:13:57,785 --> 00:13:58,284 Right. 234 00:13:58,284 --> 00:13:59,188 AUDIENCE: Omega 1? 235 00:13:59,188 --> 00:14:01,090 PROFESSOR: Omega 1. 236 00:14:01,090 --> 00:14:02,350 Good. 237 00:14:02,350 --> 00:14:07,538 So what is the answer to this piece of that derivative? 238 00:14:07,538 --> 00:14:08,442 AUDIENCE: Omega 2. 239 00:14:08,442 --> 00:14:14,980 PROFESSOR: Omega 2 dot what direction? 240 00:14:14,980 --> 00:14:16,110 Same direction, right. 241 00:14:16,110 --> 00:14:18,300 Now we want to apply this piece. 242 00:14:18,300 --> 00:14:22,050 So now the question comes down plus some omega 243 00:14:22,050 --> 00:14:23,622 cross with the q. 244 00:14:23,622 --> 00:14:25,600 And what's the q in this case? 245 00:14:30,710 --> 00:14:33,270 Two. 246 00:14:33,270 --> 00:14:36,620 So this is omega 2 j1 hat. 247 00:14:36,620 --> 00:14:38,560 That's coming from here. 248 00:14:38,560 --> 00:14:40,410 That's what this piece is. 249 00:14:40,410 --> 00:14:43,200 So now we need what's crossed with it. 250 00:14:43,200 --> 00:14:45,042 AUDIENCE: [INAUDIBLE]. 251 00:14:45,042 --> 00:14:45,625 PROFESSOR: Hm? 252 00:14:45,625 --> 00:14:47,080 AUDIENCE: Omega [INAUDIBLE]. 253 00:14:57,415 --> 00:15:03,290 PROFESSOR: I'm trying to think of a clear way to help you. 254 00:15:03,290 --> 00:15:07,780 What is causing its direction to change? 255 00:15:07,780 --> 00:15:10,500 AUDIENCE: [INAUDIBLE]. 256 00:15:10,500 --> 00:15:11,470 PROFESSOR: Right. 257 00:15:11,470 --> 00:15:13,130 That's the one you care about. 258 00:15:13,130 --> 00:15:15,500 What's causing its direction to change? 259 00:15:15,500 --> 00:15:18,320 What's causing this thing to come up? 260 00:15:18,320 --> 00:15:20,920 And its direction is changing because the platforms 261 00:15:20,920 --> 00:15:22,040 are moving. 262 00:15:22,040 --> 00:15:25,110 So it's omega 1. 263 00:15:25,110 --> 00:15:27,740 K cross j. 264 00:15:27,740 --> 00:15:31,323 So k cross j gives you? 265 00:15:31,323 --> 00:15:33,207 AUDIENCE: [INAUDIBLE]. 266 00:15:33,207 --> 00:15:39,320 PROFESSOR: K cross j negative i 1 in the rotating frame, right. 267 00:15:39,320 --> 00:15:40,650 So this whole thing. 268 00:15:40,650 --> 00:15:50,180 Omega 1 dot k plus omega 2 dot j1 plus. 269 00:15:50,180 --> 00:15:51,680 No, this is a minus, actually. 270 00:15:51,680 --> 00:15:55,160 We said, minus omega 1. 271 00:15:55,160 --> 00:15:56,320 Omega 2. 272 00:15:56,320 --> 00:15:58,690 K cross j is i 1 hat. 273 00:16:02,402 --> 00:16:03,330 OK? 274 00:16:03,330 --> 00:16:07,360 And finally, how do you deal with the fact 275 00:16:07,360 --> 00:16:11,760 if this rotor had been on an inclined angle? 276 00:16:11,760 --> 00:16:15,740 Some phi, which now this is the exact problem 277 00:16:15,740 --> 00:16:18,570 that was on homework. 278 00:16:18,570 --> 00:16:23,630 So propose a way of attacking that. 279 00:16:23,630 --> 00:16:26,984 Changes this problem just a little bit. 280 00:16:26,984 --> 00:16:29,604 AUDIENCE: Can you change the reference frame [INAUDIBLE]? 281 00:16:29,604 --> 00:16:31,770 PROFESSOR: No, you don't change the reference frame. 282 00:16:31,770 --> 00:16:33,330 You still have a coordinate system. 283 00:16:35,914 --> 00:16:37,330 Work with the coordinates you got. 284 00:16:37,330 --> 00:16:40,510 AUDIENCE: [INAUDIBLE] to its components? 285 00:16:40,510 --> 00:16:44,672 PROFESSOR: In which reference frame? 286 00:16:44,672 --> 00:16:45,604 AUDIENCE: [INAUDIBLE]. 287 00:16:45,604 --> 00:16:47,090 PROFESSOR: Yeah, the moving [? one. ?] 288 00:16:47,090 --> 00:16:48,798 That makes it easy because this thing now 289 00:16:48,798 --> 00:16:53,260 has a vector omega 2 like that. 290 00:16:53,260 --> 00:16:55,310 And you can break it up into a piece like that 291 00:16:55,310 --> 00:16:56,660 and a piece like that. 292 00:16:56,660 --> 00:16:59,590 So now this is phi. 293 00:16:59,590 --> 00:17:03,480 Omega 2 cosine phi is the side, right. 294 00:17:03,480 --> 00:17:04,900 Omega 2 sine phi is that side. 295 00:17:04,900 --> 00:17:07,369 What direction is it? 296 00:17:07,369 --> 00:17:09,069 What's its unit vector direction? 297 00:17:09,069 --> 00:17:10,020 This piece. 298 00:17:10,020 --> 00:17:10,740 AUDIENCE: K? 299 00:17:10,740 --> 00:17:12,099 PROFESSOR: K, OK. 300 00:17:12,099 --> 00:17:18,191 And we're going to end up crossing it with this term, 301 00:17:18,191 --> 00:17:18,690 right. 302 00:17:18,690 --> 00:17:21,540 So k cross k. 303 00:17:21,540 --> 00:17:22,160 Nothing. 304 00:17:22,160 --> 00:17:25,250 So in fact, this bit doesn't change in direction 305 00:17:25,250 --> 00:17:26,020 as it's rotating. 306 00:17:26,020 --> 00:17:26,690 Does it? 307 00:17:26,690 --> 00:17:29,660 So it isn't going to contribute to that second term. 308 00:17:29,660 --> 00:17:33,070 Only piece you have to be concerned with is this one. 309 00:17:33,070 --> 00:17:38,350 So down in here in this part of it, all you would do 310 00:17:38,350 --> 00:17:41,370 is say omega 1 k cross. 311 00:17:41,370 --> 00:17:46,440 And now you just put in the q full omega 312 00:17:46,440 --> 00:17:54,630 2, which is omega 2 sine k plus omega 2 cosine. 313 00:17:54,630 --> 00:17:56,410 Sorry about the writing here. 314 00:17:56,410 --> 00:17:58,534 And do the cross product. 315 00:17:58,534 --> 00:17:59,950 You find out that term disappears. 316 00:17:59,950 --> 00:18:04,830 And you get an omega 1 omega 2 cosine. 317 00:18:04,830 --> 00:18:08,480 And this term here is still the answer. 318 00:18:08,480 --> 00:18:12,560 But you end up with a cosine phi in here 319 00:18:12,560 --> 00:18:15,330 because it's just the component of omega 2 that's 320 00:18:15,330 --> 00:18:18,730 in the j1 direction now that contributes. 321 00:18:18,730 --> 00:18:21,140 OK, great. 322 00:18:21,140 --> 00:18:22,260 Good. 323 00:18:22,260 --> 00:18:22,800 That help? 324 00:18:26,590 --> 00:18:27,570 I'll leave this. 325 00:18:27,570 --> 00:18:29,320 Let's get on to the problem for the day. 326 00:18:29,320 --> 00:18:39,180 The problem for the day is you've got a rotating arm. 327 00:18:39,180 --> 00:18:41,230 You have a mass sitting on it. 328 00:18:41,230 --> 00:18:46,190 The arm is rotating at some rotation rate omega. 329 00:18:46,190 --> 00:18:52,220 You might have an acceleration omega dot theta double dot. 330 00:18:56,000 --> 00:18:56,950 This mask can slide. 331 00:18:59,490 --> 00:19:08,190 So the problem is this thing is-- I'll start here. 332 00:19:08,190 --> 00:19:10,430 So it's rotating like this. 333 00:19:10,430 --> 00:19:13,670 And eventually, the thing begins to slide. 334 00:19:13,670 --> 00:19:15,590 And in fact, if I do it, this actually 335 00:19:15,590 --> 00:19:19,580 works a little better here because I have 336 00:19:19,580 --> 00:19:20,950 a little tiny groove in this. 337 00:19:20,950 --> 00:19:22,030 It helps a little bit. 338 00:19:22,030 --> 00:19:23,320 So it's rotating like that. 339 00:19:23,320 --> 00:19:24,570 It eventually begins to slide. 340 00:19:24,570 --> 00:19:28,630 And if the rotation rate is slow, it slides down. 341 00:19:28,630 --> 00:19:33,710 If I made the rotation rate fast enough, it slides up. 342 00:19:33,710 --> 00:19:37,740 And the angle at which it happens clearly 343 00:19:37,740 --> 00:19:40,920 depends on rotation rates, accelerations, the angle, 344 00:19:40,920 --> 00:19:43,570 friction coefficients, and things of that sort. 345 00:19:43,570 --> 00:19:47,520 So what I want you to start with is here's, 346 00:19:47,520 --> 00:19:50,490 basically, a drawing of it. 347 00:19:50,490 --> 00:19:57,060 And I don't want to give you any excess information. 348 00:19:57,060 --> 00:20:00,710 But what I want you to do is each, by yourselves, 349 00:20:00,710 --> 00:20:03,180 take just a minute, and draw a free body 350 00:20:03,180 --> 00:20:05,126 diagram of this problem. 351 00:20:05,126 --> 00:20:07,500 And then, after that, I'm going to have you get in groups 352 00:20:07,500 --> 00:20:08,280 and improve on it. 353 00:20:08,280 --> 00:20:09,270 But start yourself. 354 00:20:09,270 --> 00:20:12,990 Draw a free body diagram that will describe 355 00:20:12,990 --> 00:20:17,755 this mask on this thing, plank. 356 00:20:17,755 --> 00:20:25,537 OK, we've got a sketch with a little free body diagram on it. 357 00:20:25,537 --> 00:20:27,120 What I want you to do now, we're going 358 00:20:27,120 --> 00:20:29,630 to do this two or three times today. 359 00:20:29,630 --> 00:20:31,640 Get in groups of four five here. 360 00:20:31,640 --> 00:20:36,040 Compare notes, and come up with a group free body diagram. 361 00:20:36,040 --> 00:20:38,795 So you're pretty close. 362 00:20:42,490 --> 00:20:43,100 Let me look. 363 00:20:43,100 --> 00:20:44,890 Where was your guys solution again? 364 00:20:50,940 --> 00:20:53,890 There is maybe about as good as it gets. 365 00:20:57,090 --> 00:21:02,590 So here's our mass. 366 00:21:02,590 --> 00:21:06,630 I think everybody had an mg. 367 00:21:06,630 --> 00:21:12,340 Almost everybody had a normal force. 368 00:21:12,340 --> 00:21:14,120 Everybody had a friction force. 369 00:21:14,120 --> 00:21:16,770 Some discussion about what direction the friction force 370 00:21:16,770 --> 00:21:18,330 ought to be in. 371 00:21:18,330 --> 00:21:21,980 I'll draw it uphill here. 372 00:21:21,980 --> 00:21:26,100 And I'll call it ft for tangential, here. 373 00:21:30,120 --> 00:21:32,900 So some had about that much. 374 00:21:32,900 --> 00:21:36,690 There's a very important missing piece, which 375 00:21:36,690 --> 00:21:39,340 actually several groups had. 376 00:21:39,340 --> 00:21:41,670 What is that? 377 00:21:41,670 --> 00:21:45,281 Can anybody guess what I'm-- three of your groups had 378 00:21:45,281 --> 00:21:47,280 something on here that I don't have up here yet. 379 00:21:49,870 --> 00:21:52,756 I hear coordinates. 380 00:21:52,756 --> 00:21:54,130 This one, you can almost get away 381 00:21:54,130 --> 00:21:56,254 with not putting it in the coordinate system first. 382 00:21:56,254 --> 00:21:59,330 But you, usually, need to have a coordinate system in order 383 00:21:59,330 --> 00:22:01,455 to determine the direction of the forces. 384 00:22:01,455 --> 00:22:02,560 OK? 385 00:22:02,560 --> 00:22:04,660 And we'll get to that in a second. 386 00:22:04,660 --> 00:22:09,505 So everybody that I saw had written down, I saw, 387 00:22:09,505 --> 00:22:12,380 a number of the little x, y, z frames sitting out here 388 00:22:12,380 --> 00:22:14,160 like that. 389 00:22:14,160 --> 00:22:16,270 Maybe one was lined up. 390 00:22:16,270 --> 00:22:16,820 Not sure. 391 00:22:16,820 --> 00:22:17,460 Most not. 392 00:22:17,460 --> 00:22:19,780 Most like this. 393 00:22:19,780 --> 00:22:22,720 Is polar coordinates a good choice for this problem? 394 00:22:22,720 --> 00:22:24,076 Why not? 395 00:22:24,076 --> 00:22:25,022 AUDIENCE: [INAUDIBLE]. 396 00:22:28,806 --> 00:22:32,840 PROFESSOR: So the question is whether or not r theta and z 397 00:22:32,840 --> 00:22:35,920 can completely describe the motion in the problem. 398 00:22:35,920 --> 00:22:37,560 AUDIENCE: You have tangential forces, 399 00:22:37,560 --> 00:22:39,634 so it's better if you use rectangular. 400 00:22:42,479 --> 00:22:43,895 PROFESSOR: The radius is changing. 401 00:22:43,895 --> 00:22:46,490 But that's known as r dot. 402 00:22:46,490 --> 00:22:48,790 r dot can handle the motion to this this way. 403 00:22:48,790 --> 00:22:51,880 r double dot can handle acceleration this way. 404 00:22:51,880 --> 00:22:54,320 Theta dot can handle this motion. 405 00:22:54,320 --> 00:22:57,690 Theta double dot can handle that acceleration. 406 00:22:57,690 --> 00:23:00,350 Actually, is there any z motion or z-forces? 407 00:23:00,350 --> 00:23:01,000 No. 408 00:23:01,000 --> 00:23:02,875 Actually, the polar coordinates will actually 409 00:23:02,875 --> 00:23:04,550 work here just fine. 410 00:23:04,550 --> 00:23:08,430 So you definitely have to pick a coordinate system. 411 00:23:08,430 --> 00:23:13,490 So we're going to have an that system and theta hat 412 00:23:13,490 --> 00:23:16,850 in this direction because in this problem, 413 00:23:16,850 --> 00:23:23,910 whether you use x and y as a rotating frame here or r theta, 414 00:23:23,910 --> 00:23:29,050 you want to pick your coordinate system so that these things 415 00:23:29,050 --> 00:23:30,620 break down easily. 416 00:23:30,620 --> 00:23:32,950 In this case, the normal equation 417 00:23:32,950 --> 00:23:34,680 will be in the theta hat direction. 418 00:23:34,680 --> 00:23:39,400 The sliding direction will be in just one unit vector direction. 419 00:23:39,400 --> 00:23:42,280 So you don't have multiple components in this direction. 420 00:23:42,280 --> 00:23:45,490 If you use that coordinate system to describe this motion, 421 00:23:45,490 --> 00:23:46,660 you have x and y terms. 422 00:23:46,660 --> 00:23:48,500 It makes it harder. 423 00:23:48,500 --> 00:23:50,610 So this is a pretty good coordinate system to use. 424 00:23:50,610 --> 00:23:53,060 So that's a really important missing piece 425 00:23:53,060 --> 00:23:56,460 is the coordinate system, if you don't have it. 426 00:23:56,460 --> 00:24:03,400 And as an aside about free body diagrams-- free body diagrams, 427 00:24:03,400 --> 00:24:09,870 except my d is turned around-- how 428 00:24:09,870 --> 00:24:13,375 do you know that the friction force is in that direction. 429 00:24:13,375 --> 00:24:14,770 You had to assume something. 430 00:24:18,126 --> 00:24:20,500 This is such a trivial problem you can kind of figure out 431 00:24:20,500 --> 00:24:20,832 in your head. 432 00:24:20,832 --> 00:24:21,970 I gave the demonstration. 433 00:24:21,970 --> 00:24:22,690 It slid down. 434 00:24:22,690 --> 00:24:25,460 But also, I said , you do it fast enough, it goes up. 435 00:24:25,460 --> 00:24:26,840 So you don't really know. 436 00:24:26,840 --> 00:24:31,530 So you have to have made an assumption about it. 437 00:24:31,530 --> 00:24:45,590 So the general rule is to assign or assume 438 00:24:45,590 --> 00:24:59,170 positive values for all motions. 439 00:24:59,170 --> 00:25:08,105 And then, you deduce the direction of forces. 440 00:25:11,660 --> 00:25:15,420 And I'll take another quick aside here 441 00:25:15,420 --> 00:25:19,070 to illustrate this in a problem that works better 442 00:25:19,070 --> 00:25:23,530 than the current one for this. 443 00:25:23,530 --> 00:25:27,330 An obvious coordinate system for this little mass on rollers 444 00:25:27,330 --> 00:25:28,170 inertial frame. 445 00:25:28,170 --> 00:25:30,820 Here's an x. 446 00:25:30,820 --> 00:25:35,360 And I asked you to do a free body diagram of that. 447 00:25:35,360 --> 00:25:37,390 Well you would show me a normal force. 448 00:25:37,390 --> 00:25:39,510 You'd show me an mg. 449 00:25:39,510 --> 00:25:41,290 But now I ask you to tell me the direction 450 00:25:41,290 --> 00:25:42,530 to draw the spring force. 451 00:25:46,640 --> 00:25:51,450 So if you adopt the rule that for each body 452 00:25:51,450 --> 00:25:55,770 you're working with, you assume that it has positive x, 453 00:25:55,770 --> 00:26:00,000 positive x dot, positive y, positive y dot 454 00:26:00,000 --> 00:26:03,139 then you can deduce what direction of the forces 455 00:26:03,139 --> 00:26:04,180 would result [INAUDIBLE]. 456 00:26:04,180 --> 00:26:08,900 So in this case, the only motion that generates a force up here 457 00:26:08,900 --> 00:26:10,966 is what? 458 00:26:10,966 --> 00:26:12,900 AUDIENCE: [INAUDIBLE]. 459 00:26:12,900 --> 00:26:15,990 PROFESSOR: Is x. x dot. x double dot. 460 00:26:15,990 --> 00:26:19,890 The only motion that will generate a force 461 00:26:19,890 --> 00:26:23,470 is displacement because it generates a force where? 462 00:26:23,470 --> 00:26:24,464 In the spring. 463 00:26:24,464 --> 00:26:25,880 If they add a dash [? pod ?] here, 464 00:26:25,880 --> 00:26:28,570 them velocity generates a force, too. 465 00:26:28,570 --> 00:26:34,840 But displacement, if you assume the displacement is positive, 466 00:26:34,840 --> 00:26:37,670 then which direction is the spring force? 467 00:26:37,670 --> 00:26:41,100 OK, so then you have spring force this way. 468 00:26:41,100 --> 00:26:43,691 And that's value kx. 469 00:26:43,691 --> 00:26:45,690 And when you write it in the equation of motion, 470 00:26:45,690 --> 00:26:47,750 you put a minus sign to account for the direction 471 00:26:47,750 --> 00:26:49,780 of that error. 472 00:26:49,780 --> 00:26:51,130 OK. 473 00:26:51,130 --> 00:26:55,280 So in this problem we kind of-- and if you ever 474 00:26:55,280 --> 00:26:57,376 do it wrong, if you pick this direction wrong, 475 00:26:57,376 --> 00:26:58,500 what happens in the answer? 476 00:26:58,500 --> 00:27:00,500 Will you still get the right answer? 477 00:27:00,500 --> 00:27:03,020 If you're consistent, you'll get another negative sign 478 00:27:03,020 --> 00:27:04,370 popping up to fix it. 479 00:27:04,370 --> 00:27:04,870 OK. 480 00:27:07,510 --> 00:27:09,060 Next. 481 00:27:09,060 --> 00:27:12,060 So I think we're in good shape now. 482 00:27:12,060 --> 00:27:14,010 I'm going to draw the friction force uphill 483 00:27:14,010 --> 00:27:16,610 because I'm going to assume this thing's going to slide down. 484 00:27:16,610 --> 00:27:20,070 And I'm going to simplify the problem a little bit for you. 485 00:27:20,070 --> 00:27:21,450 And I'm going to say that there's 486 00:27:21,450 --> 00:27:24,260 no angular acceleration. 487 00:27:24,260 --> 00:27:27,250 Constant angular rate. 488 00:27:27,250 --> 00:27:31,820 The distance that the mass starts off up the slope 489 00:27:31,820 --> 00:27:33,190 is at one and a half feet. 490 00:27:35,840 --> 00:27:37,440 So it's going up like this. 491 00:27:37,440 --> 00:27:40,290 And it, eventually, reaches 50 degrees. 492 00:27:40,290 --> 00:27:44,210 And at 50 degrees, it begins to slide downhill. 493 00:27:44,210 --> 00:27:47,660 So it's going up a constant rate and starts to slide. 494 00:27:47,660 --> 00:27:50,160 So there must be some friction coefficient 495 00:27:50,160 --> 00:27:54,430 that provides a system with a property such 496 00:27:54,430 --> 00:27:56,540 that it slides at 50 degrees. 497 00:27:56,540 --> 00:28:01,220 So the problem here is define the coefficient of friction. 498 00:28:01,220 --> 00:28:03,840 So now you're in your groups. 499 00:28:03,840 --> 00:28:05,070 Solve the problem. 500 00:28:05,070 --> 00:28:07,840 And what I really want you to do is come up with an expression. 501 00:28:07,840 --> 00:28:13,562 mu equals in variables and constants. 502 00:28:13,562 --> 00:28:15,270 Don't bother plugging in numbers in that. 503 00:28:18,825 --> 00:28:20,950 Well if you're comfortable using polar coordinates, 504 00:28:20,950 --> 00:28:22,074 that'd be the way to do it. 505 00:28:22,074 --> 00:28:23,890 But you need to write some equations now. 506 00:28:23,890 --> 00:28:26,800 And I suggest you need to think in terms of equations 507 00:28:26,800 --> 00:28:27,700 [? in ?] motion. 508 00:28:27,700 --> 00:28:33,420 I think it's time to come back together here and work on this. 509 00:28:37,170 --> 00:28:41,040 Two, three, or four groups are, sort of, struggling with this. 510 00:28:41,040 --> 00:28:45,340 And it's because you're not really 511 00:28:45,340 --> 00:28:51,440 going to first principles and doing the problem step by step. 512 00:28:51,440 --> 00:28:53,530 You're, sort of, jumping to the answer 513 00:28:53,530 --> 00:28:57,350 because it's a trivial, simple, Mickey Mouse problem that you 514 00:28:57,350 --> 00:29:00,030 did in high school. 515 00:29:00,030 --> 00:29:03,674 So I'm, kind of, pounding on you a little bit. 516 00:29:03,674 --> 00:29:05,590 We give you a simple problem so we can do them 517 00:29:05,590 --> 00:29:07,330 in a short period of time. 518 00:29:07,330 --> 00:29:09,980 But you need to learn to do them in the rigorous way 519 00:29:09,980 --> 00:29:14,030 so that you learn the real fundamental stuff that you 520 00:29:14,030 --> 00:29:16,860 have to know. 521 00:29:16,860 --> 00:29:19,750 So none of you-- you all are sort of thinking about, 522 00:29:19,750 --> 00:29:21,250 well, we got sum of the forces here. 523 00:29:21,250 --> 00:29:23,080 We've got an acceleration there. 524 00:29:23,080 --> 00:29:26,180 But nobody is just sitting down and saying 525 00:29:26,180 --> 00:29:31,370 that some of the forces is equal to the mass 526 00:29:31,370 --> 00:29:34,490 times the acceleration and working the problem out. 527 00:29:37,460 --> 00:29:38,550 OK, bad dog. 528 00:29:41,100 --> 00:29:44,960 Just because this problem is almost the statics problem 529 00:29:44,960 --> 00:29:48,870 doesn't mean you-- when things are statics problem, 530 00:29:48,870 --> 00:29:50,620 it just means that acceleration goes to 0. 531 00:29:50,620 --> 00:29:52,270 And the sum of the forces is now 0. 532 00:29:52,270 --> 00:29:54,350 And you solve the problem. 533 00:29:54,350 --> 00:29:59,410 Start with Newtons, in this case, it's Newtons second law. 534 00:29:59,410 --> 00:30:02,640 It helps to know an expression for acceleration, 535 00:30:02,640 --> 00:30:04,410 so you don't have to grind it out. 536 00:30:04,410 --> 00:30:08,510 So polar coordinates works pretty well in this problem. 537 00:30:08,510 --> 00:30:14,480 And I want you to commit to memory two acceleration 538 00:30:14,480 --> 00:30:15,500 equations. 539 00:30:15,500 --> 00:30:19,430 One in polar coordinates, and the one the general vector 1. 540 00:30:19,430 --> 00:30:22,640 So in polar coordinates, the one that I memorize 541 00:30:22,640 --> 00:30:32,180 is a with respect to o plus r double dot minus r theta 542 00:30:32,180 --> 00:30:44,060 dot squared r hat plus r theta double dot plus 2 omega r 543 00:30:44,060 --> 00:30:48,040 dot theta hat. 544 00:30:48,040 --> 00:30:53,500 Coriolis, Eulerian centripetal, and your linear acceleration. 545 00:30:53,500 --> 00:30:55,780 What's this term? 546 00:30:55,780 --> 00:30:58,938 What's it doing there? 547 00:30:58,938 --> 00:31:00,730 AUDIENCE: [INAUDIBLE]. 548 00:31:00,730 --> 00:31:04,780 PROFESSOR: Yeah, but the reason I remember it is, 549 00:31:04,780 --> 00:31:09,230 in this course, we break down every dynamics problem 550 00:31:09,230 --> 00:31:12,710 we're confronted with into sub problems that 551 00:31:12,710 --> 00:31:17,240 can be solved as the sum of a translation 552 00:31:17,240 --> 00:31:20,210 of a body plus a rotation. 553 00:31:20,210 --> 00:31:22,720 So this expression allows you to write down 554 00:31:22,720 --> 00:31:27,090 the acceleration of a point on a body, which is 555 00:31:27,090 --> 00:31:31,080 both translating and rotating. 556 00:31:31,080 --> 00:31:32,800 What this term accounts for. 557 00:31:36,750 --> 00:31:38,770 The acceleration contributed by what? 558 00:31:42,290 --> 00:31:45,430 So I've got this general-- let's see. 559 00:31:45,430 --> 00:31:48,570 I got a merry go round that's on wheel 560 00:31:48,570 --> 00:31:49,970 and is rolling along here. 561 00:31:49,970 --> 00:31:51,235 And I have an inertial frame. 562 00:31:54,400 --> 00:31:55,970 And I got a point a here. 563 00:31:55,970 --> 00:31:59,460 But now I've also got a system in here, 564 00:31:59,460 --> 00:32:04,880 which I'm describing with an r theta connected to this. 565 00:32:04,880 --> 00:32:07,530 Thing can have translational acceleration. 566 00:32:07,530 --> 00:32:09,182 That's that term. 567 00:32:09,182 --> 00:32:10,890 This problem doesn't happen to have that. 568 00:32:10,890 --> 00:32:11,880 It goes to 0. 569 00:32:11,880 --> 00:32:14,090 In this problem, that's a 0. 570 00:32:14,090 --> 00:32:21,160 In this problem, we can now apply this equation 571 00:32:21,160 --> 00:32:22,510 to that problem. 572 00:32:22,510 --> 00:32:26,665 How many sub equations are we going to get? 573 00:32:30,212 --> 00:32:30,920 And why no three. 574 00:32:30,920 --> 00:32:32,735 AUDIENCE: But I said two. 575 00:32:32,735 --> 00:32:34,270 PROFESSOR: I know you said two. 576 00:32:34,270 --> 00:32:35,978 But I'm asking why not three. 577 00:32:35,978 --> 00:32:37,730 AUDIENCE: Because [INAUDIBLE] theta. 578 00:32:37,730 --> 00:32:42,670 PROFESSOR: And a z, which you need because it describes theta 579 00:32:42,670 --> 00:32:44,100 dot. 580 00:32:44,100 --> 00:32:46,372 So you have a z direction. 581 00:32:46,372 --> 00:32:48,830 But is there any forces in this problem in the z direction? 582 00:32:48,830 --> 00:32:50,710 Are there any accelerations in the z direction? 583 00:32:50,710 --> 00:32:51,950 No, so you get a trivial answer there. 584 00:32:51,950 --> 00:32:53,533 So you could just write that one down. 585 00:32:53,533 --> 00:32:55,100 Mass times acceleration equals zero. 586 00:32:55,100 --> 00:32:56,016 And there's no forces. 587 00:32:56,016 --> 00:32:57,810 So there's three possible equations. 588 00:32:57,810 --> 00:33:00,590 Only two of them are meaningful in this problem. 589 00:33:00,590 --> 00:33:04,240 And we have one that we can summon the that component 590 00:33:04,240 --> 00:33:06,210 and one in the theta hat. 591 00:33:06,210 --> 00:33:11,200 So the sum of the forces in the theta hat direction 592 00:33:11,200 --> 00:33:13,445 for this problem are what? 593 00:33:17,620 --> 00:33:21,170 From your free body diagram. 594 00:33:21,170 --> 00:33:22,295 AUDIENCE: The normal force? 595 00:33:22,295 --> 00:33:24,520 PROFESSOR: The normal in the positive theta 596 00:33:24,520 --> 00:33:26,218 hat direction and? 597 00:33:26,218 --> 00:33:28,658 AUDIENCE: [INAUDIBLE]. 598 00:33:28,658 --> 00:33:32,930 PROFESSOR: Minus mg. 599 00:33:32,930 --> 00:33:35,840 And I think it's [? percosen ?] theta. 600 00:33:35,840 --> 00:33:37,200 And that's your theta hat. 601 00:33:37,200 --> 00:33:42,580 And what is the acceleration in that direction? 602 00:33:42,580 --> 00:33:44,920 Well you go to acceleration formula. 603 00:33:44,920 --> 00:33:46,550 And now it's inspect the terms. 604 00:33:46,550 --> 00:33:51,920 This one is 0 because the platform is fixed. 605 00:33:51,920 --> 00:33:53,850 r double dot? 606 00:33:53,850 --> 00:33:56,730 0 because we're just waiting. 607 00:33:56,730 --> 00:33:59,170 It's just trying to calculate when it begins to slide. 608 00:33:59,170 --> 00:34:01,490 r theta dot squared. 609 00:34:01,490 --> 00:34:03,010 Not 0. 610 00:34:03,010 --> 00:34:06,172 r theta double dot. 611 00:34:06,172 --> 00:34:06,880 Well it might be. 612 00:34:06,880 --> 00:34:11,389 But I said omega dots 0 constant. 613 00:34:11,389 --> 00:34:13,320 So that one's 0 for this problem. 614 00:34:13,320 --> 00:34:16,500 And this term, Coriolis. 615 00:34:16,500 --> 00:34:20,521 0 because our dot is 0. 616 00:34:20,521 --> 00:34:22,604 We're really treating this like a statics problem. 617 00:34:22,604 --> 00:34:24,110 It hasn't started to move yet. 618 00:34:24,110 --> 00:34:26,800 So we're only left with one term here. 619 00:34:26,800 --> 00:34:31,040 So the sum of the forces in the normal direction, which is 620 00:34:31,040 --> 00:34:36,300 the theta hat direction are 0. 621 00:34:36,300 --> 00:34:39,764 And that allows us to solve for the normal force. 622 00:34:43,530 --> 00:34:48,739 So you need Newton's law to find the normal force to start with. 623 00:34:48,739 --> 00:34:51,480 And you need the normal force to find the friction force. 624 00:34:51,480 --> 00:34:57,610 So now let's do the sum of the forces in the that direction. 625 00:34:57,610 --> 00:35:01,200 And maybe, let's do the acceleration first. 626 00:35:01,200 --> 00:35:04,930 It's the mass times the acceleration 627 00:35:04,930 --> 00:35:08,090 in the that direction. 628 00:35:08,090 --> 00:35:09,500 And that's the mass. 629 00:35:09,500 --> 00:35:13,210 And now, what's the acceleration? 630 00:35:13,210 --> 00:35:15,480 These are the r directed terms. 631 00:35:15,480 --> 00:35:19,430 We have one term, right. 632 00:35:19,430 --> 00:35:21,150 Minus r theta dot squared. 633 00:35:26,430 --> 00:35:27,520 No other terms. 634 00:35:27,520 --> 00:35:30,270 And those are going to be equal to the external forces 635 00:35:30,270 --> 00:35:31,270 in the radial direction. 636 00:35:31,270 --> 00:35:33,520 And what are they? 637 00:35:33,520 --> 00:35:36,786 From the [INAUDIBLE] diagram. 638 00:35:36,786 --> 00:35:39,719 AUDIENCE: [INAUDIBLE]. 639 00:35:39,719 --> 00:35:40,760 PROFESSOR: Plus or minus? 640 00:35:40,760 --> 00:35:41,752 AUDIENCE: Plus. 641 00:35:41,752 --> 00:35:44,360 PROFESSOR: Plus mu n. 642 00:35:44,360 --> 00:35:49,370 But we know n is mg cosine theta. 643 00:35:49,370 --> 00:35:52,480 OK, what else? 644 00:35:52,480 --> 00:35:58,250 Minus mg sine theta. 645 00:35:58,250 --> 00:35:58,956 All right. 646 00:36:07,720 --> 00:36:11,060 You know everything in this expression. 647 00:36:11,060 --> 00:36:12,330 You know given theta dot. 648 00:36:12,330 --> 00:36:13,585 You're given r. 649 00:36:13,585 --> 00:36:15,580 You know mg. 650 00:36:15,580 --> 00:36:19,130 You know [INAUDIBLE] given theta. 651 00:36:19,130 --> 00:36:20,950 You could solve this expression for mu. 652 00:36:25,860 --> 00:36:27,150 All right? 653 00:36:27,150 --> 00:36:31,970 And the mg goes to the other side. 654 00:36:31,970 --> 00:36:35,710 And you have a minus r theta squared. 655 00:36:35,710 --> 00:36:37,810 Notice the M's cancel all the way through, right. 656 00:36:42,860 --> 00:36:47,450 So mu would looks like it equals, to me, 657 00:36:47,450 --> 00:37:04,860 g sine theta minus r theta dot squared all divided by what? 658 00:37:04,860 --> 00:37:08,320 g cosine theta. 659 00:37:13,770 --> 00:37:15,720 Break these two apart, this gives me 660 00:37:15,720 --> 00:37:29,180 a tan theta minus r theta dot squared over g QoS theta. 661 00:37:29,180 --> 00:37:30,870 That has units of acceleration. 662 00:37:30,870 --> 00:37:32,540 g has units of acceleration. 663 00:37:32,540 --> 00:37:33,592 This is dimensionless. 664 00:37:33,592 --> 00:37:36,430 The answer has got to be dimensionless. 665 00:37:36,430 --> 00:37:39,780 So there's your answer. 666 00:37:39,780 --> 00:37:42,410 But you really had to use the equations in motion. 667 00:37:42,410 --> 00:37:44,020 OK now there's another thing I want 668 00:37:44,020 --> 00:37:47,130 to dress because I heard it pop up two or three times. 669 00:37:47,130 --> 00:37:55,860 Do not confuse accelerations with forces. 670 00:37:55,860 --> 00:37:59,050 Newtons second law makes it really clear 671 00:37:59,050 --> 00:38:00,870 where each one goes. 672 00:38:00,870 --> 00:38:04,310 The sum of the external forces go one side of the equal sign. 673 00:38:04,310 --> 00:38:08,610 The mass times the acceleration goes on the other. 674 00:38:08,610 --> 00:38:11,290 Don't get the two mixed up. 675 00:38:11,290 --> 00:38:15,100 Solve for the accelerations, if you can. 676 00:38:15,100 --> 00:38:15,750 Plug them in. 677 00:38:15,750 --> 00:38:16,880 And multiply them by mass. 678 00:38:16,880 --> 00:38:19,620 And now you have that one side of the equation. 679 00:38:19,620 --> 00:38:20,970 The free body diagram. 680 00:38:20,970 --> 00:38:27,760 The only vectors that should show up on a free body diagram 681 00:38:27,760 --> 00:38:30,330 are what? 682 00:38:30,330 --> 00:38:31,180 AUDIENCE: Forces. 683 00:38:31,180 --> 00:38:32,580 PROFESSOR: Forces. 684 00:38:32,580 --> 00:38:35,030 The real forces in the problem. 685 00:38:35,030 --> 00:38:37,540 So in that problem that free body diagram, 686 00:38:37,540 --> 00:38:42,340 there is no r theta dot squared term. 687 00:38:42,340 --> 00:38:44,270 It doesn't belong there. 688 00:38:44,270 --> 00:38:46,430 And it'll keep you from making these sign errors 689 00:38:46,430 --> 00:38:47,370 and things like that. 690 00:38:47,370 --> 00:38:50,020 So the business about this thing comes up minus 691 00:38:50,020 --> 00:38:54,574 because it's minus right out of the acceleration equation. 692 00:38:54,574 --> 00:39:05,825 All right, what would happen if I had not made the-- we're 693 00:39:05,825 --> 00:39:06,450 a go? 694 00:39:10,760 --> 00:39:11,615 This term 0. 695 00:39:11,615 --> 00:39:15,190 I allow this to have some angular acceleration to it. 696 00:39:15,190 --> 00:39:18,720 That's what's got to happen to. 697 00:39:18,720 --> 00:39:20,990 Well you don't have to do it. 698 00:39:20,990 --> 00:39:22,350 I can't do it constant rate. 699 00:39:22,350 --> 00:39:25,910 But if you can make this constant rate fast enough, 700 00:39:25,910 --> 00:39:28,750 it goes up. 701 00:39:28,750 --> 00:39:31,100 So even if it's constant rate fast enough, 702 00:39:31,100 --> 00:39:33,470 this thing will slide up the thing, immediately, 703 00:39:33,470 --> 00:39:35,052 when you start it. 704 00:39:35,052 --> 00:39:35,760 Almost immediate. 705 00:39:35,760 --> 00:39:36,885 You start down here, the g. 706 00:39:36,885 --> 00:39:38,360 The friction force is pretty high. 707 00:39:38,360 --> 00:39:39,190 And the friction force, of course, 708 00:39:39,190 --> 00:39:40,564 diminishes as you get further up. 709 00:39:40,564 --> 00:39:44,210 And eventually, it takes off, right. 710 00:39:44,210 --> 00:39:46,750 That's because that minus r theta 711 00:39:46,750 --> 00:39:50,681 dot squared, that's a centripetal acceleration 712 00:39:50,681 --> 00:39:51,180 inwards. 713 00:39:51,180 --> 00:39:53,080 And if you don't provide the force that 714 00:39:53,080 --> 00:39:55,720 causes that centripetal acceleration to happen, 715 00:39:55,720 --> 00:39:59,600 it says, it says I want to leave town. 716 00:39:59,600 --> 00:40:03,570 Now how about, though, if I had angular acceleration, as well? 717 00:40:03,570 --> 00:40:05,900 If I allow angular acceleration to this problem, 718 00:40:05,900 --> 00:40:09,310 how does it change the two equations that we wrote? 719 00:40:11,910 --> 00:40:13,020 What does it change? 720 00:40:13,020 --> 00:40:14,490 Does it change the forces? 721 00:40:14,490 --> 00:40:16,360 Does it change the free body diagram? 722 00:40:16,360 --> 00:40:16,910 Not a bit. 723 00:40:16,910 --> 00:40:20,530 It changes the acceleration side of the equation and what term 724 00:40:20,530 --> 00:40:26,340 it now turns up that you didn't have before. 725 00:40:26,340 --> 00:40:26,960 This one. 726 00:40:26,960 --> 00:40:31,100 So now you have a dynamic term in the theta hat equation. 727 00:40:31,100 --> 00:40:33,620 And it comes into this expression 728 00:40:33,620 --> 00:40:36,070 for some of the forces in this direction. 729 00:40:36,070 --> 00:40:42,110 And you would end up equals to m r theta double dot. 730 00:40:42,110 --> 00:40:45,680 Now it's a entirely different problem. 731 00:40:45,680 --> 00:40:48,650 It gets a little harder to solve. 732 00:40:48,650 --> 00:40:54,450 OK, but really good fundamental practices. 733 00:40:54,450 --> 00:40:55,830 f equals ma. 734 00:40:55,830 --> 00:40:57,780 And write out both sides carefully. 735 00:40:57,780 --> 00:41:00,810 And then, you won't make sign mistakes and so forth. 736 00:41:00,810 --> 00:41:01,440 Thanks. 737 00:41:01,440 --> 00:41:03,430 Good to see you.