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 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,260 at ocw.mit.edu. 8 00:00:20,872 --> 00:00:22,330 PROFESSOR: So I'm going to give you 9 00:00:22,330 --> 00:00:30,130 a quick example of what I think is a good way to do solutions. 10 00:00:30,130 --> 00:00:31,405 Our approach to solutions. 11 00:00:45,090 --> 00:00:45,980 State the problem. 12 00:00:48,910 --> 00:00:51,030 I'm going to give you a little formulaic here. 13 00:00:51,030 --> 00:00:51,720 Draw figures. 14 00:01:00,870 --> 00:01:03,440 On the first day I said I like to think of dynamics problem. 15 00:01:03,440 --> 00:01:06,500 I break them down into three categories. 16 00:01:06,500 --> 00:01:08,185 One is to describe the motion. 17 00:01:14,120 --> 00:01:15,810 What does describing the motion mean? 18 00:01:15,810 --> 00:01:21,670 Well how many-- the number of degrees of 19 00:01:21,670 --> 00:01:23,790 freedom in the problem. 20 00:01:23,790 --> 00:01:30,450 That implies the number of equations of motion 21 00:01:30,450 --> 00:01:33,360 that you're going to need to solve that problem. 22 00:01:33,360 --> 00:01:35,710 So you got to identify the number of degrees of freedom. 23 00:01:35,710 --> 00:01:41,370 It also then tells you the number of coordinates you need. 24 00:01:44,310 --> 00:01:46,730 So this is all part of describing the motion. 25 00:01:46,730 --> 00:01:50,040 It's figuring out how many coordinates, assigning them. 26 00:01:58,400 --> 00:02:01,860 So assigning the coordinates. 27 00:02:01,860 --> 00:02:05,520 And then finally, essentially all underneath this you 28 00:02:05,520 --> 00:02:06,896 essentially do the kinematics. 29 00:02:21,240 --> 00:02:27,097 And that's the velocities, accelerations, so forth. 30 00:02:27,097 --> 00:02:29,055 So once you have-- you've explained the motion. 31 00:02:34,420 --> 00:02:35,960 And this I guess is four. 32 00:02:35,960 --> 00:02:43,155 Explain the correct physical laws. 33 00:02:47,060 --> 00:02:48,460 You know how they apply. 34 00:02:48,460 --> 00:02:50,280 f equals ma. 35 00:02:50,280 --> 00:02:52,925 Newton's first, second, third law of conservation of momentum 36 00:02:52,925 --> 00:02:56,750 or whatever you want-- you think is the appropriate thing. 37 00:02:56,750 --> 00:02:59,840 So explain what the physical laws are and apply them. 38 00:03:02,620 --> 00:03:05,100 And finally, do the math. 39 00:03:10,287 --> 00:03:12,120 So if you could break problems down that way 40 00:03:12,120 --> 00:03:14,450 it'll give you a nice, logical flow. 41 00:03:14,450 --> 00:03:16,980 So I'm going to give you a bit of an example problem. 42 00:03:27,140 --> 00:03:29,640 And I'm also going to kind of pose a brain teaser to you 43 00:03:29,640 --> 00:03:31,060 at the end of this problem today. 44 00:03:31,060 --> 00:03:33,660 I want to give you something to think about. 45 00:03:33,660 --> 00:03:37,710 So I'm going to draw my problem. 46 00:03:37,710 --> 00:03:43,130 This is a block on an incline. 47 00:03:46,840 --> 00:03:50,820 It's got some scales to measure your weight. 48 00:03:50,820 --> 00:03:55,510 And you're standing on this thing. 49 00:03:55,510 --> 00:03:56,695 Riding it down the incline. 50 00:04:00,000 --> 00:04:02,870 And the first question about this 51 00:04:02,870 --> 00:04:08,490 is to find the position as a function of time. 52 00:04:26,534 --> 00:04:27,450 So that's the problem. 53 00:04:27,450 --> 00:04:28,980 That's part a. 54 00:04:32,650 --> 00:04:34,670 So state the problem, find the position. 55 00:04:45,024 --> 00:04:46,190 So how-- well what do we do? 56 00:04:46,190 --> 00:04:48,690 Well draw figures, I've started with that. 57 00:04:48,690 --> 00:04:50,240 Next, describe the motion. 58 00:04:50,240 --> 00:04:56,140 I need a free-- We have to figure out 59 00:04:56,140 --> 00:04:59,360 how many degrees of freedom this problem has. 60 00:04:59,360 --> 00:05:01,030 I'm going to just declare no rotation. 61 00:05:01,030 --> 00:05:04,180 Going to treat it as a particle. 62 00:05:04,180 --> 00:05:06,871 So a particle has how many, generally how many degrees of 63 00:05:06,871 --> 00:05:07,370 freedom? 64 00:05:07,370 --> 00:05:09,190 How many coordinates to completely describe 65 00:05:09,190 --> 00:05:10,226 where it's at? 66 00:05:10,226 --> 00:05:11,040 AUDIENCE: Three. 67 00:05:11,040 --> 00:05:12,240 PROFESSOR: Three. 68 00:05:12,240 --> 00:05:14,880 So I may need as many as three coordinates 69 00:05:14,880 --> 00:05:17,180 to describe the motion of this thing. 70 00:05:17,180 --> 00:05:19,780 And if I really doing complete equations of motion 71 00:05:19,780 --> 00:05:21,330 I need three equations of motion. 72 00:05:21,330 --> 00:05:25,560 So this is two, three-- describing the motion I 73 00:05:25,560 --> 00:05:28,576 have three degrees of freedom. 74 00:05:28,576 --> 00:05:30,075 I'm going to need three coordinates. 75 00:05:36,820 --> 00:05:42,200 So in this case here's my picture 76 00:05:42,200 --> 00:05:45,560 I'm going just to set up a Cartesian coordinate system 77 00:05:45,560 --> 00:05:49,970 aligned in a helpful way. 78 00:05:49,970 --> 00:05:53,390 X, y, z coming out. 79 00:05:53,390 --> 00:05:56,910 And this is my fixed inertial reference frame. 80 00:05:56,910 --> 00:05:58,820 And here's my center of mass. 81 00:05:58,820 --> 00:06:02,050 And basically my coordinates are describing the position 82 00:06:02,050 --> 00:06:03,130 of the center of mass. 83 00:06:19,140 --> 00:06:22,230 So that's pretty much the describing the motion, 84 00:06:22,230 --> 00:06:23,340 what I need for now. 85 00:06:23,340 --> 00:06:25,250 We'll get to the velocities and accelerations 86 00:06:25,250 --> 00:06:27,666 when we get to the math part. 87 00:06:27,666 --> 00:06:28,415 Apply the physics. 88 00:06:36,570 --> 00:06:38,380 I'm going to use Newton's second law. 89 00:06:48,570 --> 00:06:55,260 Sum of the external forces, mass times the acceleration. 90 00:06:55,260 --> 00:06:59,584 That's the physical law I'm going to apply. 91 00:06:59,584 --> 00:07:00,625 Draw a free body diagram. 92 00:07:16,800 --> 00:07:18,870 And I'm just going to consider the block 93 00:07:18,870 --> 00:07:22,709 and the person this is the whole collection, it's one thing. 94 00:07:22,709 --> 00:07:24,500 I'm just not keep drawing the person on it, 95 00:07:24,500 --> 00:07:26,840 but-- so here's my object, including 96 00:07:26,840 --> 00:07:29,540 the weight of the person. 97 00:07:29,540 --> 00:07:35,410 And it's going to have-- here's its center of mass. 98 00:07:35,410 --> 00:07:39,040 Obviously in mg, gravitational force, 99 00:07:39,040 --> 00:07:42,610 it's going to have a normal force. 100 00:07:42,610 --> 00:07:44,492 Going to have a friction force and how do I 101 00:07:44,492 --> 00:07:46,950 figure out what the friction-- which direction the friction 102 00:07:46,950 --> 00:07:47,645 is? 103 00:07:47,645 --> 00:07:51,310 I assume motion down the hill. 104 00:07:51,310 --> 00:07:53,070 Friction will oppose it. 105 00:07:53,070 --> 00:07:55,040 I draw in the arrows in the direction 106 00:07:55,040 --> 00:07:57,590 I expect the forces to act. 107 00:07:57,590 --> 00:07:58,965 Then I'll use the sign convention 108 00:07:58,965 --> 00:08:01,340 that the arrows tell me what signs. 109 00:08:01,340 --> 00:08:06,780 So since my x-coordinate is down the hill. 110 00:08:06,780 --> 00:08:08,080 This is y. 111 00:08:08,080 --> 00:08:11,320 This is x, friction x, in the minus x direction. 112 00:08:17,820 --> 00:08:23,900 And I haven't-- I've left out a key piece of information. 113 00:08:23,900 --> 00:08:26,710 Got to have the angle of the slope. 114 00:08:26,710 --> 00:08:37,270 And once you have the angle of the slope that's theta, 115 00:08:37,270 --> 00:08:38,840 this is theta. 116 00:08:38,840 --> 00:08:42,500 And I'm going to need to-- this is in a direction of one 117 00:08:42,500 --> 00:08:44,120 of my coordinates and so is this, 118 00:08:44,120 --> 00:08:47,120 but I need to break the gravitational piece 119 00:08:47,120 --> 00:08:51,520 into components lined up with my coordinates. 120 00:08:51,520 --> 00:08:55,420 And so you have a theta here as well. 121 00:08:55,420 --> 00:08:57,445 And now I can write my equation to motion. 122 00:09:02,450 --> 00:09:05,350 And the nice thing about vectors is 123 00:09:05,350 --> 00:09:08,690 that when you have three equations of motion, 124 00:09:08,690 --> 00:09:12,950 three coordinates, each that are components of vectors 125 00:09:12,950 --> 00:09:14,770 in the x, y and z direction, it gives us 126 00:09:14,770 --> 00:09:17,130 three equations immediately. 127 00:09:17,130 --> 00:09:19,590 So for example, the summation of the forces 128 00:09:19,590 --> 00:09:22,230 in this problem in the z-direction, 129 00:09:22,230 --> 00:09:25,770 the external forces are sums to? 130 00:09:25,770 --> 00:09:26,469 AUDIENCE: 0. 131 00:09:26,469 --> 00:09:27,010 PROFESSOR: 0. 132 00:09:27,010 --> 00:09:29,620 OK so we get a trivial solution out of that. 133 00:09:29,620 --> 00:09:34,340 And we don't have to go much further. 134 00:09:34,340 --> 00:09:36,760 Summation of the forces in the y-direction 135 00:09:36,760 --> 00:09:38,180 gives us some useful information. 136 00:09:42,770 --> 00:09:50,860 And then the y-directed forces I have an n and a minus mg 137 00:09:50,860 --> 00:09:53,910 and I think it's cosine theta. 138 00:09:59,320 --> 00:10:01,580 Which tells me immediately what n is. 139 00:10:01,580 --> 00:10:09,630 So from statics I get to-- and from what we know about models 140 00:10:09,630 --> 00:10:16,870 of friction then we know that we can model the friction as mu 141 00:10:16,870 --> 00:10:25,120 times n for mu mg cosine theta. 142 00:10:25,120 --> 00:10:27,726 So from the statics I learn a bunch of things 143 00:10:27,726 --> 00:10:29,225 that I need to know for the problem. 144 00:10:34,760 --> 00:10:43,389 So now I get to the real heart of the problem, 145 00:10:43,389 --> 00:10:45,430 writing my equation of motion in the x-direction. 146 00:10:56,370 --> 00:11:00,580 And the forces in the x-direction mg 147 00:11:00,580 --> 00:11:02,100 sine theta down the hill. 148 00:11:02,100 --> 00:11:04,920 So it's positive. 149 00:11:04,920 --> 00:11:08,540 Minus the friction is up the hill. 150 00:11:08,540 --> 00:11:09,500 So I get mg. 151 00:11:18,264 --> 00:11:34,280 I'm mixing up my m's here but there's no other m so-- this 152 00:11:34,280 --> 00:11:40,280 basically says, that x double dot, the m's all cancel out. 153 00:11:40,280 --> 00:11:52,470 That x double dot is g sine theta minus mu g cosine theta. 154 00:11:52,470 --> 00:11:57,140 And that just happens to be a pretty simple to solve, 155 00:11:57,140 --> 00:11:58,920 ordinary differential equation. 156 00:11:58,920 --> 00:12:01,650 This is an equation in which the acceleration in the problem 157 00:12:01,650 --> 00:12:02,290 is constant. 158 00:12:02,290 --> 00:12:04,000 The data's not changing. 159 00:12:04,000 --> 00:12:05,530 None of these things are changing 160 00:12:05,530 --> 00:12:07,497 and so you can just solve this one. 161 00:12:07,497 --> 00:12:09,205 Now we're to the third part, do the math. 162 00:12:16,110 --> 00:12:18,550 This one you can just integrate. 163 00:12:18,550 --> 00:12:21,525 And so you find out that well, x dot then 164 00:12:21,525 --> 00:12:25,370 and I'm just going to call this c some constant. 165 00:12:25,370 --> 00:12:31,340 So x dot ct plus an initial velocity, if it had one. 166 00:12:31,340 --> 00:12:34,280 And x of t, what you're looking for. 167 00:12:45,120 --> 00:12:48,370 And now so v0 and x0 are just your initial conditions. 168 00:12:48,370 --> 00:12:51,410 More than likely 0 if you set it up cleverly. 169 00:12:51,410 --> 00:12:53,890 So that's just modeling quickly what 170 00:12:53,890 --> 00:12:56,490 I think a good way to lay out a problem is. 171 00:12:56,490 --> 00:13:00,230 Describing the motion, explaining the physics, 172 00:13:00,230 --> 00:13:05,580 doing the math, drawing good pictures, stating the problem. 173 00:13:05,580 --> 00:13:07,809 All right so now the brain teaser 174 00:13:07,809 --> 00:13:09,100 that I want you to think about. 175 00:13:33,360 --> 00:13:35,725 So here's the mass of the block plus the scales. 176 00:13:39,760 --> 00:13:42,290 And here's the-- here you're standing on it. 177 00:13:45,800 --> 00:13:47,190 So mass of the person. 178 00:13:52,550 --> 00:13:53,930 You're riding this down the hill. 179 00:14:10,630 --> 00:14:11,530 So this is part b. 180 00:14:22,706 --> 00:14:24,205 Think about that when in the shower. 181 00:14:27,400 --> 00:14:30,610 If you've got a really simple way to do it, 182 00:14:30,610 --> 00:14:32,110 great write it up. 183 00:14:32,110 --> 00:14:34,920 It's not terribly hard and it's mostly-- I'll give you a hint. 184 00:14:34,920 --> 00:14:39,500 Thinking in terms of free body diagram helps a lot. 185 00:14:39,500 --> 00:14:41,930 And we'll come back to this kind of fun problem. 186 00:14:56,440 --> 00:14:58,890 OK, want to go-- that was part one. 187 00:14:58,890 --> 00:15:03,210 I want to go onto this recapping the center of mass quickly. 188 00:15:03,210 --> 00:15:11,519 We learned a couple of important things. 189 00:15:11,519 --> 00:15:13,310 We got-- we talked about the center of mass 190 00:15:13,310 --> 00:15:16,730 because we were just talking about Newton's three laws. 191 00:15:16,730 --> 00:15:18,894 From looking at the first law, found 192 00:15:18,894 --> 00:15:20,810 that it's useful in determining whether or not 193 00:15:20,810 --> 00:15:22,530 you're in a inertial frame, we used it. 194 00:15:22,530 --> 00:15:24,320 Second law we've just applied it. 195 00:15:24,320 --> 00:15:26,960 We used to do-- get equations of motion. 196 00:15:26,960 --> 00:15:31,730 Third law was about-- we used it when we're 197 00:15:31,730 --> 00:15:33,150 thinking about center of mass. 198 00:15:33,150 --> 00:15:36,040 We used it to define what the center of mass was. 199 00:15:36,040 --> 00:15:40,730 So we said the total mass of a system, 200 00:15:40,730 --> 00:15:41,890 I better draw my picture. 201 00:15:41,890 --> 00:15:43,245 Here's my system of particles. 202 00:15:45,940 --> 00:15:50,910 M1 with position vectors. 203 00:15:50,910 --> 00:15:53,420 So a whole mess of particles with their position vectors. 204 00:15:53,420 --> 00:15:56,270 This is ri with respect to O for example. 205 00:15:56,270 --> 00:15:59,480 This is my O x, y, z frame. 206 00:15:59,480 --> 00:16:03,680 We said that this total mass of the particles 207 00:16:03,680 --> 00:16:11,520 somewhere out here there's a center of mass with a position 208 00:16:11,520 --> 00:16:22,300 vector rg with respect to O. So the definition of my center 209 00:16:22,300 --> 00:16:29,140 of mass is this is equal to the summation of the m, i, r, i, o 210 00:16:29,140 --> 00:16:33,170 and these are position vectors. 211 00:16:33,170 --> 00:16:35,760 So that's the definition of my center of mass. 212 00:16:35,760 --> 00:16:43,215 If I take two time derivatives of that we arrived at mt rg 213 00:16:43,215 --> 00:16:46,760 with the respect to O double dot. 214 00:16:46,760 --> 00:16:53,500 Summation over i of my m, i, r, i, o double dots. 215 00:16:57,040 --> 00:17:01,730 And then importantly that's the summation 216 00:17:01,730 --> 00:17:07,883 of all of the external forces on each 217 00:17:07,883 --> 00:17:09,869 of these-- each one of these by itself 218 00:17:09,869 --> 00:17:12,780 satisfies Newton's law, second law, which 219 00:17:12,780 --> 00:17:14,690 he wrote about particles. 220 00:17:14,690 --> 00:17:16,105 Each one has a summation. 221 00:17:16,105 --> 00:17:17,980 I've summed these and I'm going to sum these, 222 00:17:17,980 --> 00:17:23,920 but each one has a summation of internal forces acting on it. 223 00:17:23,920 --> 00:17:27,130 These I call the f, i, j's. 224 00:17:27,130 --> 00:17:29,900 And we learn-- something about the third law 225 00:17:29,900 --> 00:17:31,530 tells us about that summation. 226 00:17:31,530 --> 00:17:34,390 The third law tells us what? 227 00:17:34,390 --> 00:17:38,650 That goes to 0 and that was that's the really powerful 228 00:17:38,650 --> 00:17:42,530 piece of the third law that we make great use of. 229 00:17:42,530 --> 00:17:46,190 Because this now essentially allows us to say, 230 00:17:46,190 --> 00:17:49,060 that the summation of the external 231 00:17:49,060 --> 00:17:53,110 forces on an assembly of particles, 232 00:17:53,110 --> 00:18:06,140 on a system of particles, is equal to the total mass 233 00:18:06,140 --> 00:18:11,820 times the acceleration of the center of mass. 234 00:18:11,820 --> 00:18:17,080 And that-- what that does in one stroke 235 00:18:17,080 --> 00:18:22,400 takes you from Newton, who's laws applied to particles 236 00:18:22,400 --> 00:18:27,487 and allows you to apply Newton's second law to rigid bodies. 237 00:18:27,487 --> 00:18:29,070 Because rigid bodies can be thought up 238 00:18:29,070 --> 00:18:33,250 a bunch of particles, which are represented 239 00:18:33,250 --> 00:18:36,520 in that simple equation. 240 00:18:36,520 --> 00:18:39,430 And that gets you from particles to rigid bodies. 241 00:18:39,430 --> 00:18:42,190 And we all know from physics that a summation 242 00:18:42,190 --> 00:18:44,200 of the external force on this thing 243 00:18:44,200 --> 00:18:45,834 is the mass times the acceleration 244 00:18:45,834 --> 00:18:46,750 of the center of mass. 245 00:18:50,000 --> 00:18:53,730 So that's actually quite an important powerful law. 246 00:18:53,730 --> 00:18:57,060 It provides this for us. 247 00:18:57,060 --> 00:19:01,900 Now I said I wanted to give you a quick, very 248 00:19:01,900 --> 00:19:04,970 useful application of thinking about center of mass. 249 00:19:04,970 --> 00:19:08,870 I showed you the other day, I had my carbon fiber tube here. 250 00:19:08,870 --> 00:19:11,120 Showed you that trick for finding the center of mass 251 00:19:11,120 --> 00:19:12,703 just by sliding your fingers along it. 252 00:19:12,703 --> 00:19:14,730 And you end up at the center of mass. 253 00:19:14,730 --> 00:19:17,190 Well as a practical matter other things-- 254 00:19:17,190 --> 00:19:19,900 you really want to be able find center of mass. 255 00:19:19,900 --> 00:19:23,290 This is a glider, a sailplane called an LS8. 256 00:19:23,290 --> 00:19:26,330 I happen-- I'm a glider flight instructor. 257 00:19:26,330 --> 00:19:29,660 Been flying gliders for 35 years or something like that. 258 00:19:29,660 --> 00:19:34,280 And I flew a glider of this type just recently. 259 00:19:34,280 --> 00:19:38,980 That machine has a 49 to one let's call it 50 to one 260 00:19:38,980 --> 00:19:41,710 to make it easy, glide ratio. 261 00:19:41,710 --> 00:19:45,030 Means if you're a mile above the ground in still air, 262 00:19:45,030 --> 00:19:47,850 you will go 50 miles before you touch the ground. 263 00:19:47,850 --> 00:19:50,840 So they're really amazing high performance machines. 264 00:19:50,840 --> 00:19:53,930 One of the things about all aircraft that you actually 265 00:19:53,930 --> 00:19:56,810 need to know is, you really need to know where 266 00:19:56,810 --> 00:19:59,630 the center of mass of it is. 267 00:19:59,630 --> 00:20:02,140 And if the center of mass is in the wrong place 268 00:20:02,140 --> 00:20:04,780 the plane will not fly properly. 269 00:20:04,780 --> 00:20:07,475 And so you can't just go throw on 50 pounds of lead 270 00:20:07,475 --> 00:20:09,300 in the tail of that plane and expect 271 00:20:09,300 --> 00:20:12,160 to survive the next flight. 272 00:20:12,160 --> 00:20:14,250 So you have to know where the center of mass 273 00:20:14,250 --> 00:20:16,560 is, and in fact, you want the center of mass 274 00:20:16,560 --> 00:20:21,040 about 25% of the-- if the wing is this wide from front 275 00:20:21,040 --> 00:20:22,890 to back it's called the cord. 276 00:20:22,890 --> 00:20:25,600 About 25% back from the leading edge 277 00:20:25,600 --> 00:20:30,560 is about where the center of lift of a wing is. 278 00:20:30,560 --> 00:20:33,110 And you want your center of mass also 279 00:20:33,110 --> 00:20:35,350 called the center of gravity in these situations, 280 00:20:35,350 --> 00:20:37,890 you want it to be pretty close to the center of lift 281 00:20:37,890 --> 00:20:41,130 so that their balance in the plane flies nicely. 282 00:20:41,130 --> 00:20:52,780 So is there a simple way to find the center of mass of something 283 00:20:52,780 --> 00:20:55,380 like a sailplane? 284 00:20:55,380 --> 00:20:58,060 So I'm going to drew-- this is exactly how you do it. 285 00:20:58,060 --> 00:21:01,490 So I'll draw a quick picture of my sailplane here. 286 00:21:11,302 --> 00:21:13,760 You usually have a little skid or a tail wheel on the back. 287 00:21:13,760 --> 00:21:17,210 And to find the center of mass you just set them on scales. 288 00:21:21,270 --> 00:21:24,410 You weigh it. 289 00:21:24,410 --> 00:21:26,340 You pick a coordinate system. 290 00:21:26,340 --> 00:21:29,200 Doesn't matter where it is, as long as it's fixed. 291 00:21:29,200 --> 00:21:32,960 The easiest place is right at the nose of the sailplane. 292 00:21:32,960 --> 00:21:37,180 So we'll make this x, make this y. 293 00:21:37,180 --> 00:21:39,080 You take it and you can take a tape measure 294 00:21:39,080 --> 00:21:45,180 and measure the distance from your reference point 295 00:21:45,180 --> 00:21:48,200 to the position where the wheel sits on the scales. 296 00:21:48,200 --> 00:21:50,720 We'll call that L1 here. 297 00:21:50,720 --> 00:21:53,820 And that's typically about five feet. 298 00:21:53,820 --> 00:21:58,230 And back here you have another position 299 00:21:58,230 --> 00:22:02,350 to where you have your second set of scales, that's L2. 300 00:22:02,350 --> 00:22:05,465 And a typical sailplane that's about 15 feet. 301 00:22:14,420 --> 00:22:16,247 Apply Newton's law. 302 00:22:16,247 --> 00:22:17,580 This thing's not going anywhere. 303 00:22:17,580 --> 00:22:19,910 Sum of the forces in the vertical direction 304 00:22:19,910 --> 00:22:23,710 is equal to 0. 305 00:22:23,710 --> 00:22:25,390 Free body diagram. 306 00:22:25,390 --> 00:22:30,990 Well you have somewhere about here, where the wing is, 307 00:22:30,990 --> 00:22:33,170 this is your center of mass. 308 00:22:33,170 --> 00:22:38,880 And you have m total times g downwards there. 309 00:22:38,880 --> 00:22:42,715 You have two weights pushing on the sailplane. 310 00:22:42,715 --> 00:22:46,920 A W2 pushing up, holding up the tail. 311 00:22:46,920 --> 00:22:51,040 And a W1 holding up the main gear. 312 00:22:51,040 --> 00:22:58,250 And so from the sum of the forces in the y-direction, 313 00:22:58,250 --> 00:22:59,960 that had better be 0. 314 00:22:59,960 --> 00:23:14,390 So you know that W1 plus W2 minus mtg is 0. 315 00:23:14,390 --> 00:23:18,260 And so you find out that the total weight of the sailplane 316 00:23:18,260 --> 00:23:23,874 is no surprise, the sum of the two weights on the scales. 317 00:23:23,874 --> 00:23:24,868 AUDIENCE: [INAUDIBLE]. 318 00:23:27,850 --> 00:23:29,530 PROFESSOR: Yeah. 319 00:23:29,530 --> 00:23:33,960 Let's do this, m, t, g. 320 00:23:33,960 --> 00:23:35,480 So the total weight times gravity 321 00:23:35,480 --> 00:23:38,820 is just the sum of the two readings on the scales. 322 00:23:38,820 --> 00:23:43,200 And the second piece that you need to do to do this problem 323 00:23:43,200 --> 00:23:45,440 is, you can have an equation that says, 324 00:23:45,440 --> 00:23:48,360 the sum of the external torques with respect 325 00:23:48,360 --> 00:23:52,930 to you're-- through a fixed point O, 326 00:23:52,930 --> 00:23:56,520 is equal to the mass moment of inertia times the angular 327 00:23:56,520 --> 00:23:58,530 acceleration, oftentimes written as alpha. 328 00:23:58,530 --> 00:24:03,560 In this case, that's going to be 0, it's going nowhere. 329 00:24:03,560 --> 00:24:06,300 So what are the external torques with respect to this point? 330 00:24:06,300 --> 00:24:08,220 Well we have a right-handed coordinate system. 331 00:24:08,220 --> 00:24:10,485 You have W1 up, times L1. 332 00:24:15,080 --> 00:24:23,210 W2 up, times L2 minus W1 plus W2, 333 00:24:23,210 --> 00:24:29,750 which is the total weight of the sailplane times 334 00:24:29,750 --> 00:24:35,760 rg, the location of the center of mass. 335 00:24:35,760 --> 00:24:37,770 That's this distance that we're looking for. 336 00:24:46,570 --> 00:24:48,860 So we know everything here, except rg. 337 00:24:48,860 --> 00:24:52,690 So we can solve for rg with respect to O 338 00:24:52,690 --> 00:25:03,680 and it's simply W1 L1 plus W2 L2 over W1 plus W2. 339 00:25:03,680 --> 00:25:15,720 And if you run the numbers, typically W1 600 pounds, 340 00:25:15,720 --> 00:25:23,560 W2 the numbers I've done here is-- might be 40 pounds. 341 00:25:26,940 --> 00:25:29,590 And I've already said five feet and 15 feet. 342 00:25:29,590 --> 00:25:31,580 And you work the answer and you come up 343 00:25:31,580 --> 00:25:40,830 with 5.62 feet, which puts this-- here is the wheel 344 00:25:40,830 --> 00:25:44,650 and you're a little bit aft of the main gear. 345 00:25:44,650 --> 00:25:47,240 And that's where you, basically where you want to be. 346 00:25:47,240 --> 00:25:53,247 That's a real practical use of knowing about centers of mass 347 00:25:53,247 --> 00:25:54,330 and how to calculate them. 348 00:26:00,970 --> 00:26:06,080 So that's the second item I wanted to talk about today. 349 00:26:06,080 --> 00:26:08,370 Essentially a recap of the center of mass. 350 00:26:08,370 --> 00:26:11,480 And now I want to move on to talking 351 00:26:11,480 --> 00:26:18,290 about a serious introduction to-- we've 352 00:26:18,290 --> 00:26:21,380 had the introduction, velocities and accelerations. 353 00:26:21,380 --> 00:26:24,510 We have to have a way of writing down 354 00:26:24,510 --> 00:26:35,170 the acceleration of a mass, a point, a dog in a rotating, 355 00:26:35,170 --> 00:26:38,740 translating, reference frame with the possibility 356 00:26:38,740 --> 00:26:41,670 that in addition to that, the dog's moving. 357 00:26:41,670 --> 00:26:44,230 So we want to have equations-- we want to have the ability 358 00:26:44,230 --> 00:26:47,310 to write down expressions for the velocity 359 00:26:47,310 --> 00:26:54,110 and acceleration of a mass moving in a translating, 360 00:26:54,110 --> 00:26:57,690 rotating, reference frame. 361 00:26:57,690 --> 00:26:58,680 So we've started this. 362 00:26:58,680 --> 00:27:02,080 We did pretty much did velocities to begin with. 363 00:27:31,740 --> 00:27:33,080 So here's my inertial frame. 364 00:27:41,030 --> 00:27:43,800 Call it O or O x, y, z. 365 00:27:43,800 --> 00:27:46,470 Here's my rigid body out there. 366 00:27:51,350 --> 00:27:57,200 It has a point a something else, b might be the dog. 367 00:27:57,200 --> 00:28:00,810 And we've described the position of this 368 00:28:00,810 --> 00:28:04,940 as the position of this point a, with respect to O. 369 00:28:04,940 --> 00:28:10,640 And at this point we're going to locate a reference frame 370 00:28:10,640 --> 00:28:13,700 attached to the rigid body. 371 00:28:13,700 --> 00:28:15,570 And so it's going to be called a, 372 00:28:15,570 --> 00:28:19,530 and I'll call it x prime, y prime, z prime. 373 00:28:19,530 --> 00:28:21,350 It's attached to the rigid body, it 374 00:28:21,350 --> 00:28:24,230 rotates with the rigid body and its attached 375 00:28:24,230 --> 00:28:26,340 at some fixed point. 376 00:28:26,340 --> 00:28:27,860 Now what would oftentimes would be 377 00:28:27,860 --> 00:28:30,950 a smart choice for that fixed point at point A? 378 00:28:35,144 --> 00:28:36,076 AUDIENCE: [INAUDIBLE]. 379 00:28:36,076 --> 00:28:38,492 PROFESSOR: If you're going to write an equation expressing 380 00:28:38,492 --> 00:28:40,280 the motion of this. 381 00:28:40,280 --> 00:28:41,832 Where would you make point a? 382 00:28:41,832 --> 00:28:42,759 [INTERPOSING VOICES] 383 00:28:42,759 --> 00:28:43,800 AUDIENCE: Center of mass. 384 00:28:43,800 --> 00:28:45,350 PROFESSOR: Center of mass. 385 00:28:45,350 --> 00:28:48,320 So very, very often, especially when objects are free 386 00:28:48,320 --> 00:28:50,820 floating around out there you're going to make smart choices 387 00:28:50,820 --> 00:28:52,770 and you're going to put this coordinate system 388 00:28:52,770 --> 00:28:53,936 right on the center of mass. 389 00:28:53,936 --> 00:28:56,040 But it doesn't have to be, but it can be. 390 00:28:59,420 --> 00:29:04,390 So we were interested in knowing things 391 00:29:04,390 --> 00:29:09,590 about the motion of this point in our inertial reference 392 00:29:09,590 --> 00:29:15,080 frame, in terms of positions of our coordinate system. 393 00:29:15,080 --> 00:29:23,190 And then also this vector here rd, with respect to a. 394 00:29:23,190 --> 00:29:29,590 Now last time we came up with expressions for the velocity 395 00:29:29,590 --> 00:29:35,285 of b with respect to O. 396 00:29:35,285 --> 00:29:39,320 We said in general it's the velocity of your-- where 397 00:29:39,320 --> 00:29:41,320 your coordinate system's located. 398 00:29:41,320 --> 00:29:46,260 The translating-- the velocity of the translating frame 399 00:29:46,260 --> 00:29:55,780 plus the derivative of rba, time derivative 400 00:29:55,780 --> 00:30:00,430 of the position as seen from, if you were sitting at a. 401 00:30:03,360 --> 00:30:06,520 And another way to say that, or this 402 00:30:06,520 --> 00:30:12,420 is a derivative taken with the rotation rate momentarily set 403 00:30:12,420 --> 00:30:14,490 equal to 0. 404 00:30:14,490 --> 00:30:16,370 Another way to think of it. 405 00:30:16,370 --> 00:30:20,180 Plus a piece that comes from rotation. 406 00:30:20,180 --> 00:30:24,150 So the rotation with respect to the fixed frame, these 407 00:30:24,150 --> 00:30:30,010 are all vectors, the rotation with respect 408 00:30:30,010 --> 00:30:36,720 to the fixed frame, cross product with r, b, a. 409 00:30:36,720 --> 00:30:38,460 And this-- and we said this is actually 410 00:30:38,460 --> 00:30:41,590 a general formula for the derivative of-- this piece is 411 00:30:41,590 --> 00:30:48,840 the derivative of a vector in a frame, in a fixed frame. 412 00:30:48,840 --> 00:30:51,970 You have two pieces, the derivative as 413 00:30:51,970 --> 00:30:55,197 seen without rotation plus the contribution that 414 00:30:55,197 --> 00:30:56,030 comes from rotation. 415 00:31:05,430 --> 00:31:08,920 When I did this center of mass thing a second ago, 416 00:31:08,920 --> 00:31:12,350 I just kind of quickly wrote down two time derivatives 417 00:31:12,350 --> 00:31:13,820 of the position vector. 418 00:31:13,820 --> 00:31:18,310 There's no omega cross O's in there right? 419 00:31:18,310 --> 00:31:20,960 Why could I do that? 420 00:31:20,960 --> 00:31:24,520 This is actually kind of an important distinct point. 421 00:31:24,520 --> 00:31:29,530 I could do that they didn't say very specifically when I did it 422 00:31:29,530 --> 00:31:31,920 was an assumption I was making. 423 00:31:31,920 --> 00:31:34,610 Except for perhaps they drew it. 424 00:31:34,610 --> 00:31:39,430 This was done in a Cartesian coordinate system. 425 00:31:39,430 --> 00:31:41,250 And my coordinates were x, y and z 426 00:31:41,250 --> 00:31:47,471 and the unit vectors were i, j, k and do they move? 427 00:31:47,471 --> 00:31:47,970 No. 428 00:31:47,970 --> 00:31:49,178 What's their time derivative? 429 00:31:49,178 --> 00:31:50,220 [INTERPOSING VOICES] 430 00:31:50,220 --> 00:31:52,761 PROFESSOR: When you don't-- when the inner vectors don't have 431 00:31:52,761 --> 00:31:54,811 time derivatives you don't get these terms. 432 00:31:54,811 --> 00:31:57,060 This is the only term that contributes so I could just 433 00:31:57,060 --> 00:31:58,390 write that equation. 434 00:31:58,390 --> 00:32:02,350 But we now have a reference frame attached to a body 435 00:32:02,350 --> 00:32:04,540 and this reference frame is rotating. 436 00:32:04,540 --> 00:32:07,940 And that means that the direction of the unit vectors 437 00:32:07,940 --> 00:32:11,710 attach-- the unit vector attached to x-prime here 438 00:32:11,710 --> 00:32:13,520 is moving, it's rotating. 439 00:32:13,520 --> 00:32:15,700 And it's going to have a time derivative. 440 00:32:15,700 --> 00:32:17,855 So we have to-- and that is given 441 00:32:17,855 --> 00:32:19,480 and you take those time derivatives you 442 00:32:19,480 --> 00:32:20,920 get this second piece. 443 00:32:24,140 --> 00:32:26,330 I'm going to give you the answer in advance. 444 00:32:26,330 --> 00:32:32,460 The acceleration of b with respect to O I'm 445 00:32:32,460 --> 00:32:37,460 going to give you the full 3D equation. 446 00:32:37,460 --> 00:32:41,240 Then we'll go back and see a bit where it comes from. 447 00:32:41,240 --> 00:32:45,660 So here's-- it's the time derivative of that velocity 448 00:32:45,660 --> 00:32:55,376 expression with respect to time taken in the inertial frame O, 449 00:32:55,376 --> 00:32:55,875 x, y, z. 450 00:33:02,310 --> 00:33:04,911 And am I going to have enough room to get this on? 451 00:33:04,911 --> 00:33:05,535 It'll be close. 452 00:33:19,090 --> 00:33:22,280 All right this has several pieces. 453 00:33:22,280 --> 00:33:28,070 It's got a contribution of the acceleration of a with respect 454 00:33:28,070 --> 00:33:30,551 to O. That's just the acceleration of this point. 455 00:33:30,551 --> 00:33:33,050 Has nothing to do with rotation, so it's just a straight out 456 00:33:33,050 --> 00:33:37,550 acceleration of my translating frame with respect to O. 457 00:33:37,550 --> 00:33:40,490 That's the first piece. 458 00:33:40,490 --> 00:33:44,690 The second piece is related-- is the derivative of this guy that 459 00:33:44,690 --> 00:33:46,620 comes from the derivative of this. 460 00:33:46,620 --> 00:33:54,010 It's the acceleration of b with respect to a as seen 461 00:33:54,010 --> 00:33:54,895 in this a frame. 462 00:34:00,000 --> 00:34:01,500 If you read the Williams book, it's 463 00:34:01,500 --> 00:34:03,940 called the relative acceleration. 464 00:34:03,940 --> 00:34:06,732 It's relative to the-- if you were sitting at point A, 465 00:34:06,732 --> 00:34:08,565 it's what you would see as the acceleration. 466 00:34:12,770 --> 00:34:25,239 Plus 2 omega cross velocity of b with respect 467 00:34:25,239 --> 00:34:36,850 to a as seen from a plus omega dot, the derivative 468 00:34:36,850 --> 00:34:57,070 of the rotation rate, cross rba plus omega cross, 469 00:34:57,070 --> 00:35:03,590 omega cross r, b, a. 470 00:35:07,536 --> 00:35:09,110 Kind of daunting right? 471 00:35:09,110 --> 00:35:11,250 A little messy. 472 00:35:11,250 --> 00:35:19,740 Basically one, two, three, four, five different terms. 473 00:35:19,740 --> 00:35:24,130 And you're going to-- and they all have names and meanings. 474 00:35:24,130 --> 00:35:27,370 And one of the things that will really help you 475 00:35:27,370 --> 00:35:30,420 is to get familiar, you really need 476 00:35:30,420 --> 00:35:36,480 to be familiar with the meaning of each one of the terms. 477 00:35:36,480 --> 00:35:38,690 And it's not terribly difficult. 478 00:35:38,690 --> 00:35:43,240 This one, just the acceleration of the translating frame. 479 00:35:43,240 --> 00:35:44,850 So if it's a merry-go-round sitting 480 00:35:44,850 --> 00:35:47,580 on a train and the train's heading down the track, 481 00:35:47,580 --> 00:35:50,320 its acceleration of the train. 482 00:35:50,320 --> 00:35:52,370 Rotating frame is attached to the merry-go-round. 483 00:35:54,990 --> 00:35:58,300 And if you've got the dog on the merry-go-round 484 00:35:58,300 --> 00:36:02,990 this is then the acceleration of the dog relative to this, 485 00:36:02,990 --> 00:36:04,460 the merry-go-round. 486 00:36:04,460 --> 00:36:07,040 This position of the coordinate system 487 00:36:07,040 --> 00:36:11,060 attached to the merry-go-round has no rotation in it. 488 00:36:11,060 --> 00:36:17,110 This is the velocity of that point, the dog, 489 00:36:17,110 --> 00:36:19,850 as seen from the A frame. 490 00:36:19,850 --> 00:36:22,100 Again, you have no sense of rotation. 491 00:36:22,100 --> 00:36:24,250 Rotation is not a part of this. 492 00:36:24,250 --> 00:36:27,050 Cross product with the rotation rate. 493 00:36:30,430 --> 00:36:31,780 This is the accelerate. 494 00:36:31,780 --> 00:36:35,810 This is the angular acceleration cross product with rba. 495 00:36:35,810 --> 00:36:38,910 Now that's a term-- what does that mean? 496 00:36:38,910 --> 00:36:44,400 I'm swinging a baseball bat and I'm accelerating this thing. 497 00:36:44,400 --> 00:36:48,540 Idealize it as just something on a radius accelerating. 498 00:36:48,540 --> 00:36:52,030 The acceleration of a point out here 499 00:36:52,030 --> 00:36:55,520 is the radius times the angular acceleration. 500 00:36:55,520 --> 00:36:57,300 So that's all this term is. 501 00:36:57,300 --> 00:36:59,870 And it's called the Euler acceleration. 502 00:36:59,870 --> 00:37:01,702 But it's just simply r theta double dot. 503 00:37:04,866 --> 00:37:08,190 This, if you multiply it out and just think about units, 504 00:37:08,190 --> 00:37:10,790 this ends up looking like r omega squared. 505 00:37:10,790 --> 00:37:13,040 Have you run into that before? 506 00:37:13,040 --> 00:37:15,290 What's that? 507 00:37:15,290 --> 00:37:16,618 Common language. 508 00:37:16,618 --> 00:37:18,630 AUDIENCE: [INAUDIBLE]. 509 00:37:18,630 --> 00:37:21,010 PROFESSOR: That's as a centrifugal-- centripetal, 510 00:37:21,010 --> 00:37:23,060 this is centripetal acceleration. 511 00:37:23,060 --> 00:37:25,600 So this is the centripetal acceleration term, 512 00:37:25,600 --> 00:37:28,060 that's the Euler acceleration term, 513 00:37:28,060 --> 00:37:30,025 this is the local acceleration. 514 00:37:30,025 --> 00:37:32,100 This is the acceleration of your frame. 515 00:37:32,100 --> 00:37:33,910 This is the strange one. 516 00:37:33,910 --> 00:37:35,685 This is the Coriolis acceleration. 517 00:37:40,390 --> 00:37:42,520 And we'll get familiar with it too. 518 00:37:42,520 --> 00:37:48,520 So that's the full blown 3D acceleration equation. 519 00:37:48,520 --> 00:37:50,700 And by the way the vector-- the velocity one 520 00:37:50,700 --> 00:37:52,730 is also perfect 3D. 521 00:37:52,730 --> 00:38:01,170 Now in this course we won't do much in the way of 3D dynamics 522 00:38:01,170 --> 00:38:01,860 problems. 523 00:38:01,860 --> 00:38:02,793 Yes. 524 00:38:02,793 --> 00:38:08,070 AUDIENCE: Does the point b on the rigid plane move? 525 00:38:08,070 --> 00:38:10,195 PROFESSOR: Does the-- it may. 526 00:38:10,195 --> 00:38:19,450 It could be this is the-- an asteroid out there in space 527 00:38:19,450 --> 00:38:22,540 and you've got-- this is home base 528 00:38:22,540 --> 00:38:26,580 and that's a guy out there in a space suit running. 529 00:38:29,650 --> 00:38:31,910 So we want to be able to describe 530 00:38:31,910 --> 00:38:38,500 the acceleration of that guy as seen from a fixed reference 531 00:38:38,500 --> 00:38:40,710 frame. 532 00:38:40,710 --> 00:38:43,720 Now why would we want to know that acceleration? 533 00:38:43,720 --> 00:38:47,180 Why do we want to know it in a fixed frame? 534 00:38:47,180 --> 00:38:53,480 Well if you want to calculate the forces on the person. 535 00:38:53,480 --> 00:38:57,200 Well how much-- what's he have to do with his feet 536 00:38:57,200 --> 00:38:59,600 to brace himself or whatever? 537 00:38:59,600 --> 00:39:01,570 What are the actual forces? 538 00:39:01,570 --> 00:39:04,970 You have to know the acceleration on the person. 539 00:39:04,970 --> 00:39:08,030 But Newton's laws, in order to say f equals ma, 540 00:39:08,030 --> 00:39:13,690 Newton's laws have to be applied in inertial reference frames. 541 00:39:13,690 --> 00:39:16,590 Is this thing out there doing this an a inertial reference 542 00:39:16,590 --> 00:39:17,571 frame? 543 00:39:17,571 --> 00:39:18,070 No. 544 00:39:18,070 --> 00:39:20,140 So you can't calculate the forces 545 00:39:20,140 --> 00:39:23,530 without having some idea of this inertial frame. 546 00:39:23,530 --> 00:39:26,750 So this is the way of getting the acceleration 547 00:39:26,750 --> 00:39:33,690 on-- at a location on a moving, rotating body with respect 548 00:39:33,690 --> 00:39:36,410 to an inertial frame. 549 00:39:36,410 --> 00:39:40,160 And with all the terms present. 550 00:39:40,160 --> 00:39:45,250 Now most discourse has generally has addressed problems 551 00:39:45,250 --> 00:39:49,740 which are in most textbooks address only planar motion 552 00:39:49,740 --> 00:39:51,270 problems. 553 00:39:51,270 --> 00:39:55,530 Planar motion basically means that we can find 554 00:39:55,530 --> 00:39:59,280 the translations to a plane. 555 00:39:59,280 --> 00:40:04,860 So imagine an x, and a y, and a z upwards coordinate system 556 00:40:04,860 --> 00:40:06,450 attached to the top this table. 557 00:40:06,450 --> 00:40:09,370 And I only allow motions that are around the table. 558 00:40:09,370 --> 00:40:12,860 And I only allow motions that have a single axis rotation. 559 00:40:12,860 --> 00:40:14,720 And that's lined up with z. 560 00:40:14,720 --> 00:40:17,210 Those are essentially planar motion problems. 561 00:40:17,210 --> 00:40:20,650 And most courses in dynamics, it lists [INAUDIBLE]. 562 00:40:20,650 --> 00:40:22,290 That's as far as they get. 563 00:40:22,290 --> 00:40:24,610 And the most of the problems that you'll do 564 00:40:24,610 --> 00:40:27,670 will be planar motion problems. 565 00:40:27,670 --> 00:40:33,970 But that equation reduces to the planar motion problem as well. 566 00:40:33,970 --> 00:40:36,270 We will do a little bit of 3D, because there's 567 00:40:36,270 --> 00:40:37,941 a class of problems that I really 568 00:40:37,941 --> 00:40:40,440 think it's important for you to understand that just come up 569 00:40:40,440 --> 00:40:43,997 all the time that require a little 3D. 570 00:40:43,997 --> 00:40:45,830 And as you want to have some things going on 571 00:40:45,830 --> 00:40:47,240 out of the plane, but we'll still 572 00:40:47,240 --> 00:40:53,420 confine the axis of rotation to a single direction. 573 00:40:53,420 --> 00:40:53,920 Yeah. 574 00:40:53,920 --> 00:40:58,330 AUDIENCE: [INAUDIBLE] in a planar, 575 00:40:58,330 --> 00:41:02,740 but then would you have three to view the freedom? [INAUDIBLE]. 576 00:41:02,740 --> 00:41:04,430 PROFESSOR: That's a great question 577 00:41:04,430 --> 00:41:05,880 she said if you had a dog running 578 00:41:05,880 --> 00:41:08,630 on the merry-go-round how many degrees of freedom do you have? 579 00:41:08,630 --> 00:41:13,510 So in general rigid bodies, each rigid body, 580 00:41:13,510 --> 00:41:16,100 each independent rigid body has-- 581 00:41:16,100 --> 00:41:18,580 you have to describe its location of its center of mass 582 00:41:18,580 --> 00:41:20,270 and that takes how many coordinates? 583 00:41:20,270 --> 00:41:23,340 How many coordinates I'll call them. 584 00:41:23,340 --> 00:41:25,172 AUDIENCE: [INAUDIBLE]. 585 00:41:25,172 --> 00:41:28,660 PROFESSOR: Well in general three. 586 00:41:28,660 --> 00:41:35,227 And it can now rotate around three different axes. 587 00:41:35,227 --> 00:41:36,310 And that takes three more. 588 00:41:36,310 --> 00:41:39,740 So rigid bodies have six degrees of freedom. 589 00:41:39,740 --> 00:41:42,620 And any problem when you go to address the problem you 590 00:41:42,620 --> 00:41:45,460 essentially for a rigid body you start with six. 591 00:41:45,460 --> 00:41:47,270 And you start applying constraints 592 00:41:47,270 --> 00:41:50,850 to reduce it down to the number of remaining 593 00:41:50,850 --> 00:41:52,810 degrees of freedom. 594 00:41:52,810 --> 00:41:55,380 So if it's confined to a plane and no z-motion 595 00:41:55,380 --> 00:41:58,250 is allowed one constraint. 596 00:41:58,250 --> 00:42:01,470 If it is on a plane and it's only 597 00:42:01,470 --> 00:42:04,640 allowed to rotate about the z-axis that 598 00:42:04,640 --> 00:42:08,920 means you've constrained its rotation in around y and x. 599 00:42:08,920 --> 00:42:10,640 So that's two more. 600 00:42:10,640 --> 00:42:16,560 And so now you're down to three degrees of freedom left, xy 601 00:42:16,560 --> 00:42:19,890 and a rotation about the z-axis. 602 00:42:19,890 --> 00:42:21,680 So planar motion problems generally 603 00:42:21,680 --> 00:42:26,450 have three degrees of freedom. 604 00:42:26,450 --> 00:42:28,230 But instant-- let's just say we're 605 00:42:28,230 --> 00:42:30,020 just interested in just something that 606 00:42:30,020 --> 00:42:33,240 rotates and doesn't translate. 607 00:42:33,240 --> 00:42:35,610 How many degrees of freedom does that have then? 608 00:42:35,610 --> 00:42:36,770 Just one. 609 00:42:36,770 --> 00:42:40,259 x and y are forced not to-- no motion, two more 610 00:42:40,259 --> 00:42:41,550 constraints you're down to one. 611 00:42:41,550 --> 00:42:45,712 So lots of problems we do are in fact single degree 612 00:42:45,712 --> 00:42:46,545 of freedom problems. 613 00:42:57,190 --> 00:43:01,640 So to do planar motion problems we oftentimes 614 00:43:01,640 --> 00:43:04,010 use polar coordinates. 615 00:43:04,010 --> 00:43:08,420 So I'm going to introduce r theta. 616 00:43:08,420 --> 00:43:10,980 And I'm actually going to call it cylindrical coordinates. 617 00:43:21,300 --> 00:43:26,298 And cylindrical coordinates then you have an r, a theta and z. 618 00:43:34,910 --> 00:43:37,900 And let's think about well let's see, I have a demo, 619 00:43:37,900 --> 00:43:40,540 a little demo here. 620 00:43:40,540 --> 00:43:45,450 So here's a problem with a single axis of rotation. 621 00:43:48,740 --> 00:43:54,570 And it's a-- there is a mass out here and just this is a rhyme. 622 00:43:54,570 --> 00:43:58,260 And so think of this think of this mass 623 00:43:58,260 --> 00:44:03,350 out here as being a bug walking out this rod. 624 00:44:03,350 --> 00:44:08,032 And the rod, this thing goes round and round. 625 00:44:08,032 --> 00:44:09,580 It's not a merry-go-round but it's 626 00:44:09,580 --> 00:44:12,660 a merry-go-round with a gang plank on it. 627 00:44:12,660 --> 00:44:14,035 It's going up at an angle and you 628 00:44:14,035 --> 00:44:16,118 can walk the gang plank while the merry-go-round's 629 00:44:16,118 --> 00:44:16,690 going around. 630 00:44:16,690 --> 00:44:19,330 So that's what we got here. 631 00:44:19,330 --> 00:44:23,540 So this is actually allowed to change position of this mass. 632 00:44:23,540 --> 00:44:28,900 So how would I describe that with cylindrical coordinates? 633 00:44:28,900 --> 00:44:34,900 Let me so it's going to take-- one would be a side view. 634 00:44:38,160 --> 00:44:40,930 So you see your vertical axis here 635 00:44:40,930 --> 00:44:43,135 and have to have a bearing to hold it in place. 636 00:44:46,790 --> 00:44:53,970 Here's the arm, here's the bug walking out the arm, 637 00:44:53,970 --> 00:44:55,480 has some rotation rate. 638 00:44:58,200 --> 00:45:13,440 Theta dot this is the z-axis and the position of this point 639 00:45:13,440 --> 00:45:22,280 is described by a r vector, in the r hat 640 00:45:22,280 --> 00:45:23,800 direction, which I think is the unit 641 00:45:23,800 --> 00:45:25,630 vectors you're used to using. 642 00:45:25,630 --> 00:45:30,375 And then this is the z-component in the k hat direction. 643 00:45:32,920 --> 00:45:45,730 And this vector here would be r of v with respect to what shall 644 00:45:45,730 --> 00:45:46,230 I call it? 645 00:45:46,230 --> 00:45:55,800 I'll make this a and over here someplace 646 00:45:55,800 --> 00:46:07,400 I have a fixed inertial let's see I got to be careful here. 647 00:46:07,400 --> 00:46:13,620 I want that to be z then I have a y going out here x, y, z 648 00:46:13,620 --> 00:46:15,180 pointing upwards. 649 00:46:15,180 --> 00:46:17,480 And my-- this fixed inertial system the unit 650 00:46:17,480 --> 00:46:25,840 vectors here this would be i hat, j hat and k hat. 651 00:46:25,840 --> 00:46:30,920 But these-- this rotating system with its unit 652 00:46:30,920 --> 00:46:35,360 vector little k and this vector they're the same, 653 00:46:35,360 --> 00:46:38,020 they're parallel. 654 00:46:38,020 --> 00:46:42,620 But looking down on this, this is my polar coordinate system. 655 00:46:42,620 --> 00:46:45,560 Now I'm going to look at my top view. 656 00:46:45,560 --> 00:46:46,870 I will see a projection. 657 00:46:49,700 --> 00:46:54,340 I'll just see the r, this is rr hat, 658 00:46:54,340 --> 00:46:57,840 this is my point B. This is theta. 659 00:47:01,600 --> 00:47:03,640 And I have a unit vector. 660 00:47:03,640 --> 00:47:08,190 So the unit vector r hat is something-- the unit long 661 00:47:08,190 --> 00:47:10,870 pointing in this direction. 662 00:47:10,870 --> 00:47:14,320 And the unit vector in this direction is theta hat. 663 00:47:14,320 --> 00:47:18,430 And it's perpendicular to that radius. 664 00:47:18,430 --> 00:47:20,330 So now I have my three unit vectors. 665 00:47:20,330 --> 00:47:24,530 One pointing in the direction of r, here's 666 00:47:24,530 --> 00:47:29,920 also my unit vector is just to make sure there's no confusion. 667 00:47:29,920 --> 00:47:32,218 This unit vector is in this direction. 668 00:47:35,360 --> 00:47:36,940 K is in that direction. 669 00:47:36,940 --> 00:47:39,050 Theta is in that direction. 670 00:47:44,450 --> 00:47:47,080 And over here you still have your-- now here's 671 00:47:47,080 --> 00:47:54,570 my x, y, z out of the board inertial frame. 672 00:47:54,570 --> 00:47:59,166 And this-- my inertial frame this might be r, b, o. 673 00:48:02,770 --> 00:48:04,180 So in my inertial frame. 674 00:48:04,180 --> 00:48:07,085 I want to know what's going on here. 675 00:48:07,085 --> 00:48:08,960 I want to be able to calculate the velocities 676 00:48:08,960 --> 00:48:09,876 and the accelerations. 677 00:48:16,970 --> 00:48:21,730 So the notation here gets-- can get a little confusing. 678 00:48:21,730 --> 00:48:26,010 The rbo notation that I've been using all along, 679 00:48:26,010 --> 00:48:30,880 that's the motion-- that's the position vector describing 680 00:48:30,880 --> 00:48:34,650 that point in my inertial frame. 681 00:48:34,650 --> 00:48:39,310 And my-- just lowercase r here, no scrub scripts or anything, 682 00:48:39,310 --> 00:48:44,850 that's just going to-- that's my polar coordinate r theta 683 00:48:44,850 --> 00:48:47,500 and z that happened to be in this case, 684 00:48:47,500 --> 00:48:50,120 this is a rotating frame. 685 00:48:50,120 --> 00:48:52,190 This is a rotating frame. 686 00:48:52,190 --> 00:48:58,870 It's the center of this coordinate system's at a. 687 00:48:58,870 --> 00:49:00,720 But this thing rotates. 688 00:49:00,720 --> 00:49:03,400 So this is a pretty simplified version 689 00:49:03,400 --> 00:49:07,760 of this general problem. 690 00:49:07,760 --> 00:49:11,250 Now because it's simplified, you can actually-- 691 00:49:11,250 --> 00:49:12,680 it's a lot easier to use. 692 00:49:12,680 --> 00:49:14,240 Also has some real limitations. 693 00:49:14,240 --> 00:49:17,230 You can only going just-- there's limited things 694 00:49:17,230 --> 00:49:18,500 that you can describe with it. 695 00:49:42,970 --> 00:49:48,070 So let's start by describing velocities in cylindrical 696 00:49:48,070 --> 00:49:48,570 coordinates. 697 00:50:23,290 --> 00:50:27,800 Remember this rba is the length of this guy here. 698 00:50:27,800 --> 00:50:31,710 And it's made up of rr hat zk hat. 699 00:50:59,900 --> 00:51:01,585 So to express the velocity we have 700 00:51:01,585 --> 00:51:03,710 to take a time derivative of this r, b, a 701 00:51:03,710 --> 00:51:07,210 and I'm going to express it in terms of r theta nz. 702 00:51:07,210 --> 00:51:09,190 And to get acceleration I have to take two time 703 00:51:09,190 --> 00:51:12,210 derivatives of this, but this is going to be expressed 704 00:51:12,210 --> 00:51:14,880 in my cylindrical coordinates. 705 00:51:14,880 --> 00:51:17,420 This is where I'm going. 706 00:51:17,420 --> 00:51:24,660 And lots of problems-- many, many of these problems 707 00:51:24,660 --> 00:51:27,240 have fixed axes of rotations and this velocity 708 00:51:27,240 --> 00:51:29,540 and this acceleration are zero. 709 00:51:29,540 --> 00:51:31,080 You just drop it out. 710 00:51:31,080 --> 00:51:34,000 I'm going to do that just to keep this-- make 711 00:51:34,000 --> 00:51:35,250 this problem a little simpler. 712 00:51:35,250 --> 00:51:37,890 So we can just focus on these terms. 713 00:51:37,890 --> 00:51:45,970 So let's just let the there be no translational of this frame. 714 00:51:45,970 --> 00:51:48,900 And that says that Va with respect 715 00:51:48,900 --> 00:51:53,679 to O the acceleration of A with respect to O over zero. 716 00:51:53,679 --> 00:51:55,470 So I want you to just focus on these terms. 717 00:51:55,470 --> 00:51:57,160 I don't lose anything, I can put these back in 718 00:51:57,160 --> 00:51:58,200 later if I need them. 719 00:51:58,200 --> 00:52:00,157 I just don't want to keep carrying them along. 720 00:52:30,530 --> 00:52:33,220 So I have my-- remember my side view. 721 00:52:37,440 --> 00:52:47,140 This is r, r hat zk hat that's my point. 722 00:52:47,140 --> 00:52:48,210 And my top view. 723 00:53:06,970 --> 00:53:12,990 This is my projection just looking down on it 724 00:53:12,990 --> 00:53:15,580 what I see is the length r. 725 00:53:15,580 --> 00:53:20,929 And what I see in my unit vector going that way r direction 726 00:53:20,929 --> 00:53:21,970 and theta that direction. 727 00:53:25,290 --> 00:53:32,870 And this is x and the i and a y with a j hat 728 00:53:32,870 --> 00:53:36,640 vector looking down on it. 729 00:53:36,640 --> 00:53:42,660 My rotation rate mega with respect to my inertial frame, 730 00:53:42,660 --> 00:53:46,100 is sum theta dot k hat. 731 00:53:50,850 --> 00:53:54,350 All right so now let's find the velocity of b with respect 732 00:53:54,350 --> 00:53:59,640 to O. Well it's 0, no translation, plus-- 733 00:53:59,640 --> 00:54:06,850 and now I need a time derivative of rb with respect to a. 734 00:54:06,850 --> 00:54:20,402 But this is then r, b, a is r, r hat plus z k hat. 735 00:54:20,402 --> 00:54:22,675 And I need the time derivative of that. 736 00:54:26,360 --> 00:54:41,980 So I get an r dot r hat plus an r r hat dot plus a z dot k. 737 00:54:41,980 --> 00:54:44,840 So this is the product of two things. 738 00:54:44,840 --> 00:54:46,760 They're both time dependent. 739 00:54:46,760 --> 00:54:51,380 So I have to get two pieces, k does not change in direction. 740 00:54:51,380 --> 00:54:53,330 So it has no time derivative. 741 00:54:53,330 --> 00:54:55,570 So I only have a z dot k. 742 00:54:55,570 --> 00:54:57,300 So this is a result of doing this, 743 00:54:57,300 --> 00:54:59,740 but I now have to figure out what 744 00:54:59,740 --> 00:55:03,205 is the time derivative of the unit vector in the r direction. 745 00:55:17,910 --> 00:55:21,260 So when I told-- when we worked out this formula the other day 746 00:55:21,260 --> 00:55:24,640 for the time derivative of a rotating vector, 747 00:55:24,640 --> 00:55:27,300 I mostly did it, it was kind of an intuitive argument. 748 00:55:27,300 --> 00:55:28,820 So on this one occasion I'm going 749 00:55:28,820 --> 00:55:32,130 to give you an example of actually figuring out 750 00:55:32,130 --> 00:55:35,390 what the derivative of this rotating vector is. 751 00:55:35,390 --> 00:55:39,120 And if you go read that kinematics handout 752 00:55:39,120 --> 00:55:42,920 and it does this in kind of full blown form for-- in general. 753 00:55:42,920 --> 00:55:45,780 So I'm just going to do it as one example. 754 00:55:45,780 --> 00:55:51,880 So here's our looking down on this, the projection on the xy 755 00:55:51,880 --> 00:55:55,100 plane, here's our r-vector. 756 00:55:55,100 --> 00:55:58,060 And here's this unit vector and it 757 00:55:58,060 --> 00:56:02,900 starts from-- I have a unit vector starting from a it's 758 00:56:02,900 --> 00:56:06,370 unit-- it's one long. 759 00:56:06,370 --> 00:56:08,930 And this is r hat. 760 00:56:08,930 --> 00:56:15,900 And it's of unit length and it's in this particular direction. 761 00:56:15,900 --> 00:56:21,050 Now in a little bit it time delta t, it moves. 762 00:56:21,050 --> 00:56:23,490 It moves to here. 763 00:56:23,490 --> 00:56:33,480 So this is delta r hat and what direction does it move? 764 00:56:33,480 --> 00:56:34,920 AUDIENCE: [INAUDIBLE]. 765 00:56:34,920 --> 00:56:38,830 PROFESSOR: Yeah, it moves in the-- moves in the theta hat 766 00:56:38,830 --> 00:56:41,090 direction. 767 00:56:41,090 --> 00:56:45,470 And the amount that it moves is the rotation rate, 768 00:56:45,470 --> 00:56:49,050 theta dot, delta t. 769 00:56:52,190 --> 00:56:59,640 So delta r hat, if I solve for this, delta t. 770 00:56:59,640 --> 00:57:04,930 And this is in the theta hat direction, 771 00:57:04,930 --> 00:57:09,185 is theta dot theta hat. 772 00:57:13,280 --> 00:57:17,810 So this is the limit as t, delta t 773 00:57:17,810 --> 00:57:24,770 goes to 0 you get the derivative of r hat with respect to time. 774 00:57:24,770 --> 00:57:27,870 Its direction is in the theta hat direction 775 00:57:27,870 --> 00:57:31,980 and its magnitude is theta dot. 776 00:57:31,980 --> 00:57:35,740 So that's the time derivative of the unit vector r hat. 777 00:57:39,390 --> 00:57:40,210 Yeah. 778 00:57:40,210 --> 00:57:43,556 AUDIENCE: How does that work with units? 779 00:57:43,556 --> 00:57:47,700 PROFESSOR: How does it work with units? 780 00:57:47,700 --> 00:57:55,640 What's left out of here is that this is unit length 781 00:57:55,640 --> 00:57:57,426 and has dimensions. 782 00:57:57,426 --> 00:58:02,270 It's unit length, one whatever unit system you're working. 783 00:58:02,270 --> 00:58:06,440 So that is implicitly in here. 784 00:58:06,440 --> 00:58:11,900 It's one meter theta dot and that theta dot, the delta t, 785 00:58:11,900 --> 00:58:12,880 the times go away. 786 00:58:12,880 --> 00:58:17,940 You're left with one meter times the magnitude of theta dot. 787 00:58:17,940 --> 00:58:19,630 So the distance it actually moves 788 00:58:19,630 --> 00:58:25,370 is r theta, the r delta theta, delta theta is theta dot delta 789 00:58:25,370 --> 00:58:27,645 t and the radius happens to be 1. 790 00:58:31,100 --> 00:58:34,830 So whatever unit system you're working in it's a unit vector. 791 00:58:34,830 --> 00:58:36,130 Has unit length. 792 00:58:36,130 --> 00:58:38,330 So its units are buried right there. 793 00:58:38,330 --> 00:58:40,890 Good question. 794 00:58:40,890 --> 00:58:45,120 OK so now we know what this is. 795 00:58:45,120 --> 00:58:50,670 So now we can come back finish our description of the velocity 796 00:58:50,670 --> 00:58:58,820 of b with respect to a then is r dot r hat 797 00:58:58,820 --> 00:59:10,795 plus z dot k hat plus theta dot times 798 00:59:10,795 --> 00:59:17,050 so r times the derivative of the unit vector r, which we just 799 00:59:17,050 --> 00:59:21,292 figured out is theta dot theta hat times r. 800 00:59:21,292 --> 00:59:25,620 R theta dot theta hat. 801 00:59:25,620 --> 00:59:28,620 So that's my velocity of b with respect to a. 802 00:59:28,620 --> 00:59:31,640 My velocity of B with respect to O all you have to add in 803 00:59:31,640 --> 00:59:33,720 is the velocity of A with perspective to O, 804 00:59:33,720 --> 00:59:35,580 which we've let be 0 for now. 805 00:59:35,580 --> 00:59:40,080 So for the moment this is also d with respect to O. 806 00:59:40,080 --> 00:59:42,910 But this is the general piece of the velocity 807 00:59:42,910 --> 00:59:46,294 of b with respect to a in polar cylindrical coordinates. 808 00:59:50,170 --> 00:59:54,050 Now we could have-- so I've actually worked it out, just 809 00:59:54,050 --> 00:59:57,090 shown you, just drew the picture and figured out the derivative. 810 00:59:57,090 --> 00:59:59,850 We could have used that magic formula. 811 00:59:59,850 --> 01:00:03,920 The formula for the derivative of a vector in a rotating 812 01:00:03,920 --> 01:00:04,420 frame. 813 01:00:15,720 --> 01:00:18,330 So I'll just do that quickly to remind you 814 01:00:18,330 --> 01:00:20,020 how we could have done this. 815 01:00:20,020 --> 01:00:27,620 rba with respect to time as seen in the O frame. 816 01:00:30,570 --> 01:00:36,990 Is the partial derivative of rba with respect 817 01:00:36,990 --> 01:00:47,577 to time as seen in the rotating frame, plus omega cross r, b, 818 01:00:47,577 --> 01:00:48,076 a. 819 01:00:55,370 --> 01:01:02,030 This term is that and that. 820 01:01:04,840 --> 01:01:08,760 The derivative of this rba as seen 821 01:01:08,760 --> 01:01:11,440 from inside of the rotating frame, 822 01:01:11,440 --> 01:01:17,090 is just the change in length, this is rba here from the side. 823 01:01:17,090 --> 01:01:21,389 So the change in length of that vector, the derivative 824 01:01:21,389 --> 01:01:22,680 of-- the time derivative of it. 825 01:01:22,680 --> 01:01:28,290 It's the vector sum of the r dot plus z dot. 826 01:01:28,290 --> 01:01:36,130 So this piece comes from this and this. 827 01:01:36,130 --> 01:01:42,840 And this piece should-- this one here, it better be this. 828 01:01:42,840 --> 01:01:44,650 Well this is-- let's figure it out. 829 01:01:44,650 --> 01:01:49,140 This is omega in the k hat direction, 830 01:01:49,140 --> 01:01:59,520 cross and rba is r r hat plus z k hat. 831 01:01:59,520 --> 01:02:03,610 k cross k is 0. 832 01:02:03,610 --> 01:02:09,460 k cross r theta hat. 833 01:02:09,460 --> 01:02:15,530 K cross-- k hat cross r hat is theta hat positive. 834 01:02:15,530 --> 01:02:21,770 r omega theta hat, same thing as r theta dot theta hat. 835 01:02:21,770 --> 01:02:25,310 So we could have just applied this formula for the derivative 836 01:02:25,310 --> 01:02:29,730 of a rotating vector and we would 837 01:02:29,730 --> 01:02:30,854 have gotten the same thing. 838 01:02:41,230 --> 01:02:43,347 OK just ran out of boards. 839 01:03:20,180 --> 01:03:22,460 Now a quick little exercise you could do on your own 840 01:03:22,460 --> 01:03:25,510 is, we're going to need to be able to calculate 841 01:03:25,510 --> 01:03:28,092 the derivative of theta hat. 842 01:03:28,092 --> 01:03:29,800 Well just plug it in that little formula. 843 01:03:36,650 --> 01:03:41,250 And the first term you'll find out the derivative of the theta 844 01:03:41,250 --> 01:03:48,800 hat, the length doesn't change in time, it's a unit vector. 845 01:03:48,800 --> 01:03:50,920 So you only have the second piece. 846 01:03:50,920 --> 01:03:54,800 So it's sum omega cross theta and you're going 847 01:03:54,800 --> 01:03:57,859 to get minus theta dot r hat. 848 01:04:17,940 --> 01:04:21,140 So I really want to get here. 849 01:04:21,140 --> 01:04:23,480 The acceleration of b and O. That's 850 01:04:23,480 --> 01:04:28,420 the real-- that's the single piece we really 851 01:04:28,420 --> 01:04:30,210 need to finish the kinematics. 852 01:04:30,210 --> 01:04:32,080 So we can do most any problems. 853 01:04:32,080 --> 01:04:35,100 Got to be able to describe the acceleration of a point 854 01:04:35,100 --> 01:04:37,770 and translating rotating frame. 855 01:04:37,770 --> 01:04:41,960 And that's going to be the acceleration of a with respect 856 01:04:41,960 --> 01:04:53,970 to O, plus a time derivative of the velocity of b with respect 857 01:04:53,970 --> 01:04:55,250 to O. 858 01:04:55,250 --> 01:04:58,790 We've calculated the velocity, we 859 01:04:58,790 --> 01:05:05,300 need to be able to essentially carry out this derivative. 860 01:05:05,300 --> 01:05:10,250 Two time derivatives of the r, b, a, or a single time 861 01:05:10,250 --> 01:05:12,870 derivative of bva. 862 01:05:12,870 --> 01:05:18,150 Well we just computed the velocity 863 01:05:18,150 --> 01:05:20,310 in this-- of this rotating frame and this 864 01:05:20,310 --> 01:05:22,150 is our final expression. 865 01:05:22,150 --> 01:05:26,860 So we need to compute the time derivative of that. 866 01:05:26,860 --> 01:05:32,650 I just-- so it's going to look like r 867 01:05:32,650 --> 01:05:43,520 dot r hat plus over here a term r theta dot theta hat 868 01:05:43,520 --> 01:05:50,250 and pardon me for doing this I think 869 01:05:50,250 --> 01:05:51,740 it'll be cleaner in the end. 870 01:05:51,740 --> 01:05:55,180 I'm going to start with my z dot k, keep it over here, 871 01:05:55,180 --> 01:06:01,500 plus r dot r hat plus this term. 872 01:06:04,440 --> 01:06:06,500 And this is going to take up a lot room. 873 01:06:06,500 --> 01:06:08,600 Just spread out this way so it-- 874 01:06:17,000 --> 01:06:19,410 So let's just-- I'm going to write down where this comes 875 01:06:19,410 --> 01:06:21,860 out, this is a little tedious, but then you'll 876 01:06:21,860 --> 01:06:26,190 have seen it once hopefully believe that it really works. 877 01:06:31,410 --> 01:06:34,350 So these terms, this first term here just 878 01:06:34,350 --> 01:06:36,650 gives you z double dot. 879 01:06:36,650 --> 01:06:37,770 So let's write her down. 880 01:06:37,770 --> 01:06:44,870 So just z double dot k hat time derivative of this, 881 01:06:44,870 --> 01:06:50,780 plus an r double dot r hat, but now I 882 01:06:50,780 --> 01:06:53,740 have to take-- do it, flip it and do the other side of it. 883 01:06:53,740 --> 01:07:05,460 So I get my-- how do I want to do this? 884 01:07:15,909 --> 01:07:16,659 Yeah, I like this. 885 01:07:26,730 --> 01:07:29,826 So this term kept-- leads to this. 886 01:07:29,826 --> 01:07:35,270 This term brings you to here. 887 01:07:35,270 --> 01:07:50,410 This term, you get r dot times theta dot theta hat plus an r 888 01:07:50,410 --> 01:07:56,660 and now you need to take a time derivative of theta dot theta 889 01:07:56,660 --> 01:07:58,550 hat. 890 01:07:58,550 --> 01:07:59,810 So that's going to expand. 891 01:07:59,810 --> 01:08:02,080 So this brings you to here. 892 01:08:28,680 --> 01:08:33,399 Now we've done this derivative, so we can put it in. 893 01:08:33,399 --> 01:08:59,930 So this gives us this term over here, 894 01:08:59,930 --> 01:09:01,449 so let's keep adding these up. 895 01:09:11,520 --> 01:09:15,319 Notice this gives me an r dot theta dot theta hat. 896 01:09:15,319 --> 01:09:20,399 This gives me an r dot theta dot theta hat. 897 01:09:23,050 --> 01:09:24,330 Two identical terms. 898 01:09:28,680 --> 01:09:38,649 Now this term gives me an r theta double dot theta hat. 899 01:09:38,649 --> 01:09:40,340 And now this-- now we need to take 900 01:09:40,340 --> 01:09:43,020 the derivative of theta hat and that gives you 901 01:09:43,020 --> 01:09:46,515 minus theta dot r hat. 902 01:09:46,515 --> 01:09:59,980 So you get a minus r but multiplied by theta dot again 903 01:09:59,980 --> 01:10:04,220 squared r hat. 904 01:10:04,220 --> 01:10:05,820 So I think we're about there. 905 01:10:05,820 --> 01:10:07,820 We're going to start collecting things together. 906 01:10:23,220 --> 01:10:30,930 Now we have a two r dot theta dot 907 01:10:30,930 --> 01:10:46,750 theta hat plus an r theta double dot theta hat, minus r theta 908 01:10:46,750 --> 01:10:49,780 dot squared r half. 909 01:10:49,780 --> 01:10:51,880 So these things, these derivatives, 910 01:10:51,880 --> 01:10:55,730 just all kind of flowed down and led to more terms. 911 01:10:58,760 --> 01:11:06,140 But now if we compare-- that we get one, two, three, four, 912 01:11:06,140 --> 01:11:08,960 five, I clump these together. 913 01:11:08,960 --> 01:11:14,250 This is the change in length of the r vector. 914 01:11:14,250 --> 01:11:19,500 Stretch in the position has a z component and a r component. 915 01:11:19,500 --> 01:11:23,630 This is the movement of the coordinate system if it moves. 916 01:11:23,630 --> 01:11:26,900 This is the Coriolis term. 917 01:11:26,900 --> 01:11:30,470 This is the Euler acceleration and this is 918 01:11:30,470 --> 01:11:33,350 the centripetal acceleration. 919 01:11:33,350 --> 01:11:36,040 So this is the-- what happens when 920 01:11:36,040 --> 01:11:41,700 you start with that messy vector thing 921 01:11:41,700 --> 01:11:47,320 and apply it, restrict it to a cylindrical coordinate problem, 922 01:11:47,320 --> 01:11:50,400 which is basically planar motion. 923 01:11:50,400 --> 01:11:53,470 But you allow some things in the z direction only. 924 01:11:53,470 --> 01:11:58,090 So polar coordinates is a more limited form of that, 925 01:11:58,090 --> 01:12:01,290 but it's-- every term comes back, 926 01:12:01,290 --> 01:12:03,270 every term is still in it. 927 01:12:03,270 --> 01:12:07,950 So acceleration of the moving coordinate 928 01:12:07,950 --> 01:12:11,080 system, change of the length of the position 929 01:12:11,080 --> 01:12:14,580 vector in the moving coordinate system, 930 01:12:14,580 --> 01:12:19,610 the Coriolis term the Euler acceleration term that angular 931 01:12:19,610 --> 01:12:25,490 acceleration speedup and finally the centripetal term. 932 01:12:25,490 --> 01:12:29,030 Now the way you go about solving problems, 933 01:12:29,030 --> 01:12:32,490 usins-- doing problems in polar coordinates. 934 01:12:32,490 --> 01:12:36,110 So now you're asked to find an equation of motion. 935 01:12:36,110 --> 01:12:40,560 This is an expression for the acceleration 936 01:12:40,560 --> 01:12:42,770 of whatever it is you're trying to describe 937 01:12:42,770 --> 01:12:46,370 in an inertial frame. 938 01:12:46,370 --> 01:12:51,690 So that when you say-- you can now say f equals ma. 939 01:12:51,690 --> 01:12:53,700 If you know the-- sometimes you're 940 01:12:53,700 --> 01:12:55,550 given the forces in problems and your asked 941 01:12:55,550 --> 01:12:59,370 to find the accelerations. 942 01:12:59,370 --> 01:13:01,140 But what if you're given the acceleration 943 01:13:01,140 --> 01:13:03,610 and you're asked to find the force? 944 01:13:03,610 --> 01:13:06,566 All right, I give you the simplest problem of this kind. 945 01:13:09,810 --> 01:13:13,030 What's the tension in the string? 946 01:13:13,030 --> 01:13:17,517 Well if I know that-- so I just say, 947 01:13:17,517 --> 01:13:19,850 the way you do these problems is how many things can you 948 01:13:19,850 --> 01:13:20,480 eliminate? 949 01:13:20,480 --> 01:13:22,710 Well I wasn't moving, this term goes away. 950 01:13:22,710 --> 01:13:23,930 That's zero. 951 01:13:23,930 --> 01:13:26,370 Z wasn't involved, it's not changing, 952 01:13:26,370 --> 01:13:30,259 it's just constant angular rotation, that term is 0. 953 01:13:30,259 --> 01:13:32,300 What's the change in length of the string while I 954 01:13:32,300 --> 01:13:33,660 was doing it? 955 01:13:33,660 --> 01:13:36,410 Now if that term's 0 we're getting easier all time. 956 01:13:36,410 --> 01:13:41,860 How far-- how fast was the length getting longer? 957 01:13:41,860 --> 01:13:43,180 That terms gone away. 958 01:13:43,180 --> 01:13:46,432 Was I speeding up, or slowing down, or constant speed? 959 01:13:46,432 --> 01:13:47,890 Well we'll say it's constant speed, 960 01:13:47,890 --> 01:13:50,060 ooh this problem's getting easier all the time. 961 01:13:50,060 --> 01:13:51,930 I'm down to one term. 962 01:13:51,930 --> 01:13:53,090 f equals ma. 963 01:14:03,730 --> 01:14:10,960 Minus r theta dot squared r hat. 964 01:14:10,960 --> 01:14:15,750 So the force that I must have been applying to the string 965 01:14:15,750 --> 01:14:18,690 was in the minus r hat direction and had 966 01:14:18,690 --> 01:14:22,380 magnitude mr omega squared. 967 01:14:25,400 --> 01:14:27,010 So that's actually all there is to it. 968 01:14:27,010 --> 01:14:29,830 We're using polar coordinates and cylindrical coordinates 969 01:14:29,830 --> 01:14:33,210 to do second law problems. 970 01:14:33,210 --> 01:14:35,410 So there's a couple of problems that you're 971 01:14:35,410 --> 01:14:37,770 doing these kind of things on the homework 972 01:14:37,770 --> 01:14:40,430 set that's being put out today. 973 01:14:40,430 --> 01:14:44,330 So give them a try. 974 01:14:44,330 --> 01:14:45,827 Have a good weekend.