1 00:00:00,080 --> 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 00:00:21,465 --> 00:00:23,840 PROFESSOR: So we were doing velocities and accelerations. 9 00:00:23,840 --> 00:00:33,070 We came up with-- I guess I ought to continue to remind us. 10 00:00:33,070 --> 00:00:36,140 We're talking about velocities and accelerations 11 00:00:36,140 --> 00:00:40,630 of a point with respect to another point, to which we've 12 00:00:40,630 --> 00:00:44,480 attached-- I'll make this point here-- 13 00:00:44,480 --> 00:00:49,760 a reference frame, x prime, y prime, z prime. 14 00:00:49,760 --> 00:00:52,750 And there's a vector that goes between these two 15 00:00:52,750 --> 00:00:56,420 and a vector that goes there so that we can say r of B 16 00:00:56,420 --> 00:01:01,660 with respect to O, my inertial frame, is r of A 17 00:01:01,660 --> 00:01:06,050 with respect to O plus r of B with respect 18 00:01:06,050 --> 00:01:07,550 to A. These are all vectors. 19 00:01:07,550 --> 00:01:10,070 And then from these, we derive velocities and acceleration 20 00:01:10,070 --> 00:01:10,570 formulas. 21 00:01:10,570 --> 00:01:15,040 And so we've come up with a couple of very handy formulas. 22 00:01:15,040 --> 00:01:21,920 The velocity formula, velocity of B with respect to O, 23 00:01:21,920 --> 00:01:23,920 is the velocity of A with respect 24 00:01:23,920 --> 00:01:31,247 to O plus you have to take a time derivative of this. 25 00:01:31,247 --> 00:01:33,580 And so I'm going to give you just the general expression 26 00:01:33,580 --> 00:01:35,870 here just as a brief reminder. 27 00:01:35,870 --> 00:01:39,760 It's the derivative of B with respect 28 00:01:39,760 --> 00:01:46,490 to A as seen in the Axyz frame plus omega with respect to O 29 00:01:46,490 --> 00:01:49,279 cross rBA. 30 00:01:49,279 --> 00:01:50,195 These are all vectors. 31 00:01:52,760 --> 00:01:55,030 So that's the velocity formula, remember? 32 00:01:55,030 --> 00:01:57,640 This is if you're in the frame rotating 33 00:01:57,640 --> 00:01:59,650 and translating with it, this is the change 34 00:01:59,650 --> 00:02:02,860 of length of the vector. 35 00:02:02,860 --> 00:02:05,430 And this, then, is the contribution to the velocity 36 00:02:05,430 --> 00:02:10,199 that you see in the fixed frame that comes from the rotation. 37 00:02:10,199 --> 00:02:12,820 And we then got into polar coordinates. 38 00:02:12,820 --> 00:02:15,750 And we found out that if you use polar coordinates, 39 00:02:15,750 --> 00:02:20,080 then you can express this as the velocity of A 40 00:02:20,080 --> 00:02:27,780 with respect to O plus r dot r hat. 41 00:02:27,780 --> 00:02:30,530 And I should really say cylindrical coordinates, 42 00:02:30,530 --> 00:02:43,250 z dot k hat plus r theta dot theta hat. 43 00:02:43,250 --> 00:02:45,070 So that's exactly the same thing. 44 00:02:45,070 --> 00:02:47,610 This is full 3D vector notation. 45 00:02:47,610 --> 00:02:50,680 This is a special case of a coordinate system which 46 00:02:50,680 --> 00:02:53,730 we call polar coordinates. 47 00:02:53,730 --> 00:02:58,010 And we came up with another formula for accelerations, 48 00:02:58,010 --> 00:03:04,030 the full 3D vector version of that, A with respect 49 00:03:04,030 --> 00:03:10,610 to O plus-- this is the acceleration of B 50 00:03:10,610 --> 00:03:14,620 with respect to A, but as seen in the Axyz frame. 51 00:03:19,000 --> 00:03:27,320 2 omega cross-- and this is v, the velocity, 52 00:03:27,320 --> 00:03:30,700 as seen in the xyz frame. 53 00:03:30,700 --> 00:03:34,100 So these things, this is no contribution from acceleration. 54 00:03:34,100 --> 00:03:36,000 This is no contribution from acceleration. 55 00:03:39,020 --> 00:03:47,810 Plus omega naught cross rBA plus omega cross 56 00:03:47,810 --> 00:03:55,750 omega cross rBA-- all vectors. 57 00:03:59,190 --> 00:04:03,130 This is a movement of the frame, the acceleration of it 58 00:04:03,130 --> 00:04:04,010 with respect to this. 59 00:04:04,010 --> 00:04:06,330 This is pure translation. 60 00:04:06,330 --> 00:04:10,800 This is the acceleration of the dog on the merry go 61 00:04:10,800 --> 00:04:14,535 round with respect to the center of the coordinate system there. 62 00:04:14,535 --> 00:04:17,170 It does not involve rotation. 63 00:04:17,170 --> 00:04:21,050 This is this term that we call Coriolis. 64 00:04:21,050 --> 00:04:22,650 This is a term we call Euler. 65 00:04:22,650 --> 00:04:26,080 This is the angular speed up of the system, the angular 66 00:04:26,080 --> 00:04:27,280 acceleration. 67 00:04:27,280 --> 00:04:28,450 And this is centripetal. 68 00:04:32,100 --> 00:04:33,900 And if you do these, if you want to go 69 00:04:33,900 --> 00:04:37,240 to cylindrical coordinates-- and what we're going to do next 70 00:04:37,240 --> 00:04:39,610 is just do some applications here. 71 00:04:39,610 --> 00:04:42,490 The acceleration of B with respect to O and cylindrical 72 00:04:42,490 --> 00:04:46,670 coordinates-- it's a translation piece. 73 00:04:46,670 --> 00:04:49,970 Plus now in cylindrical, it's useful to group these. 74 00:04:56,550 --> 00:05:02,650 So I'm going to put together this term and this term. 75 00:05:02,650 --> 00:05:05,024 Because they're both in the r hat direction. 76 00:05:05,024 --> 00:05:06,940 And I'm going to put together these two terms. 77 00:05:06,940 --> 00:05:11,090 Because they're both in the theta hat direction. 78 00:05:11,090 --> 00:05:16,990 And I need a z double dot k, because it's cylindrical. 79 00:05:16,990 --> 00:05:20,000 And then I have the theta hat piece, 80 00:05:20,000 --> 00:05:29,180 r theta double dot plus 2r dot theta dot theta hat. 81 00:05:29,180 --> 00:05:32,880 This is the same thing as this. 82 00:05:32,880 --> 00:05:37,587 Except this is expressed in cylindrical coordinates. 83 00:05:37,587 --> 00:05:39,420 And cylindrical coordinates are particularly 84 00:05:39,420 --> 00:05:42,740 good for doing the kind of problems that are mostly 85 00:05:42,740 --> 00:05:45,780 done at the level of this course, which we call planar 86 00:05:45,780 --> 00:05:50,190 motion problems, confined to an x, y plane, 87 00:05:50,190 --> 00:05:54,650 and confined to single axis rotation in z. 88 00:05:54,650 --> 00:05:57,620 This is a coordinate system that's ideally suited 89 00:05:57,620 --> 00:05:58,810 to do problems like that. 90 00:05:58,810 --> 00:06:00,410 And that's why we use it. 91 00:06:00,410 --> 00:06:04,550 So now let's do some examples. 92 00:06:04,550 --> 00:06:11,550 This is really quite a powerful-- 93 00:06:11,550 --> 00:06:13,260 now that you have these two equations, 94 00:06:13,260 --> 00:06:15,625 you can do a lot of kinematics and dynamics. 95 00:06:25,270 --> 00:06:31,010 So we started last time right at the very end. 96 00:06:31,010 --> 00:06:36,100 I said, OK, let's do this problem really quickly, right? 97 00:06:36,100 --> 00:06:39,650 Constant rotation rate, constant radius, 98 00:06:39,650 --> 00:06:42,030 no angular acceleration, no change 99 00:06:42,030 --> 00:06:44,340 in length of the thing-- pretty simple problem. 100 00:06:44,340 --> 00:06:46,630 And we zipped it off really fast. 101 00:06:46,630 --> 00:06:50,870 And I wanted to start there, do that really quickly. 102 00:06:50,870 --> 00:06:53,760 So this is my case one. 103 00:06:53,760 --> 00:06:55,495 It's my ball on the string. 104 00:07:01,260 --> 00:07:08,060 Here's point A. Here's point B. Here's the theta hat 105 00:07:08,060 --> 00:07:12,930 direction, r hat direction. 106 00:07:12,930 --> 00:07:17,880 And it has some constant length R here. 107 00:07:17,880 --> 00:07:24,140 So r dot r double dot are 0. 108 00:07:24,140 --> 00:07:33,370 There's no also z, z dot, z double dot, no z motion at all. 109 00:07:33,370 --> 00:07:36,090 Those are all 0. 110 00:07:36,090 --> 00:07:39,980 Theta dot is a constant. 111 00:07:39,980 --> 00:07:45,460 I'll call it cap omega in the k hat direction, right hand rule. 112 00:07:45,460 --> 00:07:50,080 Theta double dot is 0. 113 00:07:50,080 --> 00:07:53,177 So the easy way to use these formulas 114 00:07:53,177 --> 00:07:55,010 is you just start knocking out all the terms 115 00:07:55,010 --> 00:07:56,490 that you don't need. 116 00:07:56,490 --> 00:08:00,800 But let's, just to give you a little quick review of how 117 00:08:00,800 --> 00:08:05,560 to use these things, use the vector full 118 00:08:05,560 --> 00:08:08,770 3D version of the velocity equation for a second. 119 00:08:08,770 --> 00:08:11,930 The full 3D version says the velocity of A 120 00:08:11,930 --> 00:08:16,250 with respect to O, what's that in this problem? 121 00:08:16,250 --> 00:08:19,630 That's the translating term. 122 00:08:19,630 --> 00:08:22,060 So I was standing still, not going anywhere. 123 00:08:22,060 --> 00:08:25,070 But if I were walking along, I'd still spin that thing. 124 00:08:25,070 --> 00:08:28,070 OK, so this problem, this term is 0. 125 00:08:28,070 --> 00:08:33,409 This problem, r dot-- let's go here. 126 00:08:33,409 --> 00:08:35,580 This is the rate of change of the length 127 00:08:35,580 --> 00:08:40,321 of the string in the coordinate system of the string walking 128 00:08:40,321 --> 00:08:40,820 along. 129 00:08:40,820 --> 00:08:43,110 So what's r dot? 130 00:08:43,110 --> 00:08:45,180 OK, so that term is 0. 131 00:08:45,180 --> 00:08:51,170 And omega cross rBA-- well, r is in the R hat, capital R R hat, 132 00:08:51,170 --> 00:08:56,030 cross with omega k. 133 00:08:56,030 --> 00:09:01,883 k cross R is-- k cross of R hat? 134 00:09:01,883 --> 00:09:02,630 STUDENT: Theta. 135 00:09:02,630 --> 00:09:03,505 PROFESSOR: Theta hat. 136 00:09:03,505 --> 00:09:07,230 So we get our familiar-- this first term is 0. 137 00:09:07,230 --> 00:09:08,860 The second term is 0. 138 00:09:08,860 --> 00:09:12,830 The third term, we just get our R omega 139 00:09:12,830 --> 00:09:14,920 in the theta hat direction. 140 00:09:14,920 --> 00:09:16,050 We know that to be true. 141 00:09:16,050 --> 00:09:17,633 The point I'm just trying to make here 142 00:09:17,633 --> 00:09:21,530 is you can always just fall back and use the vector formula, 143 00:09:21,530 --> 00:09:23,230 full 3D formulas of both. 144 00:09:23,230 --> 00:09:26,370 Just plug it in, and everything will just drop out. 145 00:09:26,370 --> 00:09:30,270 You can also then go to the cylindrical coordinate terms 146 00:09:30,270 --> 00:09:32,750 when you happen to have it all expressed 147 00:09:32,750 --> 00:09:37,710 in coordinates of that kind to make it simpler for you. 148 00:09:37,710 --> 00:09:43,400 OK, so now let's quickly then do the acceleration of B 149 00:09:43,400 --> 00:09:49,300 with respect to O. And let's use the formulation already 150 00:09:49,300 --> 00:09:50,890 in cylindrical coordinates. 151 00:09:50,890 --> 00:09:53,760 What's A with respect to O? 152 00:09:53,760 --> 00:09:54,720 0. 153 00:09:54,720 --> 00:09:58,200 What's r double dot? 154 00:09:58,200 --> 00:10:01,470 How about the r theta dot squared term, 0 or not? 155 00:10:01,470 --> 00:10:02,290 Nope. 156 00:10:02,290 --> 00:10:04,243 Let's go on-- z double dot? 157 00:10:04,243 --> 00:10:04,861 STUDENT: 0. 158 00:10:04,861 --> 00:10:06,110 PROFESSOR: r theta double dot? 159 00:10:06,110 --> 00:10:07,082 STUDENT: 0. 160 00:10:07,082 --> 00:10:08,311 PROFESSOR: r dot theta dot? 161 00:10:08,311 --> 00:10:08,810 STUDENT: 0 162 00:10:08,810 --> 00:10:10,540 PROFESSOR: OK, we only end up with one term. 163 00:10:10,540 --> 00:10:12,770 Actually, I'm going to keep this one for a second-- A 164 00:10:12,770 --> 00:10:16,030 with respect to O minus. 165 00:10:16,030 --> 00:10:25,360 And we came up with our R theta dot squared r hat term. 166 00:10:25,360 --> 00:10:28,390 I leave this in here because I actually don't have to say. 167 00:10:28,390 --> 00:10:33,000 If I wanted now to do this problem, 168 00:10:33,000 --> 00:10:35,880 if I asked you to do a problem where I'm doing this, 169 00:10:35,880 --> 00:10:40,039 and I start accelerating, do you know how to do it? 170 00:10:40,039 --> 00:10:41,705 There's the answer-- still there, right? 171 00:10:46,437 --> 00:11:06,155 All right, let's do a quick free body diagram. 172 00:11:11,660 --> 00:11:18,920 Here's our mass, my master coordinate system out here. 173 00:11:18,920 --> 00:11:23,735 Let's draw-- let's say it's right here at 90 degrees. 174 00:11:26,570 --> 00:11:29,300 What are the external forces on the mass in this problem? 175 00:11:33,196 --> 00:11:34,660 STUDENT: [INAUDIBLE]. 176 00:11:34,660 --> 00:11:37,120 PROFESSOR: So there's tension in the string, right? 177 00:11:37,120 --> 00:11:41,700 OK, so now this is a really trivially simple problem. 178 00:11:41,700 --> 00:11:45,660 So the emphasize here is on the concept. 179 00:11:45,660 --> 00:11:50,100 So now when you're asked to come up with an equation of motion 180 00:11:50,100 --> 00:11:55,520 or compute the forces on a mass, use Newton's second law, 181 00:11:55,520 --> 00:11:58,040 F equals mass times acceleration. 182 00:11:58,040 --> 00:12:01,870 You now have the complete 3D vector formulation 183 00:12:01,870 --> 00:12:08,290 for acceleration of a particle in a translating rotating 184 00:12:08,290 --> 00:12:09,870 coordinate system. 185 00:12:09,870 --> 00:12:12,780 That's all you need to compute accelerations for lots 186 00:12:12,780 --> 00:12:14,850 and lots of difficult problems. 187 00:12:14,850 --> 00:12:18,670 And so if you can write down the acceleration, 188 00:12:18,670 --> 00:12:23,960 you say it's equal to what? 189 00:12:23,960 --> 00:12:26,870 Mass-- if you multiply mass times that acceleration, 190 00:12:26,870 --> 00:12:28,269 what's that equal to? 191 00:12:28,269 --> 00:12:29,060 STUDENT: The force. 192 00:12:29,060 --> 00:12:31,393 PROFESSOR: The forces that must be acting on the system. 193 00:12:31,393 --> 00:12:32,860 And that's the point here. 194 00:12:32,860 --> 00:12:37,090 So now I want to know the forces on the system, the summation 195 00:12:37,090 --> 00:12:38,650 of the external forces. 196 00:12:38,650 --> 00:12:39,940 And then these are vectors. 197 00:12:39,940 --> 00:12:42,100 And you can do them component by component. 198 00:12:42,100 --> 00:12:47,010 Some of the external forces in the r hat direction 199 00:12:47,010 --> 00:12:49,810 must be equal to the mass-- in this case, just 200 00:12:49,810 --> 00:12:56,360 a particle-- times the acceleration of that particle. 201 00:12:56,360 --> 00:12:58,050 And in this problem, then that would 202 00:12:58,050 --> 00:13:01,670 be the mass times the acceleration of A with respect 203 00:13:01,670 --> 00:13:10,540 to O minus R theta dot squared r hat. 204 00:13:10,540 --> 00:13:15,690 And this would have to be the r hat component of this thing. 205 00:13:15,690 --> 00:13:17,990 I said I just want the R component. 206 00:13:17,990 --> 00:13:20,180 I'd have to figure out if I was running along, 207 00:13:20,180 --> 00:13:23,110 if I had it in the same direction as r hat. 208 00:13:23,110 --> 00:13:25,490 What part of that acceleration is in that direction? 209 00:13:25,490 --> 00:13:26,700 That would come here. 210 00:13:26,700 --> 00:13:29,040 If there's 0, you just make it 0. 211 00:13:29,040 --> 00:13:34,610 So let's just let the acceleration of A 212 00:13:34,610 --> 00:13:36,830 with respect to O be 0. 213 00:13:36,830 --> 00:13:41,690 Then that says the sum of the forces in the r direction 214 00:13:41,690 --> 00:13:49,120 is equal to the mass minus mR theta dot squared. 215 00:13:49,120 --> 00:13:52,210 And if we were to draw a free body diagram, 216 00:13:52,210 --> 00:13:53,730 we would find out that, ahh, there 217 00:13:53,730 --> 00:13:55,980 must be a tension on the string pulling 218 00:13:55,980 --> 00:14:00,120 in on the mass sufficient to give it 219 00:14:00,120 --> 00:14:03,000 the acceleration that you've computed. 220 00:14:03,000 --> 00:14:04,460 So every problem, when you're asked 221 00:14:04,460 --> 00:14:06,790 to compute the force, or the next step up, 222 00:14:06,790 --> 00:14:09,250 find the equation of motion, the equation of motion 223 00:14:09,250 --> 00:14:13,115 is just writing this thing out. 224 00:14:20,250 --> 00:14:30,600 OK, now I want to move on to a more interesting problem. 225 00:14:55,510 --> 00:15:02,260 All right, I'm going to do this problem. 226 00:15:02,260 --> 00:15:05,540 So it's a hollow tube. 227 00:15:05,540 --> 00:15:09,230 And we're going to look at things like, I put a ping pong 228 00:15:09,230 --> 00:15:11,010 ball in it. 229 00:15:11,010 --> 00:15:15,345 And if I swing this tube around, the ping pong ball 230 00:15:15,345 --> 00:15:16,220 is going to come out. 231 00:15:23,290 --> 00:15:25,710 OK, there must be some forces on that thing 232 00:15:25,710 --> 00:15:27,405 to cause it to come out. 233 00:15:27,405 --> 00:15:29,640 There must be some accelerations on them. 234 00:15:29,640 --> 00:15:31,960 And so I could conceivably have an R, 235 00:15:31,960 --> 00:15:35,070 a theta double dot acceleration. 236 00:15:35,070 --> 00:15:39,130 It certainly is going to have theta dot rotation rates. 237 00:15:39,130 --> 00:15:40,520 The ball is allowed to move. 238 00:15:40,520 --> 00:15:44,375 So there can be nonzero r dots, r double dots-- 239 00:15:44,375 --> 00:15:47,390 a lot going on inside of this simple little tube. 240 00:15:47,390 --> 00:15:49,500 So that's what I want to figure out. 241 00:15:49,500 --> 00:15:52,990 Let's see if we can come up with a model for this problem. 242 00:16:08,090 --> 00:16:09,620 So here's my z-axis. 243 00:16:09,620 --> 00:16:14,980 I have a rotation around it, some theta dot k hat direction. 244 00:16:14,980 --> 00:16:17,950 Here's my tube. 245 00:16:17,950 --> 00:16:20,020 It's rotating around. 246 00:16:20,020 --> 00:16:22,980 So this is sort of a side view, your view of the tube. 247 00:16:22,980 --> 00:16:26,940 Here's that ping pong ball in there. 248 00:16:26,940 --> 00:16:35,110 And I'm going to idealize that ping pong 249 00:16:35,110 --> 00:16:38,970 ball for a minute, a little more general problem. 250 00:16:38,970 --> 00:16:42,570 Let's say I have kind of a nut on this thing, a disk, 251 00:16:42,570 --> 00:16:45,280 something hanging onto the outside. 252 00:16:45,280 --> 00:16:49,380 And I can control the rate, the speed 253 00:16:49,380 --> 00:16:51,650 at which this thing goes out. 254 00:16:51,650 --> 00:16:54,330 The ping pong ball, this is going 255 00:16:54,330 --> 00:16:55,610 to be an application of this. 256 00:16:55,610 --> 00:16:57,735 But I want to be able to do several other versions, 257 00:16:57,735 --> 00:17:02,120 like make the speed constant for a second. 258 00:17:02,120 --> 00:17:06,690 OK, and looking down on this thing, 259 00:17:06,690 --> 00:17:13,000 top view, here's our inertial frame, 260 00:17:13,000 --> 00:17:14,579 maybe out here like this. 261 00:17:14,579 --> 00:17:16,800 Here's my mass. 262 00:17:16,800 --> 00:17:20,000 So in polar coordinates, here's your theta. 263 00:17:20,000 --> 00:17:24,192 The r is this. 264 00:17:24,192 --> 00:17:27,150 This is your r hat. 265 00:17:27,150 --> 00:17:28,810 This is your theta hat directions. 266 00:17:33,750 --> 00:17:39,650 I'm going to let the velocity of A with respect to O be 0, 267 00:17:39,650 --> 00:17:43,380 so there's no translational of this system. 268 00:17:43,380 --> 00:17:49,510 And z, z dot, z double dot, those are all 0. 269 00:17:49,510 --> 00:17:53,280 So nothing's happening in the z direction. 270 00:17:53,280 --> 00:17:57,090 So I want to first compute the velocity of B 271 00:17:57,090 --> 00:17:59,740 with respect to O. 272 00:17:59,740 --> 00:18:04,970 And you ought to be able to do that sort of by inspection. 273 00:18:04,970 --> 00:18:10,830 It comes only from-- oh, I haven't told you enough. 274 00:18:10,830 --> 00:18:12,790 What do I want to make happen in this problem? 275 00:18:12,790 --> 00:18:20,993 I want to let r dot-- it's going to be some vr, 276 00:18:20,993 --> 00:18:21,743 and it's constant. 277 00:18:24,660 --> 00:18:27,259 I'm not going to go quite to my ping pong shooter here yet. 278 00:18:27,259 --> 00:18:29,300 I'm going to do a slightly simpler problem first. 279 00:18:29,300 --> 00:18:30,260 So this is a constant. 280 00:18:30,260 --> 00:18:32,980 That means r double dot is 0. 281 00:18:32,980 --> 00:18:34,860 So this thing is just-- let's say 282 00:18:34,860 --> 00:18:37,490 you had threads on this thing, and it's a screw, 283 00:18:37,490 --> 00:18:41,950 and it's just moving its way out at a constant rate. 284 00:18:41,950 --> 00:18:47,980 And I'm going to have constant angular, so theta dot. 285 00:18:47,980 --> 00:18:51,350 I'll call that cap omega k hat. 286 00:18:51,350 --> 00:18:55,450 So the angular rate is also constant. 287 00:18:55,450 --> 00:18:58,070 All right, if that's the case, can you tell me, 288 00:18:58,070 --> 00:19:04,030 what's v of B with respect to my fixed frame O? 289 00:19:07,800 --> 00:19:11,480 Well, any time you're not sure, you go back to this formula, 290 00:19:11,480 --> 00:19:12,560 throw out terms. 291 00:19:12,560 --> 00:19:14,640 There's no z dot term. 292 00:19:14,640 --> 00:19:15,670 This is constant. 293 00:19:15,670 --> 00:19:18,390 This is 0, 0, 0. 294 00:19:18,390 --> 00:19:19,850 You have this term. 295 00:19:19,850 --> 00:19:22,560 It's some vr in the r hat direction. 296 00:19:22,560 --> 00:19:26,520 You have this term, wherever it happens to be in the theta hat 297 00:19:26,520 --> 00:19:27,020 direction. 298 00:19:39,372 --> 00:19:43,950 r dot in the r hat direction plus r theta 299 00:19:43,950 --> 00:19:46,450 dot in the theta hat direction-- OK, 300 00:19:46,450 --> 00:19:50,850 we need acceleration next, B with respect to O. 301 00:19:50,850 --> 00:19:53,110 And now you can crank through the terms again. 302 00:19:53,110 --> 00:19:57,290 This time, the first term is 0. 303 00:19:57,290 --> 00:19:59,880 The second term is 0, because it's constant. 304 00:19:59,880 --> 00:20:02,560 The third term is definitely not 0. 305 00:20:02,560 --> 00:20:03,660 The fourth term is 0. 306 00:20:03,660 --> 00:20:04,280 Fifth term? 307 00:20:08,040 --> 00:20:09,270 0. 308 00:20:09,270 --> 00:20:11,140 This term? 309 00:20:11,140 --> 00:20:13,020 Not 0. 310 00:20:13,020 --> 00:20:14,160 Let me just write them. 311 00:20:18,830 --> 00:20:23,600 So you get a minus r theta dot squared r hat-- that's 312 00:20:23,600 --> 00:20:30,904 the radial direction term-- plus 2r dot theta dot theta hat. 313 00:20:30,904 --> 00:20:31,945 That's the accelerations. 314 00:20:36,000 --> 00:20:38,530 So you have an acceleration now in the r hat direction 315 00:20:38,530 --> 00:20:40,640 and in the theta hat direction. 316 00:20:40,640 --> 00:20:42,610 If there's acceleration to those directions, 317 00:20:42,610 --> 00:20:44,665 there must be forces. 318 00:20:47,680 --> 00:20:51,720 Newton's second law now says, again, the force 319 00:20:51,720 --> 00:20:55,200 is the mass times acceleration. 320 00:20:55,200 --> 00:20:56,930 And this is a vector. 321 00:20:56,930 --> 00:20:58,390 This is a vector. 322 00:20:58,390 --> 00:20:59,470 It has two components. 323 00:20:59,470 --> 00:21:02,585 And the nice thing about these Newton's laws and vectors 324 00:21:02,585 --> 00:21:05,180 is you can break the problems down into their vector 325 00:21:05,180 --> 00:21:07,070 components and treat the r direction 326 00:21:07,070 --> 00:21:09,300 as one equation of motion, and the theta 327 00:21:09,300 --> 00:21:11,310 direction as a separate one. 328 00:21:11,310 --> 00:21:15,240 So we might want to draw a free body diagram. 329 00:21:15,240 --> 00:21:20,260 Here's this block working its way out. 330 00:21:20,260 --> 00:21:22,660 We know that there's probably some axial force. 331 00:21:22,660 --> 00:21:24,700 I'm just going to call it T. And there's 332 00:21:24,700 --> 00:21:27,700 some other unknown force here in the theta hat direction. 333 00:21:27,700 --> 00:21:30,704 So this is my theta hat. 334 00:21:30,704 --> 00:21:32,870 I've drawn this just intentionally in the positive r 335 00:21:32,870 --> 00:21:33,490 hat direction. 336 00:21:33,490 --> 00:21:35,210 The sign that comes out will tell us 337 00:21:35,210 --> 00:21:38,490 which direction it really is if you're not certain. 338 00:21:38,490 --> 00:21:42,250 Just draw it positive, in the positive r hat direction. 339 00:21:42,250 --> 00:21:45,520 And that's your free body diagram. 340 00:21:45,520 --> 00:21:48,180 If I wanted to put gravity in there, I might have. 341 00:21:48,180 --> 00:21:51,200 But we're doing this in the horizontal plane. 342 00:21:51,200 --> 00:21:52,560 Gravity is in and out this way. 343 00:21:52,560 --> 00:21:53,670 It's in the z direction. 344 00:21:53,670 --> 00:21:57,740 And we know it's constrained, can't move. z double dot is 0. 345 00:21:57,740 --> 00:22:01,710 So there's certainly a support force that picks up the weight. 346 00:22:01,710 --> 00:22:04,170 But this is in our horizontal plane. 347 00:22:04,170 --> 00:22:06,170 There's your free body diagram. 348 00:22:06,170 --> 00:22:09,310 And we can write two equations to solve for these things. 349 00:22:09,310 --> 00:22:16,050 So the sum of the forces in the r hat direction is T. 350 00:22:16,050 --> 00:22:19,640 And that must be equal to the mass times the acceleration 351 00:22:19,640 --> 00:22:27,690 in the r, minus mr theta dot squared in the r hat direction. 352 00:22:27,690 --> 00:22:31,890 Sure enough, the tension has to pull inwards 353 00:22:31,890 --> 00:22:33,820 in the minus r hat direction. 354 00:22:33,820 --> 00:22:39,115 And that's the full result. 355 00:22:39,115 --> 00:22:44,350 STUDENT: Where are the T and F forces exactly? 356 00:22:44,350 --> 00:22:45,620 PROFESSOR: Where are they? 357 00:22:45,620 --> 00:22:53,450 OK, so I'm going to bring the ball to the outside 358 00:22:53,450 --> 00:22:54,590 where you can see it. 359 00:22:54,590 --> 00:22:58,260 This thing is going in this direction, horizontal plane, x, 360 00:22:58,260 --> 00:22:59,380 y plane. 361 00:22:59,380 --> 00:23:03,520 It's moving its fixed rate out. 362 00:23:03,520 --> 00:23:05,550 So its speed when it's in here is r 363 00:23:05,550 --> 00:23:07,370 over 2 omega in that direction. 364 00:23:07,370 --> 00:23:10,730 And the speed when it's out here is r omega in that direction. 365 00:23:10,730 --> 00:23:12,630 So clearly it's picking up speed. 366 00:23:12,630 --> 00:23:17,320 If it's picking up speed, is it picking up kinetic energy? 367 00:23:17,320 --> 00:23:19,200 Is there work being done on it somehow 368 00:23:19,200 --> 00:23:20,570 to build up that energy? 369 00:23:20,570 --> 00:23:22,970 So there must be some forces at play. 370 00:23:22,970 --> 00:23:26,090 So there's a normal force from the wall 371 00:23:26,090 --> 00:23:29,670 of this thing pushing this ball sideways to speed it up, 372 00:23:29,670 --> 00:23:30,250 for sure. 373 00:23:30,250 --> 00:23:31,650 That's one force. 374 00:23:31,650 --> 00:23:35,810 And the other force is because I'm not allowing this thing 375 00:23:35,810 --> 00:23:38,180 just to go freely out. 376 00:23:38,180 --> 00:23:41,310 I'm constraining it to constant speed out. 377 00:23:41,310 --> 00:23:43,900 It would really like to go a lot faster than that. 378 00:23:43,900 --> 00:23:46,600 So what's holding it back? 379 00:23:46,600 --> 00:23:49,510 At any instance in time, it has centripetal acceleration. 380 00:23:49,510 --> 00:23:53,290 And what's making it go in the circle 381 00:23:53,290 --> 00:23:56,060 is a force that is-- in this case, 382 00:23:56,060 --> 00:24:00,570 if that were a nut with threads, in the threads are applying 383 00:24:00,570 --> 00:24:03,368 to the nut to keep it from running away. 384 00:24:03,368 --> 00:24:05,360 STUDENT: In that free body diagram, 385 00:24:05,360 --> 00:24:08,965 F, doesn't F act similar between theta and r hat? 386 00:24:08,965 --> 00:24:10,340 And then there's like a component 387 00:24:10,340 --> 00:24:13,830 of theta hat in there? 388 00:24:13,830 --> 00:24:15,700 PROFESSOR: Well, I've broken down. 389 00:24:15,700 --> 00:24:18,080 I've chosen. 390 00:24:18,080 --> 00:24:21,200 The total force acting on this thing 391 00:24:21,200 --> 00:24:23,890 is some combination of a force in that direction 392 00:24:23,890 --> 00:24:25,760 and a combination in the axial direction. 393 00:24:25,760 --> 00:24:31,810 So it has some net direction that's neither this nor that. 394 00:24:31,810 --> 00:24:34,735 STUDENT: From that equation, it looks like what you drew, 395 00:24:34,735 --> 00:24:39,640 F has an r hat component. 396 00:24:39,640 --> 00:24:42,830 PROFESSOR: So this thing is rotating 397 00:24:42,830 --> 00:24:45,560 about some center over here. 398 00:24:45,560 --> 00:24:47,460 So this is theta dot. 399 00:24:47,460 --> 00:24:50,690 Theta is going in this direction, theta dot. 400 00:24:50,690 --> 00:24:55,050 And we've chosen a coordinate system that has unit vectors r 401 00:24:55,050 --> 00:24:56,510 hat and theta hat. 402 00:24:56,510 --> 00:24:57,560 And so it makes sense. 403 00:24:57,560 --> 00:24:59,840 We can express the acceleration in terms 404 00:24:59,840 --> 00:25:01,100 of those two components. 405 00:25:01,100 --> 00:25:05,270 It makes sense to express the forces in the same direction. 406 00:25:05,270 --> 00:25:08,620 So I've just arbitrarily said, I have some unknown force 407 00:25:08,620 --> 00:25:09,690 that's in this direction. 408 00:25:09,690 --> 00:25:11,814 And I have another unknown force in that direction. 409 00:25:14,340 --> 00:25:17,230 Then I'm saying, they account for all forces 410 00:25:17,230 --> 00:25:19,770 in this direction, whatever their source. 411 00:25:19,770 --> 00:25:23,000 That tells me that the sum of the external forces 412 00:25:23,000 --> 00:25:29,090 in the theta hat direction in this case is this unknown F. 413 00:25:29,090 --> 00:25:31,610 But I know from Newton that that's 414 00:25:31,610 --> 00:25:33,930 got to be equal to the mass times the acceleration 415 00:25:33,930 --> 00:25:35,380 in that direction. 416 00:25:35,380 --> 00:25:42,520 In this case, then, that is 2mr dot theta dot theta hat. 417 00:25:45,630 --> 00:25:50,250 So just from applying the equation 418 00:25:50,250 --> 00:25:52,580 and applying Newton's second law, 419 00:25:52,580 --> 00:25:56,359 I can find out what that force must be. 420 00:25:56,359 --> 00:25:57,900 It would've been a lot more work if I 421 00:25:57,900 --> 00:26:00,286 had drawn it in some arbitrary in between direction. 422 00:26:00,286 --> 00:26:02,660 Because then I'd have to break it down into its compounds 423 00:26:02,660 --> 00:26:03,690 to write this. 424 00:26:03,690 --> 00:26:07,060 So I've made it as easy as possible for myself. 425 00:26:07,060 --> 00:26:08,850 So there's a force like this. 426 00:26:08,850 --> 00:26:10,840 And there is another force like that. 427 00:26:10,840 --> 00:26:14,550 This one is caused by the centripetal acceleration. 428 00:26:14,550 --> 00:26:19,820 Or this force causes the centripetal acceleration. 429 00:26:19,820 --> 00:26:22,080 In order to make something go in a circular path, 430 00:26:22,080 --> 00:26:24,940 you have to exert a force on it. 431 00:26:24,940 --> 00:26:26,900 That's the force. 432 00:26:26,900 --> 00:26:30,160 In order to accelerate something angularly, 433 00:26:30,160 --> 00:26:31,500 you have to apply a force. 434 00:26:31,500 --> 00:26:33,760 That comes from the Euler term. 435 00:26:33,760 --> 00:26:36,830 And this is a curious term. 436 00:26:36,830 --> 00:26:38,250 This is the Coriolis term. 437 00:26:41,290 --> 00:26:42,580 So where does it come from? 438 00:26:45,870 --> 00:26:47,900 That's the crux of the matter here. 439 00:26:47,900 --> 00:26:48,900 Where does it come from? 440 00:26:56,600 --> 00:27:01,420 So part of the reading that you need to do now is Chapter 15. 441 00:27:01,420 --> 00:27:04,700 Chapter 15, most of it is going to be complete review. 442 00:27:04,700 --> 00:27:07,810 It just says the conservation of a linear momentum, impulse 443 00:27:07,810 --> 00:27:08,355 and momentum. 444 00:27:08,355 --> 00:27:10,900 But it also gets into angular momentum. 445 00:27:10,900 --> 00:27:14,040 So we're going to talk quite a lot about angular momentum. 446 00:27:17,660 --> 00:27:28,240 And I want to do a very brief little review right now so 447 00:27:28,240 --> 00:27:29,700 that it applies to this problem. 448 00:27:32,320 --> 00:27:34,110 So we've come up with an expression 449 00:27:34,110 --> 00:27:36,025 that the force in the theta direction 450 00:27:36,025 --> 00:27:43,420 here comes-- there's got to be that, 2mr dot theta dot. 451 00:27:52,820 --> 00:27:56,400 So here's my point O, and A for that matter. 452 00:27:56,400 --> 00:28:00,200 But looking down on our problem, here's 453 00:28:00,200 --> 00:28:03,160 my mass at some instant in time. 454 00:28:03,160 --> 00:28:09,400 My rotation rate is theta dot-- or actually it's constant. 455 00:28:09,400 --> 00:28:15,170 So it's k hat like that, cap omega k hat. 456 00:28:15,170 --> 00:28:20,520 This is my r hat direction, theta hat. 457 00:28:20,520 --> 00:28:23,710 Now, I'm going to treat this as a particle. 458 00:28:23,710 --> 00:28:26,530 Not long-- we're going to be talking about the dynamics 459 00:28:26,530 --> 00:28:27,560 of rigid bodies. 460 00:28:27,560 --> 00:28:29,510 We're just doing particles for the moment. 461 00:28:29,510 --> 00:28:31,860 We think of them just as point masses 462 00:28:31,860 --> 00:28:34,660 and don't deal with their finite extent. 463 00:28:34,660 --> 00:28:37,410 So we're still thinking of this as a particle. 464 00:28:37,410 --> 00:28:40,310 And I'm going to write down the definition of the angular 465 00:28:40,310 --> 00:28:42,390 momentum of a particle. 466 00:28:42,390 --> 00:28:47,590 This is B out here with respect to my fixed frame here at O. 467 00:28:47,590 --> 00:28:49,600 And I'm going to use a lowercase h to describe 468 00:28:49,600 --> 00:28:51,410 angular momentum of particles. 469 00:28:51,410 --> 00:28:54,640 And I'll use capital H to describe the angular 470 00:28:54,640 --> 00:28:57,670 momentum of rigid bodies. 471 00:28:57,670 --> 00:29:00,920 So this is a particle, the definition 472 00:29:00,920 --> 00:29:03,940 of the angular momentum of a particle with respect 473 00:29:03,940 --> 00:29:07,166 to a fixed point. 474 00:29:07,166 --> 00:29:08,540 We're going to come back to that. 475 00:29:08,540 --> 00:29:10,670 That'll turn out to have some significance. 476 00:29:10,670 --> 00:29:17,960 The definition of this is it's RBO, the position vector, 477 00:29:17,960 --> 00:29:25,535 crossed with the linear momentum evaluated in the fixed frame. 478 00:29:25,535 --> 00:29:27,660 So that's the definition of angular momentum, which 479 00:29:27,660 --> 00:29:29,470 is you have linear momentum, and it's 480 00:29:29,470 --> 00:29:32,810 the cross product with the position vector out to it. 481 00:29:32,810 --> 00:29:38,845 So in this case, the RBO is capital R R hat. 482 00:29:44,560 --> 00:29:47,300 Well, it's not-- this varies, excuse me. 483 00:29:47,300 --> 00:29:50,750 So I'm not going to use-- this I use as a constant. 484 00:29:50,750 --> 00:29:53,090 I better keep it as the variable. 485 00:29:53,090 --> 00:29:57,140 Pardon that, so it's r, whatever the local position is, 486 00:29:57,140 --> 00:30:01,970 in the r hat direction crossed with the linear momentum. 487 00:30:01,970 --> 00:30:06,270 What's the linear momentum of that particle? 488 00:30:06,270 --> 00:30:09,050 Mass times velocity. 489 00:30:09,050 --> 00:30:09,983 What's the velocity? 490 00:30:09,983 --> 00:30:10,858 STUDENT: [INAUDIBLE]. 491 00:30:17,182 --> 00:30:19,390 PROFESSOR: Theta dot-- where did we write it up here? 492 00:30:19,390 --> 00:30:31,790 Someplace-- the total velocity is r dot r hat plus r theta 493 00:30:31,790 --> 00:30:33,300 dot theta hat. 494 00:30:36,780 --> 00:30:38,640 And we need a mass in here. 495 00:30:38,640 --> 00:30:44,530 So mass times velocity would be the momentum. 496 00:30:44,530 --> 00:30:46,670 And we need the cross products of those. 497 00:30:46,670 --> 00:30:49,310 r hat cross r hat, you get nothing from that. 498 00:30:49,310 --> 00:30:56,500 And r hat cross theta hat-- positive or negative? 499 00:30:56,500 --> 00:30:59,040 Positive k, right? 500 00:30:59,040 --> 00:31:13,280 So this becomes mr squared theta dot in the k hat direction. 501 00:31:19,200 --> 00:31:21,685 And this one is a constant. 502 00:31:21,685 --> 00:31:23,491 We'll write this as cap omega. 503 00:31:27,340 --> 00:31:32,970 So this is my angular momentum of my particle with respect 504 00:31:32,970 --> 00:31:35,480 to this fixed reference frame. 505 00:31:35,480 --> 00:31:38,354 And so one final step that you know 506 00:31:38,354 --> 00:31:43,220 from your previous physics is-- how's torque 507 00:31:43,220 --> 00:31:44,400 related to angular momentum? 508 00:31:49,710 --> 00:31:50,505 Take a guess. 509 00:31:50,505 --> 00:31:52,600 What do you remember? 510 00:31:52,600 --> 00:32:01,870 Time derivative-- so d by dt of hBO here. 511 00:32:01,870 --> 00:32:04,380 The time rate of change of this vector 512 00:32:04,380 --> 00:32:11,190 is the torque with respect to this point, with respect to O. 513 00:32:11,190 --> 00:32:12,920 So what are constants in this thing? 514 00:32:12,920 --> 00:32:15,590 We don't have to deal with their derivatives. 515 00:32:15,590 --> 00:32:22,640 k hat is-- does the direction of the angular momentum change? 516 00:32:22,640 --> 00:32:23,960 It's upwards. 517 00:32:23,960 --> 00:32:26,290 The derivative of k, the unit vector k, 518 00:32:26,290 --> 00:32:28,430 does it change in this problem? 519 00:32:28,430 --> 00:32:30,930 No, so its time derivative is 0. 520 00:32:30,930 --> 00:32:32,490 Its time derivative is 0. 521 00:32:32,490 --> 00:32:33,999 The m time derivative is 0. 522 00:32:33,999 --> 00:32:36,040 The only thing you have to take the derivative of 523 00:32:36,040 --> 00:32:37,605 is r, so 2rr dot. 524 00:32:54,350 --> 00:32:59,380 So the torque is 2mrr dot cap omega. 525 00:32:59,380 --> 00:33:01,410 And it's all in the k direction. 526 00:33:01,410 --> 00:33:11,750 And that had better be rBO cross F. Torque comes 527 00:33:11,750 --> 00:33:14,290 as a force times a moment arm, right? 528 00:33:14,290 --> 00:33:18,000 And we computed here a force in the theta direction. 529 00:33:18,000 --> 00:33:20,610 So what would give us a torque in the k direction? 530 00:33:20,610 --> 00:33:23,090 An r cross a theta. 531 00:33:23,090 --> 00:33:26,440 r hat cross theta hat gives you a k. 532 00:33:26,440 --> 00:33:29,800 r cross-- and this force, the part of the force. 533 00:33:29,800 --> 00:33:32,450 We have two forces, one in the r direction 534 00:33:32,450 --> 00:33:34,470 and one in the theta direction. 535 00:33:34,470 --> 00:33:38,540 The cross product, this term up here, the T gives you nothing, 536 00:33:38,540 --> 00:33:41,620 r cross r hat cross r hat. 537 00:33:41,620 --> 00:33:46,420 So the only force that matters here is this Coriolis one. 538 00:33:46,420 --> 00:33:55,800 And so that force is 2mr dot theta dot. 539 00:33:55,800 --> 00:33:59,280 And we have the r cross. 540 00:33:59,280 --> 00:34:01,800 We know this comes out in the k direction. 541 00:34:01,800 --> 00:34:05,150 And we get an r in here when we compute this product. 542 00:34:05,150 --> 00:34:09,880 And this, to me, looks an awful lot like this. 543 00:34:09,880 --> 00:34:11,719 Except I missed an r squared. 544 00:34:11,719 --> 00:34:12,469 How did I do that? 545 00:34:19,049 --> 00:34:19,924 STUDENT: [INAUDIBLE]. 546 00:34:25,900 --> 00:34:27,460 PROFESSOR: Oh no, it's not r squared. 547 00:34:27,460 --> 00:34:29,650 It started off over here as the angular momentum has an r 548 00:34:29,650 --> 00:34:30,858 squared, took the derivative. 549 00:34:30,858 --> 00:34:33,109 It dropped down to 2rr dot. 550 00:34:33,109 --> 00:34:36,150 And that's right. 551 00:34:36,150 --> 00:34:39,650 So the Coriolis force, what's it got 552 00:34:39,650 --> 00:34:41,489 to do with the angular momentum? 553 00:34:46,460 --> 00:34:48,520 That's kind of the point of the exercise here. 554 00:34:48,520 --> 00:34:57,700 In order for this ping pong ball to be accelerated, 555 00:34:57,700 --> 00:35:00,760 as it goes slowly out this tube, the angular momentum 556 00:35:00,760 --> 00:35:02,630 is increasing. 557 00:35:02,630 --> 00:35:05,030 In order to change angular momentum with time, 558 00:35:05,030 --> 00:35:06,910 you have to apply a torque. 559 00:35:06,910 --> 00:35:09,060 The torque that you have to apply 560 00:35:09,060 --> 00:35:12,700 is 2mrr dot in the k direction. 561 00:35:12,700 --> 00:35:16,670 And that is r cross the Coriolis force. 562 00:35:16,670 --> 00:35:19,240 So the Coriolis force, in this case, 563 00:35:19,240 --> 00:35:21,990 is the force that's necessary to increase the angular 564 00:35:21,990 --> 00:35:24,630 momentum of a system. 565 00:35:24,630 --> 00:35:26,760 That's very often the reason-- that's 566 00:35:26,760 --> 00:35:30,110 what Coriolis force is about. 567 00:35:30,110 --> 00:35:36,100 So when you shoot an artillery piece on the Earth, 568 00:35:36,100 --> 00:35:38,410 you've got that projectile going out there. 569 00:35:38,410 --> 00:35:41,760 It has angular momentum with respect to the Earth. 570 00:35:41,760 --> 00:35:44,960 And you'll find out that this little term pops up. 571 00:35:44,960 --> 00:35:47,920 And in fact, the projectile doesn't go straight. 572 00:35:47,920 --> 00:35:48,590 It curves. 573 00:35:51,550 --> 00:35:54,020 There's lots of things that because of conservation 574 00:35:54,020 --> 00:35:59,810 of angular momentum, you end up with this term popping up. 575 00:35:59,810 --> 00:36:01,810 In this case, angular momentum is not conserved. 576 00:36:01,810 --> 00:36:02,750 It's increasing. 577 00:36:02,750 --> 00:36:04,860 So you have this force to make it happen. 578 00:36:04,860 --> 00:36:06,260 Any questions about this? 579 00:36:06,260 --> 00:36:08,090 You're going to use this one a lot. 580 00:36:08,090 --> 00:36:09,548 You're going to work with it a lot. 581 00:36:17,200 --> 00:36:20,390 So anytime you see changes in angular momentum happening 582 00:36:20,390 --> 00:36:25,990 in a problem, in these problems with circular motion, 583 00:36:25,990 --> 00:36:30,060 velocities of parts increasing in radius, 584 00:36:30,060 --> 00:36:32,860 you'll almost always see this term pop up. 585 00:36:32,860 --> 00:36:35,200 Any time you see these changes in angular momentum, 586 00:36:35,200 --> 00:36:38,030 you'll often see the Coriolis term. 587 00:36:38,030 --> 00:36:46,070 All right, now we're going to do another interesting problem-- 588 00:36:46,070 --> 00:36:48,040 simple but interesting. 589 00:36:48,040 --> 00:36:52,790 And that is really to go-- let's go do this problem 590 00:36:52,790 --> 00:36:55,170 where the thing is really allowed to come freely out. 591 00:37:02,079 --> 00:37:03,745 All right, you ready to defend yourself? 592 00:37:12,050 --> 00:37:15,040 A little short stick is pretty effective at throwing candy. 593 00:37:19,640 --> 00:37:20,950 You got your safety glasses on? 594 00:37:20,950 --> 00:37:24,400 You want me to see if I can get one up there? 595 00:37:24,400 --> 00:37:27,600 Aww, also, I obviously haven't practiced this. 596 00:37:31,210 --> 00:37:32,550 All right, last one. 597 00:37:38,380 --> 00:37:40,840 Actually, there's two more. 598 00:37:48,400 --> 00:37:51,300 What makes this work? 599 00:37:51,300 --> 00:37:52,880 Let's do this problem. 600 00:37:56,490 --> 00:37:58,505 So let's look at our candy shooter. 601 00:38:17,870 --> 00:38:20,460 So I'm whipping this thing around. 602 00:38:20,460 --> 00:38:21,690 Candy is coming out of here. 603 00:38:21,690 --> 00:38:24,670 It's at B. 604 00:38:24,670 --> 00:38:28,970 Again, I get no z, not mess with the z part. 605 00:38:28,970 --> 00:38:31,170 The z part is trivial to usually deal with. 606 00:38:31,170 --> 00:38:33,636 Because it's totally independent, 607 00:38:33,636 --> 00:38:34,635 just separate equations. 608 00:38:34,635 --> 00:38:36,630 It doesn't complicate things much at all, 609 00:38:36,630 --> 00:38:38,542 even when you have z. 610 00:38:38,542 --> 00:38:42,340 Now, in this case, the velocity-- and here's 611 00:38:42,340 --> 00:38:45,230 my O frame here. 612 00:38:45,230 --> 00:38:47,110 The velocity B with respect to O, 613 00:38:47,110 --> 00:38:52,520 well, now it's got-- I'm not going to move. 614 00:38:52,520 --> 00:38:55,080 I'm standing still when I do it. 615 00:38:55,080 --> 00:38:57,660 So the first term is 0. 616 00:38:57,660 --> 00:39:02,350 It's got an r term, r dot in the r hat direction. 617 00:39:02,350 --> 00:39:08,570 And it has an r theta dot theta hat term. 618 00:39:13,810 --> 00:39:17,330 And these can now-- this might be changing. 619 00:39:17,330 --> 00:39:19,380 This might have a time derivative, r double dot. 620 00:39:19,380 --> 00:39:20,130 It certainly does. 621 00:39:20,130 --> 00:39:22,000 It's accelerating coming out of that tube. 622 00:39:22,000 --> 00:39:25,670 And my theta, angular motion, it can be accelerating, too. 623 00:39:25,670 --> 00:39:28,985 So we're going to have to deal with those. 624 00:39:28,985 --> 00:39:30,360 I need to know the accelerations. 625 00:39:36,100 --> 00:39:39,060 Now, I haven't emphasized it till now, 626 00:39:39,060 --> 00:39:42,990 but I find it conceptually useful to-- when 627 00:39:42,990 --> 00:39:45,750 you work with polar coordinates, you 628 00:39:45,750 --> 00:39:48,810 can have this ability to aggregate 629 00:39:48,810 --> 00:39:57,910 the terms in these two component directions. 630 00:39:57,910 --> 00:39:59,240 So you have this. 631 00:39:59,240 --> 00:40:01,640 All the r terms go together. 632 00:40:01,640 --> 00:40:06,870 And we've let the z double dot term-- 633 00:40:06,870 --> 00:40:08,270 it's just separate by itself. 634 00:40:08,270 --> 00:40:10,380 It drops out easily. 635 00:40:10,380 --> 00:40:12,040 And you have the theta hat term. 636 00:40:12,040 --> 00:40:25,890 And that's the r theta double dot plus 2r dot theta dot. 637 00:40:25,890 --> 00:40:27,900 And these are in the theta hat direction. 638 00:40:27,900 --> 00:40:30,530 There's your Coriolis term, your Euler acceleration 639 00:40:30,530 --> 00:40:33,500 term, centripetal term. 640 00:40:33,500 --> 00:40:34,039 Yeah? 641 00:40:34,039 --> 00:40:34,914 STUDENT: [INAUDIBLE]. 642 00:40:38,964 --> 00:40:40,630 PROFESSOR: Well, I don't know if I ought 643 00:40:40,630 --> 00:40:42,692 to tell you secrets about me. 644 00:40:42,692 --> 00:40:45,025 Because it's going to give you an advantage on the quiz. 645 00:40:48,300 --> 00:40:50,840 But I've almost never, ever been known 646 00:40:50,840 --> 00:40:55,630 to ask a question that says, "derive." 647 00:40:55,630 --> 00:40:59,920 But I'll sure ask you concept questions. 648 00:40:59,920 --> 00:41:01,930 I really want you to understand the principles. 649 00:41:01,930 --> 00:41:04,625 I don't get real hung up on having you do 650 00:41:04,625 --> 00:41:07,029 the grungy grind it out things. 651 00:41:07,029 --> 00:41:08,570 Do I want you to remember the formula 652 00:41:08,570 --> 00:41:13,130 for how to take the derivative of a vector in rotating frame? 653 00:41:13,130 --> 00:41:15,150 Yeah, that's where these have come from. 654 00:41:18,250 --> 00:41:21,830 You had better remember this. 655 00:41:21,830 --> 00:41:24,850 These two formulas, the velocity formula and this, 656 00:41:24,850 --> 00:41:27,820 the acceleration formula, are just core to this course. 657 00:41:27,820 --> 00:41:29,670 Now, the way quizzes are done-- first quiz, 658 00:41:29,670 --> 00:41:31,900 you come in, one sheet of paper. 659 00:41:31,900 --> 00:41:35,580 What had better be on your paper? 660 00:41:35,580 --> 00:41:37,030 OK? 661 00:41:37,030 --> 00:41:39,470 And second quiz, two sheets, final, three sheets, 662 00:41:39,470 --> 00:41:40,870 that kind of thing. 663 00:41:40,870 --> 00:41:43,290 But conceptually, you've got forces in the r, 664 00:41:43,290 --> 00:41:47,000 forces in the z, accelerations in r theta 665 00:41:47,000 --> 00:41:49,455 and z, forces r theta and z. 666 00:41:52,210 --> 00:41:53,890 And for these planar motion problems, 667 00:41:53,890 --> 00:41:57,190 this one is sure easy to use. 668 00:41:57,190 --> 00:42:00,980 So let's think about this problem. 669 00:42:00,980 --> 00:42:06,650 I'm going to let this be frictionless just 670 00:42:06,650 --> 00:42:09,410 to make it easy. 671 00:42:09,410 --> 00:42:12,130 All right, so what possible forces act 672 00:42:12,130 --> 00:42:14,870 on the hunk of candy? 673 00:42:14,870 --> 00:42:17,210 Let's do a free body diagram of the hunk of candy 674 00:42:17,210 --> 00:42:19,680 coming out of here. 675 00:42:19,680 --> 00:42:22,010 What are the forces? 676 00:42:22,010 --> 00:42:23,640 And let's keep it planar. 677 00:42:23,640 --> 00:42:25,080 We can get gravity into this. 678 00:42:25,080 --> 00:42:26,770 But let's just do it in a plane. 679 00:42:26,770 --> 00:42:29,500 So I'm just going horizontal and slinging this thing. 680 00:42:29,500 --> 00:42:31,160 Gravity's in the z direction, and I've 681 00:42:31,160 --> 00:42:32,430 constrained it in the z. 682 00:42:32,430 --> 00:42:37,140 So it's something supporting the gravity in the tube. 683 00:42:37,140 --> 00:42:40,540 So definitely there's an mg on this thing downwards. 684 00:42:40,540 --> 00:42:42,990 But it's in the z direction, and we're not 685 00:42:42,990 --> 00:42:45,200 letting it move in the z. 686 00:42:45,200 --> 00:42:49,290 What about the horizontal, this direction? 687 00:42:49,290 --> 00:42:53,750 The r-- this is my r hat direction. 688 00:42:53,750 --> 00:42:56,960 This is my theta hat direction. 689 00:42:56,960 --> 00:42:59,516 Whoops, not either-- theta is going this way. 690 00:42:59,516 --> 00:43:03,260 So I've drawn this as a side view. 691 00:43:03,260 --> 00:43:06,430 What's the force in the r hat direction? 692 00:43:06,430 --> 00:43:07,305 STUDENT: [INAUDIBLE]. 693 00:43:10,600 --> 00:43:12,350 PROFESSOR: OK, in order to do the problem, 694 00:43:12,350 --> 00:43:14,980 you have to figure that out. 695 00:43:14,980 --> 00:43:15,850 What are the forces? 696 00:43:19,030 --> 00:43:22,150 What are the source of forces in the r direction? 697 00:43:22,150 --> 00:43:24,016 Here's the r direction. 698 00:43:24,016 --> 00:43:24,890 STUDENT: [INAUDIBLE]. 699 00:43:24,890 --> 00:43:25,723 PROFESSOR: Say what? 700 00:43:28,230 --> 00:43:29,140 Come on, you guys. 701 00:43:29,140 --> 00:43:31,158 Somebody be-- yeah. 702 00:43:31,158 --> 00:43:33,650 STUDENT: [INAUDIBLE]. 703 00:43:33,650 --> 00:43:34,860 PROFESSOR: There are not any. 704 00:43:34,860 --> 00:43:35,776 Is that what you said? 705 00:43:35,776 --> 00:43:36,710 There aren't any. 706 00:43:36,710 --> 00:43:38,025 And why's that? 707 00:43:38,025 --> 00:43:38,900 STUDENT: [INAUDIBLE]. 708 00:43:38,900 --> 00:43:41,780 PROFESSOR: Right, so there's no forces in the r direction. 709 00:43:41,780 --> 00:43:44,980 So there's no forces on this thing in the r. 710 00:43:44,980 --> 00:43:48,260 And so then this is a side view. 711 00:43:48,260 --> 00:43:50,260 We could do the top view looking down. 712 00:43:50,260 --> 00:43:53,260 Top view, you've got your x, y. 713 00:43:53,260 --> 00:43:56,580 You also have your-- here's the ball. 714 00:43:56,580 --> 00:43:58,310 Here's the r hat. 715 00:43:58,310 --> 00:44:00,040 Here's the theta hat. 716 00:44:00,040 --> 00:44:04,800 Now, free body diagram in the top view-- 717 00:44:04,800 --> 00:44:10,937 well, there's some force here probably. 718 00:44:10,937 --> 00:44:14,290 STUDENT: I was going to ask that the fact that we 719 00:44:14,290 --> 00:44:16,685 have no forces in the r hat direction, but we do 720 00:44:16,685 --> 00:44:18,529 have acceleration. 721 00:44:18,529 --> 00:44:19,445 PROFESSOR: Absolutely. 722 00:44:19,445 --> 00:44:22,030 She's commenting that we do have accelerations 723 00:44:22,030 --> 00:44:23,240 in the r direction, right? 724 00:44:23,240 --> 00:44:24,490 STUDENT: And we have no force. 725 00:44:24,490 --> 00:44:25,910 PROFESSOR: And no force. 726 00:44:25,910 --> 00:44:28,765 So that's the conundrum of this problem. 727 00:44:28,765 --> 00:44:30,140 That's the point of this problem. 728 00:44:30,140 --> 00:44:30,931 So let me continue. 729 00:44:34,070 --> 00:44:36,830 Free body diagram-- I'm looking down on it. 730 00:44:36,830 --> 00:44:38,080 I'm allowing for some force. 731 00:44:38,080 --> 00:44:42,590 It's the normal force that comes from the pipe exerting 732 00:44:42,590 --> 00:44:44,420 the force on the candy. 733 00:44:44,420 --> 00:44:47,410 And since it's frictionless, it can only be normal to the pipe. 734 00:44:47,410 --> 00:44:54,270 OK, so there's the free body diagram in the top view. 735 00:44:54,270 --> 00:44:56,466 So we can write two equations. 736 00:44:56,466 --> 00:44:58,510 We can write three questions-- this one 737 00:44:58,510 --> 00:45:01,286 equal to 0, this one in the r hat direction, 738 00:45:01,286 --> 00:45:02,785 this one in the theta hat direction. 739 00:45:22,600 --> 00:45:27,350 Sum of the forces r hat direction must be 0. 740 00:45:27,350 --> 00:45:28,667 And we have a mass. 741 00:45:28,667 --> 00:45:29,666 We have an acceleration. 742 00:45:40,170 --> 00:45:42,120 Solve for r double dot. 743 00:45:50,860 --> 00:45:58,930 Remarkable-- there's no force in the r hat direction. 744 00:45:58,930 --> 00:46:05,220 The position of the object, the r 745 00:46:05,220 --> 00:46:08,490 coordinate, it has a velocity. 746 00:46:08,490 --> 00:46:10,250 It has an acceleration. 747 00:46:14,380 --> 00:46:18,045 But the total acceleration in the r hat directions 748 00:46:18,045 --> 00:46:19,590 are actually 0. 749 00:46:19,590 --> 00:46:25,790 The rate of change of the velocity 750 00:46:25,790 --> 00:46:28,860 of this thing in the radial direction, r double dot, 751 00:46:28,860 --> 00:46:30,870 is nonzero. 752 00:46:30,870 --> 00:46:32,151 But there are no forces on it. 753 00:46:37,340 --> 00:46:42,150 I have pondered another way to explain this. 754 00:46:42,150 --> 00:46:43,390 I'm still thinking about it. 755 00:46:43,390 --> 00:46:45,010 And you think about this, too. 756 00:46:45,010 --> 00:46:48,300 How do you explain this in the absence of forces? 757 00:46:48,300 --> 00:46:49,960 And it's partly where the concept 758 00:46:49,960 --> 00:46:53,760 comes from of fictitious forces, that centrifugal force is 759 00:46:53,760 --> 00:46:54,580 a force. 760 00:46:54,580 --> 00:46:55,140 It's not. 761 00:46:55,140 --> 00:46:56,600 It's an acceleration. 762 00:46:56,600 --> 00:47:01,330 This is just saying that-- let's go back to this problem. 763 00:47:01,330 --> 00:47:05,170 In order to make something go in a circle, 764 00:47:05,170 --> 00:47:08,200 you have to put a force on it to cause 765 00:47:08,200 --> 00:47:11,280 the centripetal acceleration. 766 00:47:11,280 --> 00:47:16,090 You have to allow the thing to go out if you're not 767 00:47:16,090 --> 00:47:18,190 forcing it to go in a circle. 768 00:47:18,190 --> 00:47:20,940 You have to allow it to go out at that rate in order 769 00:47:20,940 --> 00:47:25,020 for there to be no centripetal accelerations on the object. 770 00:47:25,020 --> 00:47:25,521 Yeah? 771 00:47:25,521 --> 00:47:27,603 STUDENT: So I was going to say, when you start it, 772 00:47:27,603 --> 00:47:29,026 you push it in the y direction. 773 00:47:29,026 --> 00:47:29,930 So that's a force. 774 00:47:29,930 --> 00:47:32,620 It's not in the r hat, but it's in the y. 775 00:47:32,620 --> 00:47:33,830 PROFESSOR: To get it started. 776 00:47:33,830 --> 00:47:35,958 STUDENT: Right, and there's no force opposing it 777 00:47:35,958 --> 00:47:38,090 in that direction [INAUDIBLE]. 778 00:47:38,090 --> 00:47:45,990 PROFESSOR: Does it experience centripetal acceleration? 779 00:47:45,990 --> 00:47:47,550 So there's no rotation. 780 00:47:47,550 --> 00:47:50,125 Does it experience centripetal acceleration? 781 00:47:53,100 --> 00:47:54,560 What do you think? 782 00:47:54,560 --> 00:47:58,450 Yeah, because it goes through curved motion. 783 00:47:58,450 --> 00:48:00,360 At any instance in time when it's doing that, 784 00:48:00,360 --> 00:48:02,279 there is a radius of curvature. 785 00:48:02,279 --> 00:48:04,070 You can at that instant in time think of it 786 00:48:04,070 --> 00:48:07,050 as being in a circular path. 787 00:48:07,050 --> 00:48:09,840 And sometimes it's just really easy 788 00:48:09,840 --> 00:48:13,510 to do these kind of problems with normal and tangential 789 00:48:13,510 --> 00:48:14,970 coordinates. 790 00:48:14,970 --> 00:48:23,670 So I'm looking down on the x, y plane. 791 00:48:23,670 --> 00:48:25,330 Gravity is into the Earth. 792 00:48:25,330 --> 00:48:27,790 So I'm looking down on a vehicle, a car. 793 00:48:27,790 --> 00:48:31,940 And that car, the guy is kind of drunk. 794 00:48:31,940 --> 00:48:35,100 He's going down the road like this. 795 00:48:35,100 --> 00:48:36,440 So here's my y. 796 00:48:36,440 --> 00:48:37,770 Here's my x. 797 00:48:37,770 --> 00:48:49,270 And y equals some A sine 2 pi over the wavelength, 798 00:48:49,270 --> 00:48:58,150 2 pi over lambda, times x at some A sine kx. 799 00:48:58,150 --> 00:49:00,380 And as he drives down-- if you're in a car, 800 00:49:00,380 --> 00:49:03,490 and you're doing that, you get thrown side to side in the car. 801 00:49:03,490 --> 00:49:04,997 So you are being accelerated. 802 00:49:04,997 --> 00:49:06,580 And so we want to be able to calculate 803 00:49:06,580 --> 00:49:09,850 the acceleration due to the fact that you're going 804 00:49:09,850 --> 00:49:12,480 down and doing a curved path. 805 00:49:12,480 --> 00:49:16,390 And we deal with these things sometimes 806 00:49:16,390 --> 00:49:19,260 with a convenient little set of coordinates 807 00:49:19,260 --> 00:49:23,810 that are our normal and tangential unit 808 00:49:23,810 --> 00:49:30,140 vectors, u normal and u tangential, 809 00:49:30,140 --> 00:49:31,350 at any instant in time. 810 00:49:34,610 --> 00:49:38,990 And we know that if this is along the path, 811 00:49:38,990 --> 00:49:41,700 at any instant in time you're right here, 812 00:49:41,700 --> 00:49:45,070 what direction is your velocity? 813 00:49:45,070 --> 00:49:52,170 Just definition of velocity-- tangent to the path, right? 814 00:49:52,170 --> 00:49:54,170 So at any instant in time, the velocity 815 00:49:54,170 --> 00:49:58,000 has got to be tangent to the path at that moment. 816 00:49:58,000 --> 00:50:09,170 So velocity, the vector, has a magnitude and a unit vector 817 00:50:09,170 --> 00:50:12,606 uT here I'll call it, tangent. 818 00:50:12,606 --> 00:50:15,020 That's all there is to it. 819 00:50:15,020 --> 00:50:20,320 And the acceleration of this thing 820 00:50:20,320 --> 00:50:22,435 is your time derivative of this. 821 00:50:26,330 --> 00:50:34,000 And that's going to give you a v dot uT hat plus a v. 822 00:50:34,000 --> 00:50:39,290 And now you need a time derivative of this guy. 823 00:50:39,290 --> 00:50:41,904 But this is a unit length vector. 824 00:50:41,904 --> 00:50:43,320 You can plug it into that equation 825 00:50:43,320 --> 00:50:46,700 for the derivative of a rotating vector 826 00:50:46,700 --> 00:50:48,440 and calculate what this should be. 827 00:50:48,440 --> 00:50:50,210 You could also just draw it out. 828 00:50:50,210 --> 00:50:53,060 So I'll draw this for you. 829 00:50:58,880 --> 00:51:01,750 How are we doing on time? 830 00:51:01,750 --> 00:51:04,070 I should just be able to finish this. 831 00:51:11,500 --> 00:51:16,500 Here's my unit vector in the tangential direction. 832 00:51:16,500 --> 00:51:23,320 As I'm going around this curve, this 833 00:51:23,320 --> 00:51:27,650 is my tangential direction. 834 00:51:27,650 --> 00:51:29,280 There's some instant I have a radius. 835 00:51:29,280 --> 00:51:31,050 We call that rho. 836 00:51:31,050 --> 00:51:33,655 And in the little time, delta t, I 837 00:51:33,655 --> 00:51:38,300 go through an angle delta theta in delta t. 838 00:51:38,300 --> 00:51:42,070 And this is my uT vector here. 839 00:51:42,070 --> 00:51:46,120 It changes by a little bit. 840 00:51:46,120 --> 00:51:53,290 That's the change in the uT unit vector in this time, delta t. 841 00:51:53,290 --> 00:51:55,870 And it goes perpendicular. 842 00:51:55,870 --> 00:51:58,652 And it goes in the positive un direction. 843 00:52:01,700 --> 00:52:06,050 So delta-- what's the easiest way to write this one? 844 00:52:12,030 --> 00:52:26,600 Delta uT equals some theta dot delta t-- that's 845 00:52:26,600 --> 00:52:35,390 the angle-- times the length of the unit vector, 1. 846 00:52:35,390 --> 00:52:38,410 That's the distance it goes, so 1. 847 00:52:38,410 --> 00:52:42,990 So r omega, 1 theta dot is the distance 848 00:52:42,990 --> 00:52:48,500 that this unit vector goes through in delta t. 849 00:52:48,500 --> 00:52:54,940 And the direction it goes in is u normal hat. 850 00:52:54,940 --> 00:53:01,820 So delta uT over delta t limit as t 851 00:53:01,820 --> 00:53:11,280 goes to 0, you get theta dot un, just like before. 852 00:53:11,280 --> 00:53:13,700 So the time derivative of this unit vector 853 00:53:13,700 --> 00:53:17,080 in the tangential direction is just theta dot 854 00:53:17,080 --> 00:53:19,770 in the normal direction. 855 00:53:19,770 --> 00:53:24,090 And then from that, we can very quickly 856 00:53:24,090 --> 00:53:27,630 derive the rest of this acceleration. 857 00:53:27,630 --> 00:53:30,820 The acceleration then is that plus this. 858 00:53:30,820 --> 00:53:41,119 We now know an expression for-- this is my uT dot term. 859 00:53:41,119 --> 00:53:42,410 I'm going to plug that in here. 860 00:53:42,410 --> 00:53:45,510 This then, we need an expression for v. What's v? 861 00:53:48,770 --> 00:53:52,000 Well, at that instant in time, it has some radius rho. 862 00:53:52,000 --> 00:53:55,100 It has an angular velocity theta dot. 863 00:53:55,100 --> 00:53:57,930 So rho theta dot would be the v here. 864 00:54:07,980 --> 00:54:13,240 So I'm looking for an expression for vuT dot. 865 00:54:18,300 --> 00:54:24,928 So that's v theta dot un. 866 00:54:24,928 --> 00:54:28,550 But theta dot is v over rho. 867 00:54:31,150 --> 00:54:37,210 v squared-- sorry about this-- over rho un. 868 00:54:37,210 --> 00:54:45,160 So this guy up here, this is v dot uT plus v squared-- 869 00:54:45,160 --> 00:54:47,150 I'll rewrite it-- over rho un. 870 00:54:50,082 --> 00:54:51,540 So if you're speeding up, if you're 871 00:54:51,540 --> 00:54:54,250 going from 30 miles an hour to 40 miles an hour, 872 00:54:54,250 --> 00:54:55,520 that's your tangential. 873 00:54:55,520 --> 00:54:57,360 That's your speed along the path. 874 00:54:57,360 --> 00:54:58,901 That's this term. 875 00:54:58,901 --> 00:55:00,650 But because you're going around the curve, 876 00:55:00,650 --> 00:55:02,720 you have an acceleration of v squared over rho. 877 00:55:02,720 --> 00:55:05,010 You've run into this before in physics. 878 00:55:05,010 --> 00:55:07,760 This is where it comes from. 879 00:55:07,760 --> 00:55:12,490 This is a centripetal acceleration like term. 880 00:55:12,490 --> 00:55:15,210 If you replace v with rho theta dot, 881 00:55:15,210 --> 00:55:17,557 you'd get r theta dot squared. 882 00:55:17,557 --> 00:55:19,890 So you can either put it in terms of v squared over rho, 883 00:55:19,890 --> 00:55:22,820 or you can put it in terms of rho theta dot squared. 884 00:55:22,820 --> 00:55:28,350 And rho theta dot squared sounds a lot like my acceleration term 885 00:55:28,350 --> 00:55:28,890 right here. 886 00:55:32,280 --> 00:55:35,500 So with that simple little formula, 887 00:55:35,500 --> 00:55:38,580 you can do-- you need one other thing. 888 00:55:41,160 --> 00:55:44,960 And you just go look it up in the book. 889 00:55:44,960 --> 00:55:48,680 There is an expression from calculus for the radius 890 00:55:48,680 --> 00:55:52,570 of curvature of a path. 891 00:55:52,570 --> 00:55:56,480 And it has first and second derivatives of y with respect 892 00:55:56,480 --> 00:55:56,980 to x. 893 00:55:56,980 --> 00:56:00,710 So you need a dy/dx and a d2y dx squared. 894 00:56:00,710 --> 00:56:03,650 From a sine function, you can calculate that. 895 00:56:03,650 --> 00:56:05,960 So you calculate rho from a formula 896 00:56:05,960 --> 00:56:07,910 that's in the book for calculating 897 00:56:07,910 --> 00:56:09,240 radius of curvature. 898 00:56:09,240 --> 00:56:11,100 And then you're done, all right? 899 00:56:11,100 --> 00:56:15,520 So see you on Thursday.