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 --> 00:00:17,297 at ocw.mit.edu. 8 00:00:22,070 --> 00:00:24,590 PROFESSOR: The technical topic for today, 9 00:00:24,590 --> 00:00:26,210 we start-- I rushed, right at the end, 10 00:00:26,210 --> 00:00:28,380 a little bit about tangent and normal unit vectors. 11 00:00:28,380 --> 00:00:31,120 I'm going to just recap that quickly. 12 00:00:31,120 --> 00:00:35,070 And then we're going to go on really a review, which 13 00:00:35,070 --> 00:00:37,790 by a review, this is the sort of stuff that, for the most part, 14 00:00:37,790 --> 00:00:41,760 I'm sure you've seen in 801 Physics and other Physics 15 00:00:41,760 --> 00:00:42,550 you've had before. 16 00:00:42,550 --> 00:00:45,650 And that's impulse-- linear momentum, and impulse. 17 00:00:45,650 --> 00:00:47,580 So that'll be a quick review. 18 00:00:47,580 --> 00:00:51,280 And then the third subject is one that's much deeper, 19 00:00:51,280 --> 00:00:53,160 and this is angular momentum. 20 00:00:53,160 --> 00:00:55,530 And angular momentum with respect 21 00:00:55,530 --> 00:00:57,210 to moving points, which you probably 22 00:00:57,210 --> 00:00:59,290 haven't encountered before. 23 00:00:59,290 --> 00:01:02,130 So those are the three topics for the day. 24 00:01:02,130 --> 00:01:03,160 Let's get started. 25 00:01:06,070 --> 00:01:12,000 So last time, the piece that I rushed 26 00:01:12,000 --> 00:01:19,105 a bit is this notion of tangent and normal coordinates. 27 00:01:22,400 --> 00:01:24,180 So I gave this-- had this example. 28 00:01:24,180 --> 00:01:32,856 You're driving down the road, you're drunk or whatever. 29 00:01:35,770 --> 00:01:39,100 And when you're at this point, this point, and this point, 30 00:01:39,100 --> 00:01:41,580 I'd like to know what the accelerations are. 31 00:01:41,580 --> 00:01:43,840 So I'll call this 1, 2, 3. 32 00:01:43,840 --> 00:01:50,225 And this curve, y, is of the form some y of some f of x. 33 00:01:52,730 --> 00:01:59,600 And in this case, it's A sine kx. 34 00:01:59,600 --> 00:02:02,700 And k is what's known as wave number. 35 00:02:02,700 --> 00:02:07,170 This is 2 pi over the wave length, 2 pi over lambda. 36 00:02:07,170 --> 00:02:10,830 And the wave length then is, for example, from here to here. 37 00:02:17,520 --> 00:02:20,000 Now, the velocity-- I'm going to pick a point here. 38 00:02:20,000 --> 00:02:23,270 The velocity at any point, we know, 39 00:02:23,270 --> 00:02:26,267 is just the tangent to the path. 40 00:02:26,267 --> 00:02:28,100 So this is the path, and this is horizontal. 41 00:02:28,100 --> 00:02:32,100 So you're driving-- driving down the road like this. 42 00:02:32,100 --> 00:02:34,700 That's what we're trying to do here. 43 00:02:34,700 --> 00:02:36,390 The gravity's down into the board. 44 00:02:36,390 --> 00:02:38,300 It doesn't really come into the problem. 45 00:02:41,260 --> 00:02:45,930 So the velocity, at any point-- at anytime-- 46 00:02:45,930 --> 00:02:50,860 it's a vector-- we can describe as a magnitude. 47 00:02:50,860 --> 00:02:54,370 And in a unit vector, we'll call ut which 48 00:02:54,370 --> 00:02:56,610 is the tangent unit vector. 49 00:02:56,610 --> 00:02:59,750 And at any instant in time, it's just aligned with the tangent 50 00:02:59,750 --> 00:03:01,980 to the curve. 51 00:03:01,980 --> 00:03:08,730 And it's perpendicular partner is a normal unit vector, which 52 00:03:08,730 --> 00:03:11,450 it points inward on the curve. 53 00:03:11,450 --> 00:03:14,480 And this will be un. 54 00:03:14,480 --> 00:03:16,480 Now, we're interested in accelerations, 55 00:03:16,480 --> 00:03:18,890 so we'll need to take a derivative of this. 56 00:03:27,630 --> 00:03:29,960 So I was going to say, taken aside. 57 00:03:29,960 --> 00:03:30,792 But I won't. 58 00:03:30,792 --> 00:03:32,250 So the acceleration vector, we have 59 00:03:32,250 --> 00:03:33,541 to take the derivative of this. 60 00:03:33,541 --> 00:03:43,500 Well, it's a v dot ut plus a v ut dot. 61 00:03:43,500 --> 00:03:44,530 It's a unit vector. 62 00:03:44,530 --> 00:03:46,080 And we've encountered this problem 63 00:03:46,080 --> 00:03:49,780 before because if this rotates, the unit vector 64 00:03:49,780 --> 00:03:52,885 has a non-zero derivative because of this rotation. 65 00:03:57,020 --> 00:04:03,920 And this derivative of the unit vector 66 00:04:03,920 --> 00:04:13,380 is given by theta dot u m hat Now, what's theta dot? 67 00:04:13,380 --> 00:04:17,490 Well, on this curve, at any instant in time, 68 00:04:17,490 --> 00:04:21,410 here's u-- can't draw arrows today. 69 00:04:21,410 --> 00:04:23,200 Here's your ut. 70 00:04:23,200 --> 00:04:27,970 At any instant when you're traveling along on a curve, 71 00:04:27,970 --> 00:04:32,690 you are going around the circle of some radius we'll call rho. 72 00:04:32,690 --> 00:04:34,720 So there's some radius of curvature 73 00:04:34,720 --> 00:04:37,070 at any instant in time. 74 00:04:37,070 --> 00:04:44,780 And that radius, as you drive along in a little time delta t, 75 00:04:44,780 --> 00:04:51,990 you go forward an amount, rho theta dot, 76 00:04:51,990 --> 00:04:59,820 is the velocity at which you're traveling tangent to the curve. 77 00:04:59,820 --> 00:05:03,420 There's an angle in here, so in here, there's some delta. 78 00:05:03,420 --> 00:05:09,320 You advance some delta theta, so our omega is a velocity, right? 79 00:05:09,320 --> 00:05:12,530 So rho theta dot its velocity, and that gives you 80 00:05:12,530 --> 00:05:20,234 this velocity magnitude, v. And we've 81 00:05:20,234 --> 00:05:21,650 gone through-- I'm not going to go 82 00:05:21,650 --> 00:05:24,191 through that little derivative-- that little argument before, 83 00:05:24,191 --> 00:05:28,260 but the change in direction of this unit vector, 84 00:05:28,260 --> 00:05:35,560 ut, when you go forward a little bit is actually inward, un, 85 00:05:35,560 --> 00:05:42,080 by an amount rho theta dot-- by an amount theta dot one times 86 00:05:42,080 --> 00:05:42,874 theta dot. 87 00:05:42,874 --> 00:05:44,290 So we've done- I'm not going to do 88 00:05:44,290 --> 00:05:46,081 that piece of the derivative we did before. 89 00:05:46,081 --> 00:05:49,340 But here is this derivative of the unit vector. 90 00:05:49,340 --> 00:05:51,780 It's theta dot un. 91 00:05:51,780 --> 00:05:54,070 So if we substitute those back in here, 92 00:05:54,070 --> 00:05:57,625 we can get an expression for our acceleration. 93 00:06:02,600 --> 00:06:11,760 This is just the acceleration along the path plus v theta dot 94 00:06:11,760 --> 00:06:14,420 un. 95 00:06:14,420 --> 00:06:17,550 And then we need to take into account 96 00:06:17,550 --> 00:06:22,280 the fact that we know that v, the magnitude of velocity-- 97 00:06:22,280 --> 00:06:25,650 the speed in other words-- is rho theta dot. 98 00:06:28,300 --> 00:06:34,800 And that means theta dot is v over rho. 99 00:06:34,800 --> 00:06:37,670 So if we plug that in up here, then 100 00:06:37,670 --> 00:06:42,600 we get a final expression for our acceleration, 101 00:06:42,600 --> 00:06:52,610 v dot ut plus v squared over rho un. 102 00:06:52,610 --> 00:06:56,030 And that v squared over rho-- this 103 00:06:56,030 --> 00:06:59,850 is a centripetal acceleration term. 104 00:06:59,850 --> 00:07:01,640 You're going around a curve. 105 00:07:01,640 --> 00:07:04,050 There is an acceleration inward. 106 00:07:04,050 --> 00:07:07,280 It's just like as we did from polar coordinates. 107 00:07:07,280 --> 00:07:09,422 It comes from rotating things. 108 00:07:09,422 --> 00:07:11,130 Well, as soon as you go around the curve, 109 00:07:11,130 --> 00:07:14,030 you're going to generate an acceleration that 110 00:07:14,030 --> 00:07:16,340 is of the same kind, but now we-- 111 00:07:16,340 --> 00:07:19,720 because you tend to know speed, it's 112 00:07:19,720 --> 00:07:21,730 easier to express it this way when you're doing 113 00:07:21,730 --> 00:07:23,760 these tangent normal problems. 114 00:07:23,760 --> 00:07:27,120 And this is just the acceleration along a path, 115 00:07:27,120 --> 00:07:30,190 the usual hit the gas petal and speed up. 116 00:07:30,190 --> 00:07:31,115 That's this term. 117 00:07:34,240 --> 00:07:37,190 So the piece that I didn't have time 118 00:07:37,190 --> 00:07:47,760 to put up last time, which is in the book as 119 00:07:47,760 --> 00:07:50,010 is this little derivative-- the derivation of that 120 00:07:50,010 --> 00:07:53,310 is one page in one of the early chapters of the book. 121 00:07:53,310 --> 00:07:55,700 How do you get rho? 122 00:07:55,700 --> 00:07:58,900 And rho, when you have y as a function of x, 123 00:07:58,900 --> 00:08:03,470 there is just this formula from mathematics-- from calculus-- 124 00:08:03,470 --> 00:08:12,700 that says rho is dy dx 1 plus dy dx quantity squared to the 3/2 125 00:08:12,700 --> 00:08:22,710 power all over the magnitude of d2y dx squared. 126 00:08:22,710 --> 00:08:25,200 And you just calculate these quantities and plug them in. 127 00:08:25,200 --> 00:08:42,850 So for y equals a sine kx, then dy dx is Ak cosine kx 128 00:08:42,850 --> 00:08:56,720 and d2y dx squared is minus a k squared sine kx. 129 00:08:56,720 --> 00:08:59,580 So we now have these two quantities. 130 00:08:59,580 --> 00:09:01,550 We pick a value of x which we want 131 00:09:01,550 --> 00:09:04,720 to know the answer, like 0.1. 132 00:09:04,720 --> 00:09:10,510 So at 0.1, kx is pi over 2 because sine is a maximum. 133 00:09:17,330 --> 00:09:21,290 X is lambda over 4. 134 00:09:21,290 --> 00:09:22,970 kx is pi over 2. 135 00:09:26,370 --> 00:09:30,030 Sine kx is 1. 136 00:09:30,030 --> 00:09:34,400 Cosine kx is 0. 137 00:09:34,400 --> 00:09:36,280 And so we can calculate rho. 138 00:09:39,450 --> 00:09:47,440 1 plus and dy dx the derivative, is cosine. 139 00:09:47,440 --> 00:09:48,170 That's 0. 140 00:09:48,170 --> 00:09:55,560 So this term in here is 0. 141 00:09:55,560 --> 00:09:56,780 And this is to the 3/2. 142 00:09:56,780 --> 00:09:58,470 That's pretty easy to calculate. 143 00:09:58,470 --> 00:10:03,480 This term down here, d2y dx squared, well, sine of kx is 1. 144 00:10:03,480 --> 00:10:08,120 So that's minus Ak squared, but this is an absolute value sign. 145 00:10:12,650 --> 00:10:19,700 So this just turns out to be 1 over Ak squared. 146 00:10:19,700 --> 00:10:22,970 And you plug-in some numbers. 147 00:10:22,970 --> 00:10:31,036 I'm going to let my lambda be 150 meters. 148 00:10:31,036 --> 00:10:34,340 My amplitude here that I'm swerving back and forth, 149 00:10:34,340 --> 00:10:35,650 let's make that 5 meters. 150 00:10:38,790 --> 00:10:44,340 We need a-- that's all we need to get rho. 151 00:10:44,340 --> 00:10:50,110 So rho, in this case, works out to be 114 meters. 152 00:10:50,110 --> 00:10:53,360 So the radius of curvature, this road that I'm driving down, 153 00:10:53,360 --> 00:10:58,900 at that point right at the peak in the curve is 114 meters. 154 00:11:05,710 --> 00:11:08,890 So the acceleration that I'm looking for, 155 00:11:08,890 --> 00:11:13,600 a, well, it's v dot ut. 156 00:11:13,600 --> 00:11:15,330 And if I'm not accelerating-- I'm not 157 00:11:15,330 --> 00:11:17,010 hitting the gas petal at constant speed, 158 00:11:17,010 --> 00:11:19,500 I'll let that would be 0. 159 00:11:19,500 --> 00:11:21,870 And the other one, then, is v squared 160 00:11:21,870 --> 00:11:26,120 over rho of the un direction. 161 00:11:26,120 --> 00:11:32,140 And let's let v equal-- whatever I have here in my example-- 162 00:11:32,140 --> 00:11:38,600 20 meters per second, which is about 40 knots. 163 00:11:38,600 --> 00:11:42,060 And a knot is 15% more than a mile per hour, 164 00:11:42,060 --> 00:11:44,985 so it's somewhere about 45 miles an hour, so 165 00:11:44,985 --> 00:11:46,100 a typical road speed. 166 00:11:46,100 --> 00:11:48,650 Driving down the road, 20 meters per second, 167 00:11:48,650 --> 00:11:49,810 now you can plug-in here. 168 00:11:49,810 --> 00:11:51,360 You can plug-in here, and you'll find 169 00:11:51,360 --> 00:11:54,525 that the acceleration that I worked out here 170 00:11:54,525 --> 00:12:00,120 is 3.51 meters per second squared 171 00:12:00,120 --> 00:12:01,670 and in what direction is it? 172 00:12:07,470 --> 00:12:09,636 What direction is that acceleration? 173 00:12:12,384 --> 00:12:13,300 AUDIENCE: The normal. 174 00:12:13,300 --> 00:12:14,670 PROFESSOR: It's in the normal direction. 175 00:12:14,670 --> 00:12:17,200 Is it to the inside of the curve or the outside of the curve? 176 00:12:17,200 --> 00:12:17,690 AUDIENCE: Inside. 177 00:12:17,690 --> 00:12:18,440 PROFESSOR: Inside. 178 00:12:18,440 --> 00:12:20,266 It's always to the inside of the curve. 179 00:12:20,266 --> 00:12:23,800 So these tangent normal coordinates 180 00:12:23,800 --> 00:12:26,190 are really simple coordinates. 181 00:12:26,190 --> 00:12:28,620 They're meant for a particular kind of simple problem 182 00:12:28,620 --> 00:12:30,650 when you know the path. 183 00:12:30,650 --> 00:12:34,350 And it's just defined, the normal un is positive, 184 00:12:34,350 --> 00:12:36,620 always inward to the center of the curve, 185 00:12:36,620 --> 00:12:38,910 pointing toward the center of rotation 186 00:12:38,910 --> 00:12:41,450 where that radius of-- where your radius of curve 187 00:12:41,450 --> 00:12:44,640 is, always pointing at the origin 188 00:12:44,640 --> 00:12:48,920 of that radius of your curve. 189 00:12:48,920 --> 00:12:52,820 So g is order of 10 meters per second squared, 190 00:12:52,820 --> 00:12:56,265 so that's about a third of a g. 191 00:12:56,265 --> 00:13:00,690 Is 1/3 of a g enough to notice? 192 00:13:00,690 --> 00:13:01,737 Yeah, absolutely! 193 00:13:01,737 --> 00:13:03,820 You'll get-- when you push to the side of the car, 194 00:13:03,820 --> 00:13:04,896 you do that . 195 00:13:08,030 --> 00:13:10,440 So let's move-- any questions about that? 196 00:13:10,440 --> 00:13:11,640 I'm going to move on next. 197 00:13:11,640 --> 00:13:14,435 If not, I'm going to move on to linear impulse and momentum. 198 00:13:23,140 --> 00:13:28,860 And this will be a quick review because this 199 00:13:28,860 --> 00:13:35,104 consists of the kind of physics that you've done lots of times. 200 00:13:35,104 --> 00:13:37,020 We're going to hit it quickly and then move on 201 00:13:37,020 --> 00:13:37,853 to angular momentum. 202 00:13:41,390 --> 00:13:43,880 So we know for a particle-- this is what Newton told us. 203 00:13:43,880 --> 00:13:47,830 We've got a particle here and some mass, m. 204 00:13:47,830 --> 00:13:51,220 We know for a particle-- and it has some external forces on it. 205 00:13:54,480 --> 00:13:59,410 We know for a particle that the sum of these external forces-- 206 00:13:59,410 --> 00:14:03,710 it's a vector sum-- is equal to the mass 207 00:14:03,710 --> 00:14:07,280 times the acceleration for the particle, Newton's Second Law. 208 00:14:13,010 --> 00:14:17,880 And that's the mass times the derivative of the velocity 209 00:14:17,880 --> 00:14:19,000 with respect to time. 210 00:14:19,000 --> 00:14:21,830 That's where we get acceleration. 211 00:14:21,830 --> 00:14:24,810 So this formula we can just rearrange the dt's 212 00:14:24,810 --> 00:14:25,400 a little bit. 213 00:14:25,400 --> 00:14:35,360 So the summation of the forces, dt, is m dv. 214 00:14:35,360 --> 00:14:36,650 And I want to integrate these. 215 00:14:39,410 --> 00:14:44,670 So if I integrate this over time, from t1 to t2, 216 00:14:44,670 --> 00:14:49,150 this equality says we're going to find some change in velocity 217 00:14:49,150 --> 00:14:53,650 from v1 to v2. 218 00:14:53,650 --> 00:14:55,820 And this is our impulse-- this is 219 00:14:55,820 --> 00:14:59,530 the beginnings of our impulse-momentum relationship. 220 00:14:59,530 --> 00:15:03,620 We tend to call the integration of forces over time, impulse. 221 00:15:03,620 --> 00:15:07,280 And if you have a nonzero impulse, 222 00:15:07,280 --> 00:15:10,190 that leads to a change in the linear momentum. 223 00:15:13,660 --> 00:15:19,520 So when you carry it out so writing 224 00:15:19,520 --> 00:15:22,370 this a little bit different way, you can do summation 225 00:15:22,370 --> 00:15:25,940 because these are vectors, and we have rules about vectors. 226 00:15:25,940 --> 00:15:29,930 You can bring the summation outside of the integral. 227 00:15:29,930 --> 00:15:33,930 So that this says that the summation of the integral 228 00:15:33,930 --> 00:15:41,340 from t1 to t2 of these external forces 229 00:15:41,340 --> 00:15:46,853 is just mv2 minus mv1 because the integral 230 00:15:46,853 --> 00:15:51,390 on the right-hand side is really simple. 231 00:15:51,390 --> 00:15:56,280 And we'll finally state this in a way 232 00:15:56,280 --> 00:16:00,490 that we most commonly use it, moving this term to the left. 233 00:16:00,490 --> 00:16:04,950 So you start off with an initial momentum, mv1. 234 00:16:04,950 --> 00:16:11,750 You add to it the summation of the impulses that 235 00:16:11,750 --> 00:16:18,820 occur over time, t1, t2 of the external forces, 236 00:16:18,820 --> 00:16:22,670 and you get the final momentum mv2. 237 00:16:22,670 --> 00:16:27,544 And this is the way you usually use the formula. 238 00:16:27,544 --> 00:16:29,460 So what happens if there's no external forces? 239 00:16:32,680 --> 00:16:36,700 That's where we get the law of conservation of momentum. 240 00:16:36,700 --> 00:16:39,940 If there are no external forces on the particle, 241 00:16:39,940 --> 00:16:44,220 the momentum doesn't change and mv1 has got to be equal to mv2. 242 00:17:02,260 --> 00:17:07,390 So just a really trivial example drawing for something 243 00:17:07,390 --> 00:17:08,380 that we've done before. 244 00:17:08,380 --> 00:17:12,069 You got the block sliding down the hill. 245 00:17:12,069 --> 00:17:15,444 Draw your free body diagram. 246 00:17:15,444 --> 00:17:22,120 You got friction, got gravity, got a normal force. 247 00:17:27,010 --> 00:17:29,930 And we'll set ourselves up with a coordinate system aligned 248 00:17:29,930 --> 00:17:31,590 with the motion we're interested in. 249 00:17:31,590 --> 00:17:34,730 So this is an inertial system, x,y. 250 00:17:37,810 --> 00:17:39,380 And we've done this problem before. 251 00:17:39,380 --> 00:17:41,600 So we said the summation-- we found out 252 00:17:41,600 --> 00:17:46,160 that the summation of the forces in the x direction-- and this 253 00:17:46,160 --> 00:17:49,460 is a good moment to take an aside for a second. 254 00:17:49,460 --> 00:17:53,160 This was a vector expression. 255 00:17:53,160 --> 00:17:56,456 But you can break it-- you can implement it-- 256 00:17:56,456 --> 00:18:00,140 you can break it down into its individual vector components. 257 00:18:00,140 --> 00:18:02,220 So that equation gives you-- for particles-- 258 00:18:02,220 --> 00:18:06,220 gives you three sub-equations one in the x, one in the y, 259 00:18:06,220 --> 00:18:06,870 one in the z. 260 00:18:06,870 --> 00:18:09,030 And you can use them each independently. 261 00:18:09,030 --> 00:18:13,020 So in this case, we're going to use-- only need the x component 262 00:18:13,020 --> 00:18:14,320 in order to do the problem. 263 00:18:14,320 --> 00:18:17,400 So the sum of the forces in the x direction, 264 00:18:17,400 --> 00:18:20,860 in this problem which we had done before, 265 00:18:20,860 --> 00:18:28,260 is mg sine theta, that component of gravity pulling it 266 00:18:28,260 --> 00:18:35,360 down the hill, minus mu mg cosine theta. 267 00:18:35,360 --> 00:18:37,270 And that's the friction. 268 00:18:37,270 --> 00:18:41,490 And none of these are functions of time. 269 00:18:41,490 --> 00:18:44,681 So that whole thing is just some constant. 270 00:18:44,681 --> 00:18:47,980 I'll just call it k. 271 00:18:47,980 --> 00:18:51,149 So the total forces on the system in the x direction 272 00:18:51,149 --> 00:18:52,065 is just some constant. 273 00:18:52,065 --> 00:18:54,360 And that makes this integral up here pretty trivial 274 00:18:54,360 --> 00:18:56,140 to implement. 275 00:18:56,140 --> 00:19:03,530 So now we can say that mv1-- I'll make it v1x to emphasize 276 00:19:03,530 --> 00:19:07,020 that integral from 0, t1 to t2.. 277 00:19:07,020 --> 00:19:10,470 But I'm going to let t1 be 0 because it 278 00:19:10,470 --> 00:19:11,700 makes the problems easier. 279 00:19:11,700 --> 00:19:21,320 So from 0 to t of k dt equals mv2-- 280 00:19:21,320 --> 00:19:25,580 I can't write this morning-- in the x direction. 281 00:19:25,580 --> 00:19:34,390 And at v1, x is 0, starts off at 0 time and 0 velocity 282 00:19:34,390 --> 00:19:36,490 when you let it go to start with. 283 00:19:36,490 --> 00:19:38,350 Then, this term will go away. 284 00:19:38,350 --> 00:19:42,310 And we just implement this integral. 285 00:19:42,310 --> 00:19:46,960 And the integral of k dt is? 286 00:19:46,960 --> 00:19:47,940 kt, right? 287 00:19:47,940 --> 00:19:55,270 So this says then that kt evaluated from 0 to some time 288 00:19:55,270 --> 00:20:01,810 that we want to know the answer is mv2 in the x direction. 289 00:20:01,810 --> 00:20:05,930 And so let's let t equal 3 seconds. 290 00:20:08,470 --> 00:20:18,618 You find out that v 2x is 3k over m. 291 00:20:22,610 --> 00:20:30,760 And in this problem, then that looks like 3g sine theta 292 00:20:30,760 --> 00:20:35,670 minus mu cosine theta. 293 00:20:35,670 --> 00:20:42,870 So it's really an almost trivial example. 294 00:20:42,870 --> 00:20:44,990 It's not a hard example, but it emphasizes 295 00:20:44,990 --> 00:20:47,750 all of the key points in the problem. 296 00:20:47,750 --> 00:20:49,600 Start off with a vector equation. 297 00:20:49,600 --> 00:20:53,520 You can apply it in any one of the three vector component 298 00:20:53,520 --> 00:20:54,830 directions. 299 00:20:54,830 --> 00:20:56,870 We integrate the forces-- the sum 300 00:20:56,870 --> 00:21:00,160 of the forces on the object in that direction over time. 301 00:21:00,160 --> 00:21:02,860 And you apply the impulse-momentum formula, 302 00:21:02,860 --> 00:21:04,590 and you get the answer. 303 00:21:04,590 --> 00:21:05,320 You can do that. 304 00:21:05,320 --> 00:21:07,290 So that basic step-by-step process 305 00:21:07,290 --> 00:21:11,670 is how basically most impulse-momentum problems 306 00:21:11,670 --> 00:21:12,670 are done. 307 00:21:12,670 --> 00:21:14,390 And if there's no forces, then you 308 00:21:14,390 --> 00:21:15,750 have conservation of momentum. 309 00:21:19,800 --> 00:21:23,030 So let's do-- this was for particles. 310 00:21:37,390 --> 00:21:41,780 Just a quick reminder of we need to ask 311 00:21:41,780 --> 00:21:45,260 ourselves does this apply to groups of particles, systems 312 00:21:45,260 --> 00:21:46,360 of particles. 313 00:21:46,360 --> 00:21:48,590 Well, remember, that we said if you've got a bunch 314 00:21:48,590 --> 00:21:54,210 of particles, a system-- all of these are 315 00:21:54,210 --> 00:21:59,750 the mi''s-- we've already figured out that the total mass 316 00:21:59,750 --> 00:22:05,070 of the system times the velocity of the center of mass-- 317 00:22:05,070 --> 00:22:14,990 so g is my center of mass with respect to my inertial frame. 318 00:22:14,990 --> 00:22:19,330 So this is the momentum of the system that was just 319 00:22:19,330 --> 00:22:23,960 a summation of the individual mivi's with respect 320 00:22:23,960 --> 00:22:30,130 to-- And furthermore, we took the derivative of this, 321 00:22:30,130 --> 00:22:32,220 so this is the momentum of the system. 322 00:22:32,220 --> 00:22:34,240 The time derivative of the momentum 323 00:22:34,240 --> 00:22:38,810 should be the external forces so that we were able to-- 324 00:22:38,810 --> 00:22:43,810 by taking that derivative, you get mt, the total mass, 325 00:22:43,810 --> 00:22:47,950 times vg with respect to o dot. 326 00:22:47,950 --> 00:22:55,160 And that had better give you v equal to the summations 327 00:22:55,160 --> 00:23:00,466 of the external forces on all the particles 328 00:23:00,466 --> 00:23:02,340 because you can do it one particle at a time. 329 00:23:02,340 --> 00:23:05,890 And they all add-- each of these particles has forces. 330 00:23:05,890 --> 00:23:07,370 You sum them all up. 331 00:23:07,370 --> 00:23:12,920 You get these forces, and this allows 332 00:23:12,920 --> 00:23:17,290 you to say that the sum of all of the external forces 333 00:23:17,290 --> 00:23:19,540 on the system is equal to the total mass 334 00:23:19,540 --> 00:23:21,260 of the system times the acceleration 335 00:23:21,260 --> 00:23:23,860 of the center of mass. 336 00:23:23,860 --> 00:23:27,420 So this formula now can-- it looks 337 00:23:27,420 --> 00:23:31,380 very similar to where we started with this little derivation 338 00:23:31,380 --> 00:23:32,810 for a particle. 339 00:23:32,810 --> 00:23:36,350 You can now say that the summation-- this all 340 00:23:36,350 --> 00:23:41,830 leads to the summation of the integral 341 00:23:41,830 --> 00:23:52,570 from t1 to t2 of all of these courses over time-- 342 00:23:52,570 --> 00:23:58,030 gives you the change in the total linear momentum 343 00:23:58,030 --> 00:24:05,910 of the system, vg with respect to o1, 2 rather, 344 00:24:05,910 --> 00:24:12,160 minus vg with respect to o2, the second one. 345 00:24:12,160 --> 00:24:14,540 So this is exactly the same kind of formulation. 346 00:24:14,540 --> 00:24:20,660 The change in the linear momentum of the system 347 00:24:20,660 --> 00:24:24,730 is equal to the integral of the forces on the system over time, 348 00:24:24,730 --> 00:24:26,570 so the impulse to the system. 349 00:24:26,570 --> 00:24:30,410 So this allows you to do problems that you might not 350 00:24:30,410 --> 00:24:33,700 have thought about before. 351 00:24:33,700 --> 00:24:36,550 So I think about something like this. 352 00:24:36,550 --> 00:24:45,345 I've got an old Revolutionary War earlier period canon. 353 00:24:50,280 --> 00:24:51,910 I'm trying to draw it 354 00:24:51,910 --> 00:24:54,740 So here's my cannon barrel. 355 00:24:54,740 --> 00:25:03,870 And back in those days-- I'll do a little better job here. 356 00:25:03,870 --> 00:25:05,610 So we got this cannon sitting here. 357 00:25:05,610 --> 00:25:07,870 And in the barrel, you've got a bunch 358 00:25:07,870 --> 00:25:13,800 of what's known as-- in the old days, known as grapeshot. 359 00:25:13,800 --> 00:25:17,620 So sometimes, for anti personnel stuff, 360 00:25:17,620 --> 00:25:20,000 they would just throw a whole bunch of metal and junk 361 00:25:20,000 --> 00:25:20,530 into it. 362 00:25:20,530 --> 00:25:23,090 But just imagine a bunch of iron balls 363 00:25:23,090 --> 00:25:26,274 or lead ball rocks or whatever you want to put in there. 364 00:25:26,274 --> 00:25:28,940 And you put a whole mess of them in here, and you have a charge. 365 00:25:28,940 --> 00:25:31,807 You set it off, and it shoots them out the end of the gun. 366 00:25:31,807 --> 00:25:33,265 I want to know what's the reaction. 367 00:25:46,190 --> 00:25:48,210 What's the reaction force on the gun? 368 00:25:51,409 --> 00:25:53,200 When that gun-- when that charge goes off-- 369 00:25:53,200 --> 00:25:55,970 if you go down to the Constitution, which 370 00:25:55,970 --> 00:26:00,710 is the oldest commissioned warship in the US Navy 371 00:26:00,710 --> 00:26:02,850 still afloat-- I don't know if any of you 372 00:26:02,850 --> 00:26:05,130 have seen it down here in the dock in Chelsea. 373 00:26:05,130 --> 00:26:07,860 It was built around 1799. 374 00:26:07,860 --> 00:26:11,160 But when they shot one of those guns, 375 00:26:11,160 --> 00:26:15,600 the gun would roll back several feet and get dragged to a stop 376 00:26:15,600 --> 00:26:19,220 by a bunch of restraining lines and then dragged back up 377 00:26:19,220 --> 00:26:19,940 for another load. 378 00:26:19,940 --> 00:26:22,920 So the reaction force, the force pushing back-- 379 00:26:22,920 --> 00:26:25,640 these things were usually sitting on wheels-- 380 00:26:25,640 --> 00:26:26,642 could be pretty large. 381 00:26:26,642 --> 00:26:27,850 Let's take an estimate of it. 382 00:26:27,850 --> 00:26:29,845 So this is going to be a relatively small gun. 383 00:26:34,180 --> 00:26:40,270 The total mass of the shot here times g 384 00:26:40,270 --> 00:26:44,300 is 10 pounds, so 10 pounds a shot. 385 00:26:44,300 --> 00:26:45,840 A gallon of milk weighs 8 pounds, 386 00:26:45,840 --> 00:26:49,650 so this isn't a very big load. 387 00:26:49,650 --> 00:26:53,700 And let's say that this barrel here, from here to here, 388 00:26:53,700 --> 00:26:58,840 the acceleration length is 7 feet. 389 00:26:58,840 --> 00:27:02,220 So that shot is accelerated out of the barrel 390 00:27:02,220 --> 00:27:05,060 over a distance of 7 feet. 391 00:27:05,060 --> 00:27:16,270 And it has a muzzle velocity, an exit velocity, 392 00:27:16,270 --> 00:27:19,750 of 700 feet per second. 393 00:27:22,740 --> 00:27:26,140 Most guns, subsonic, supersonic, projectiles 394 00:27:26,140 --> 00:27:28,530 when they come out of guns, have you thought about that? 395 00:27:28,530 --> 00:27:32,240 I'm just asking-- seeing if you have a feeling for speed. 396 00:27:32,240 --> 00:27:36,362 And how close to the speed of sound is that? 397 00:27:36,362 --> 00:27:37,278 AUDIENCE: [INAUDIBLE]. 398 00:27:37,278 --> 00:27:38,736 PROFESSOR: So the speed of sound is 399 00:27:38,736 --> 00:27:43,100 about 1,100 feet per second, 340 meters per second. 400 00:27:43,100 --> 00:27:46,030 So this is order of mach 0.6 or something like that. 401 00:27:52,170 --> 00:27:56,540 Something like 0.6, not 0.060, 0.6. 402 00:27:56,540 --> 00:27:58,610 So about 60% of the speed of sound. 403 00:27:58,610 --> 00:28:00,490 That's slow actually as guns go. 404 00:28:03,800 --> 00:28:12,510 So the initial velocity-- the shot's just sitting theree-- 405 00:28:12,510 --> 00:28:13,640 is 0. 406 00:28:13,640 --> 00:28:21,350 The final velocity is 700 feet per second. 407 00:28:21,350 --> 00:28:23,935 The average velocity, which I'm going to need for a second 408 00:28:23,935 --> 00:28:25,810 because I'm just making some estimates here-- 409 00:28:25,810 --> 00:28:28,940 the average velocity, average of 0 plus 700 410 00:28:28,940 --> 00:28:30,690 is 350 feet per second. 411 00:28:33,650 --> 00:28:36,392 The reason I need this is I need to estimate the delta t. 412 00:28:36,392 --> 00:28:38,725 How long does it take to get the shot out of the barrel? 413 00:28:42,120 --> 00:28:45,120 So distance equals rate times time, right? 414 00:28:45,120 --> 00:28:56,260 So the v average times delta t equals 7 feet. 415 00:28:56,260 --> 00:28:59,050 At 350 feet per second, you find out 416 00:28:59,050 --> 00:29:09,050 delta t is about 0.02 seconds. 417 00:29:09,050 --> 00:29:12,950 So it gets out the barrel pretty quickly, 0.02 seconds, 418 00:29:12,950 --> 00:29:15,110 20 milliseconds. 419 00:29:15,110 --> 00:29:17,870 Now, that powder, when it goes off, 420 00:29:17,870 --> 00:29:21,420 it's putting a lot of forces on the shot. 421 00:29:21,420 --> 00:29:24,900 So if the total force on the shot is 5,000 pounds, 422 00:29:24,900 --> 00:29:27,030 what's the reaction force on the cannon? 423 00:29:35,040 --> 00:29:38,520 So we're going to apply a principle here 424 00:29:38,520 --> 00:29:40,890 that we talked about earlier. 425 00:29:40,890 --> 00:29:43,990 Newton had three laws. 426 00:29:43,990 --> 00:29:44,980 This is a group. 427 00:29:44,980 --> 00:29:49,970 This is just massive balls in there. 428 00:29:49,970 --> 00:29:53,190 The hot exploding gases are pushing them out the barrel. 429 00:29:53,190 --> 00:29:57,920 What must be the push of the same gas on the canon that's 430 00:29:57,920 --> 00:29:58,630 containing it? 431 00:29:58,630 --> 00:29:59,710 AUDIENCE: Equal and opposite. 432 00:29:59,710 --> 00:30:01,470 PROFESSOR: Equal and opposite, Newton's Third Law. 433 00:30:01,470 --> 00:30:01,970 Right. 434 00:30:01,970 --> 00:30:08,290 So the reaction force that we're looking for is minus the force 435 00:30:08,290 --> 00:30:10,290 that it takes to get the shot out of the barrel. 436 00:30:13,260 --> 00:30:19,890 So the force on the balls, on the shot, 437 00:30:19,890 --> 00:30:22,440 minus the reaction force. 438 00:30:22,440 --> 00:30:24,040 And that's Newton's Third Law. 439 00:30:31,400 --> 00:30:33,840 So we're almost there. 440 00:30:33,840 --> 00:30:35,760 We're just applying now this concept 441 00:30:35,760 --> 00:30:36,910 of impulse and momentum. 442 00:30:40,990 --> 00:30:44,860 So now we can say the integral-- and this is a summation. 443 00:30:44,860 --> 00:30:47,380 You got a bunch of balls in there. 444 00:30:47,380 --> 00:30:49,170 They all got forces on them, but we're 445 00:30:49,170 --> 00:30:51,660 treating the group of balls as a system, 446 00:30:51,660 --> 00:30:55,370 so we can use this notion of the total mass 447 00:30:55,370 --> 00:31:00,250 times the velocity of the center of gravity to do this problem. 448 00:31:00,250 --> 00:31:11,280 So this is the force on all of these little balls integrated 449 00:31:11,280 --> 00:31:11,970 over time. 450 00:31:14,560 --> 00:31:19,970 It's going to be the mass total of the balls times the change 451 00:31:19,970 --> 00:31:24,000 in the velocity of the center of gravity, the final velocity 452 00:31:24,000 --> 00:31:25,860 minus the initial velocity. 453 00:31:28,460 --> 00:31:29,225 This is 0. 454 00:31:32,800 --> 00:31:34,010 We know this. 455 00:31:34,010 --> 00:31:36,900 We know this. 456 00:31:36,900 --> 00:31:41,570 So this mt is the 10 pounds over g. 457 00:31:41,570 --> 00:31:44,470 We know this is 700 feet per second. 458 00:31:44,470 --> 00:31:46,980 We know that all of this mounts up 459 00:31:46,980 --> 00:31:54,700 to the force external on the group delta t. 460 00:31:58,700 --> 00:32:01,620 And this is the weight of the ball, w/g. 461 00:32:04,330 --> 00:32:10,430 And this is 700 feet per second. 462 00:32:10,430 --> 00:32:15,680 That's 10 pounds divided by gravity, 32.12. 463 00:32:15,680 --> 00:32:18,210 And we need to divide by delta t. 464 00:32:18,210 --> 00:32:21,990 So the external force on the balls 465 00:32:21,990 --> 00:32:32,710 is w/g delta t times 700 feet per second. 466 00:32:32,710 --> 00:32:45,170 And if you work out those numbers, you get 10,870 pounds. 467 00:32:45,170 --> 00:32:56,180 Now, this is an average force because we-- 468 00:32:56,180 --> 00:32:59,280 that delta t is the length of time 469 00:32:59,280 --> 00:33:04,130 it took, assuming we had some-- we assumed an average velocity 470 00:33:04,130 --> 00:33:05,250 going down the barrel. 471 00:33:05,250 --> 00:33:07,820 So the peak force is probably higher than this, 472 00:33:07,820 --> 00:33:10,340 and the pressure in the barrel probably isn't constant. 473 00:33:10,340 --> 00:33:12,610 This actually isn't a bad estimate. 474 00:33:12,610 --> 00:33:15,610 If you know the muzzle velocity, which you can figure out 475 00:33:15,610 --> 00:33:18,436 probably from the distance it goes and things like that, 476 00:33:18,436 --> 00:33:19,560 you can make this estimate. 477 00:33:19,560 --> 00:33:24,870 So 10,000 pounds of reaction force is quite a lot. 478 00:33:24,870 --> 00:33:27,020 This is a dinky gun. 479 00:33:27,020 --> 00:33:29,955 10 pounds a shot is not much. 480 00:33:29,955 --> 00:33:32,850 This is the force to get the balls out the barrel. 481 00:33:32,850 --> 00:33:35,100 If it's positive, what's the reaction force? 482 00:33:39,630 --> 00:33:42,340 Minus 10,870. 483 00:33:42,340 --> 00:33:45,080 And all through this, I've been doing this in a single vector 484 00:33:45,080 --> 00:33:46,920 direction, so I would have set up 485 00:33:46,920 --> 00:33:49,460 my coordinate system like here's o. 486 00:33:49,460 --> 00:33:52,780 This is the x direction, y direction. 487 00:33:52,780 --> 00:33:54,700 Obviously, in the positive x direction 488 00:33:54,700 --> 00:33:57,305 is what I've lined up my positive velocities 489 00:33:57,305 --> 00:33:59,670 and positive forces to be. 490 00:33:59,670 --> 00:34:01,241 So this is a really simple way. 491 00:34:01,241 --> 00:34:02,740 One of the kind of ways in which you 492 00:34:02,740 --> 00:34:05,610 use this notion of impulse and momentum 493 00:34:05,610 --> 00:34:07,290 that you can use to make estimates too. 494 00:34:12,830 --> 00:34:16,921 Any of you ever fired a shot gun. 495 00:34:16,921 --> 00:34:17,420 Not many. 496 00:34:17,420 --> 00:34:18,909 There's-- yeah. 497 00:34:18,909 --> 00:34:20,131 Any kick? 498 00:34:20,131 --> 00:34:20,630 Right. 499 00:34:20,630 --> 00:34:23,240 So what gauge shotgun did you fire? 500 00:34:23,240 --> 00:34:23,909 You don't know. 501 00:34:23,909 --> 00:34:24,409 OK. 502 00:34:24,409 --> 00:34:25,786 A 12-gauge shot gun. 503 00:34:25,786 --> 00:34:29,734 12-gauge shot gun, the cartridge is just about 3/4 504 00:34:29,734 --> 00:34:31,900 of an inch diameter and probably has a little amount 505 00:34:31,900 --> 00:34:32,750 of powder in there. 506 00:34:32,750 --> 00:34:34,747 It amounts up to about that much stuff. 507 00:34:34,747 --> 00:34:36,580 But it can give you a bruise in the shoulder 508 00:34:36,580 --> 00:34:38,739 if you don't hold it right. 509 00:34:38,739 --> 00:34:40,570 So that's what this is about. 510 00:34:43,610 --> 00:34:46,860 So this quick review, you've done 511 00:34:46,860 --> 00:34:49,710 lots of conservation of momentum problems in your time. 512 00:34:49,710 --> 00:34:53,690 The homework set this time has two or three problems on it 513 00:34:53,690 --> 00:34:56,850 that are conservation of momentum, linear momentum. 514 00:34:56,850 --> 00:35:01,990 But now I want to move on to talking about angular momentum. 515 00:35:01,990 --> 00:35:05,480 And angular momentum, in a way in which you probably 516 00:35:05,480 --> 00:35:09,060 haven't done angular momentum problems before. 517 00:35:09,060 --> 00:35:13,290 So anything about last thing on this. 518 00:35:13,290 --> 00:35:16,040 So mostly this, I want you to read the chapter. 519 00:35:16,040 --> 00:35:19,926 I think it's chapter 15. 520 00:35:19,926 --> 00:35:22,540 The first few sections of it are on linear momentum, 521 00:35:22,540 --> 00:35:24,880 the last few sections on angular momentum. 522 00:35:24,880 --> 00:35:28,160 Go read them and just work the problems. 523 00:35:28,160 --> 00:35:30,630 That's where you'll get most of your refresher 524 00:35:30,630 --> 00:35:33,620 is doing the practice problems. 525 00:35:33,620 --> 00:35:34,440 Angular momentum. 526 00:36:02,462 --> 00:36:03,670 So we'll start with particle. 527 00:36:03,670 --> 00:36:06,560 We're very rapidly going to get to rigid bodies. 528 00:36:11,990 --> 00:36:15,250 Now many times, you've done angular momentum problems 529 00:36:15,250 --> 00:36:18,610 before, mostly rotations about fixed axes. 530 00:36:18,610 --> 00:36:24,300 So here's our inertial frame, fixed x,y and you have 531 00:36:24,300 --> 00:36:27,170 a particle out here. 532 00:36:27,170 --> 00:36:31,240 And it just has some mass, m. 533 00:36:31,240 --> 00:36:35,009 And it has some total force on it, the force. 534 00:36:35,009 --> 00:36:36,550 And this says the particle is located 535 00:36:36,550 --> 00:36:41,180 at B here, so the force at B, total force at B, 536 00:36:41,180 --> 00:36:44,370 just some vector. 537 00:36:44,370 --> 00:36:47,580 It also is traveling with some velocity at that instant 538 00:36:47,580 --> 00:36:50,720 in time, so it has momentum. 539 00:36:50,720 --> 00:36:58,480 Momentum is p of this particle at B with respect to o. 540 00:37:02,630 --> 00:37:05,640 And I'd like to-- the standard expression 541 00:37:05,640 --> 00:37:13,960 then for the angular momentum of this particle at B 542 00:37:13,960 --> 00:37:23,700 with respect to o-- we have a position vector here, 543 00:37:23,700 --> 00:37:29,190 r Bo, which we've used lots of times by now, 544 00:37:29,190 --> 00:37:35,610 is just the cross product of the position vector 545 00:37:35,610 --> 00:37:37,540 with the linear momentum. 546 00:37:43,520 --> 00:37:46,400 And that's the definition of angular momentum. 547 00:37:46,400 --> 00:37:48,630 I'm using a lowercase h here, so they're 548 00:37:48,630 --> 00:37:50,970 going to do that to indicate single particles. 549 00:37:50,970 --> 00:37:53,200 And I'll use a capital H later on when we're 550 00:37:53,200 --> 00:37:55,010 referring to rigid bodies. 551 00:37:55,010 --> 00:37:58,000 So the angular momentum of this particle, 552 00:37:58,000 --> 00:38:01,234 with respect to this point, is given by that. 553 00:38:05,170 --> 00:38:07,710 And where you've used this before, 554 00:38:07,710 --> 00:38:11,630 then, it's the derivative of hB with respect 555 00:38:11,630 --> 00:38:17,830 to o dt is-- what's that? 556 00:38:17,830 --> 00:38:19,400 Remember, where does this give you? 557 00:38:19,400 --> 00:38:22,340 The time rate of change of the angular momentum is the? 558 00:38:22,340 --> 00:38:23,007 AUDIENCE: Torque 559 00:38:23,007 --> 00:38:23,756 PROFESSOR: Torque. 560 00:38:23,756 --> 00:38:24,330 Right. 561 00:38:24,330 --> 00:38:25,240 And it's the torque. 562 00:38:25,240 --> 00:38:29,150 It's a sum of the torques applied 563 00:38:29,150 --> 00:38:33,710 on that object with respect to the coordinate system 564 00:38:33,710 --> 00:38:36,540 in which you're computing the angular momentum. 565 00:38:36,540 --> 00:38:38,990 So that's-- we've used this formula many times. 566 00:38:38,990 --> 00:38:41,360 And in planar motion where you only 567 00:38:41,360 --> 00:38:46,410 have one axis of rotation using the z-axis of rotation, 568 00:38:46,410 --> 00:38:50,830 you usually would write this as I theta double dot. 569 00:38:50,830 --> 00:38:52,850 In simplest form, if it's a rigid body, 570 00:38:52,850 --> 00:38:55,260 it has a mass moment of inertia times 571 00:38:55,260 --> 00:38:58,190 theta double dot is equal to the sum of the external torque. 572 00:38:58,190 --> 00:39:00,337 So that's where you've met this before. 573 00:39:00,337 --> 00:39:02,170 But for the moment, this is just a particle. 574 00:39:02,170 --> 00:39:05,480 Let's stick with a particle. 575 00:39:05,480 --> 00:39:09,080 So the piece that's new here-- probably new 576 00:39:09,080 --> 00:39:18,440 for you-- is that what if I want to know the angular momentum 577 00:39:18,440 --> 00:39:21,630 with respect to another point? 578 00:39:21,630 --> 00:39:32,830 So here's a point A. I'd like to compute hB with respect to A. 579 00:39:32,830 --> 00:39:45,120 Well, that's rB with respect to A cross PB with respect to o. 580 00:39:45,120 --> 00:39:47,460 And this is really easy to forget. 581 00:39:47,460 --> 00:39:50,530 The momentum is always calculated with respect 582 00:39:50,530 --> 00:39:53,670 to your inertial frame. 583 00:39:53,670 --> 00:39:56,470 And that's why I keep in this with respect to 584 00:39:56,470 --> 00:39:59,710 and telling you what the frame is is pretty important. 585 00:39:59,710 --> 00:40:02,310 But we're out here at some arbitrary point, 586 00:40:02,310 --> 00:40:06,660 computing the cross product of the position vector 587 00:40:06,660 --> 00:40:10,870 from this arbitrary point to this moving mass. 588 00:40:10,870 --> 00:40:13,070 And we're defining-- it's just a definition 589 00:40:13,070 --> 00:40:16,910 defining the angular momentum with respect to this point A 590 00:40:16,910 --> 00:40:21,200 as r B/A cross PB with respect to o. 591 00:40:31,570 --> 00:40:42,680 And what I want to get to is now is the torque on this system, 592 00:40:42,680 --> 00:40:49,810 around this particle, B with respect to A 593 00:40:49,810 --> 00:40:57,260 is the time rate of change of h B/A with respect to t. 594 00:40:57,260 --> 00:41:00,250 But now it's a little more complicated. 595 00:41:00,250 --> 00:41:02,770 Plus, the velocity of A with respect 596 00:41:02,770 --> 00:41:12,900 to o-- these are all vectors-- cross PB with respect to o. 597 00:41:12,900 --> 00:41:13,700 Messy term. 598 00:41:17,352 --> 00:41:18,810 I was trying to think of an example 599 00:41:18,810 --> 00:41:23,630 where you might want to do this. 600 00:41:31,810 --> 00:41:38,320 So imagine that you've got an arm which can rotate. 601 00:41:38,320 --> 00:41:40,710 It might be on a robot or something like that. 602 00:41:40,710 --> 00:41:49,510 And it has attached to it another arm with a mass on it. 603 00:41:49,510 --> 00:41:54,850 And you've got a motor here, which can make this rotate. 604 00:41:54,850 --> 00:41:57,290 And you're trying to design-- the motor has 605 00:41:57,290 --> 00:42:00,240 to be able to put out a certain amount of torque. 606 00:42:00,240 --> 00:42:02,150 So this is o. 607 00:42:02,150 --> 00:42:05,930 This is B. This is A. And you actually-- 608 00:42:05,930 --> 00:42:09,880 I want to know the torque required in this motor 609 00:42:09,880 --> 00:42:12,990 to drive this thing around. 610 00:42:12,990 --> 00:42:15,380 But the motor is here, and it only 611 00:42:15,380 --> 00:42:19,420 cares about what it feels, so the torque at this point 612 00:42:19,420 --> 00:42:21,240 to drive this thing. 613 00:42:21,240 --> 00:42:24,550 But this whole system is now in motion. 614 00:42:24,550 --> 00:42:25,980 You'd have to use this formula. 615 00:42:28,900 --> 00:42:30,380 So there are practical times when 616 00:42:30,380 --> 00:42:33,360 you'd like to be able to calculate something like this. 617 00:42:42,870 --> 00:42:44,930 So there's a-- we're going to show 618 00:42:44,930 --> 00:42:52,530 you a very brief derivation of this 619 00:42:52,530 --> 00:42:54,620 just so you can get a feeling for where this comes 620 00:42:54,620 --> 00:42:56,450 from because there's a couple of outcomes 621 00:42:56,450 --> 00:42:57,890 that are very important to it. 622 00:43:10,020 --> 00:43:18,480 So the sum of the forces there at-- those forces at B, 623 00:43:18,480 --> 00:43:24,360 give you the time rate of change of the momentum 624 00:43:24,360 --> 00:43:26,080 at B with respect to o. 625 00:43:26,080 --> 00:43:29,930 This is the momentum vector of our particle. 626 00:43:29,930 --> 00:43:33,882 And this fB, this is the total external forces 627 00:43:33,882 --> 00:43:34,840 acting on the particle. 628 00:43:39,460 --> 00:43:42,715 And we know that's m. 629 00:43:42,715 --> 00:43:55,525 It's a single particle times the velocity of That 630 00:43:55,525 --> 00:43:56,830 certainly could be right. 631 00:44:03,920 --> 00:44:08,280 That's our familiar formula for F equals ma. 632 00:44:08,280 --> 00:44:11,970 So the time derivative of a linear momentum gives you this. 633 00:44:11,970 --> 00:44:17,690 And the torque of B with respect to A 634 00:44:17,690 --> 00:44:25,350 is r B/A cross-- I'm going let this just be a total vector. 635 00:44:25,350 --> 00:44:28,460 I'll call it fB. 636 00:44:28,460 --> 00:44:30,120 It's a vector. 637 00:44:30,120 --> 00:44:32,160 The torque with respect to A is just 638 00:44:32,160 --> 00:44:34,590 r B/A cross this total external force. 639 00:44:40,940 --> 00:44:53,283 r B/A cross time derivative of P B/o. 640 00:44:56,220 --> 00:44:58,680 Because the forces give us r if it's 641 00:44:58,680 --> 00:45:01,730 from the time rate of change of the linear momentum 642 00:45:01,730 --> 00:45:04,356 of that particle. 643 00:45:04,356 --> 00:45:06,630 So I can say it like this. 644 00:45:06,630 --> 00:45:12,640 But now there is just a little vector identity 645 00:45:12,640 --> 00:45:17,340 for products of vectors that I'm going to take advantage. 646 00:45:17,340 --> 00:45:31,620 And I'll call this Q. So Q is of the form-- 647 00:45:31,620 --> 00:45:35,290 and I'm going to say this is a quantity of vector A, 648 00:45:35,290 --> 00:45:38,600 and this is a quantity of vector-- time 649 00:45:38,600 --> 00:45:42,950 derivative of a vector B. It's A cross dB dt. 650 00:45:42,950 --> 00:45:47,080 So there's a little identity that you can use. 651 00:45:47,080 --> 00:45:55,330 It says A cross dB dt. 652 00:45:55,330 --> 00:46:01,510 You can alternatively write that as the time derivative 653 00:46:01,510 --> 00:46:14,499 of A cross B minus time derivative of A cross B. 654 00:46:14,499 --> 00:46:16,040 We're going to take advantage of this 655 00:46:16,040 --> 00:46:23,325 and just re-construct this formula using this expression. 656 00:46:30,540 --> 00:46:37,860 So that says torque of B with respect to A 657 00:46:37,860 --> 00:46:46,695 is the time derivative of r B/A cross P 658 00:46:46,695 --> 00:47:02,240 B/o minus the derivative of r B/A cross P B/o. 659 00:47:02,240 --> 00:47:05,360 So we've just made this substitution down here in terms 660 00:47:05,360 --> 00:47:09,375 of r B/A and P B/o all vectors. 661 00:47:12,540 --> 00:47:19,520 We know that r B/A, from all the previous work we've done, 662 00:47:19,520 --> 00:47:27,620 is just r B/o minus r A/o so the derivative 663 00:47:27,620 --> 00:47:43,900 of r B/A with respect to time is v B/o minus v A/o 664 00:47:43,900 --> 00:47:45,210 We're almost there. 665 00:47:45,210 --> 00:47:46,457 We're almost there. 666 00:47:46,457 --> 00:47:47,790 I'm going to need another board. 667 00:48:08,210 --> 00:48:15,700 This quantity here, this is just h B/A. 668 00:48:15,700 --> 00:48:19,510 This is the angular momentum of the particle with respect to A. 669 00:48:19,510 --> 00:48:22,145 It's just r cross B, if you recall. 670 00:48:52,360 --> 00:48:56,840 So then we can rewrite this expression for the torque 671 00:48:56,840 --> 00:49:02,800 as the time derivative of h of B with respect 672 00:49:02,800 --> 00:49:08,190 to A. That's because of what I pointed out there. 673 00:49:08,190 --> 00:49:19,217 Now, this is minus v B/o minus v A/o cross P B/o. 674 00:49:28,210 --> 00:49:31,670 So P of B with respect to o is just 675 00:49:31,670 --> 00:49:38,740 m velocity of B with respect to o. 676 00:49:38,740 --> 00:49:42,980 So v B/o cross P B/o gives you what? 677 00:49:42,980 --> 00:49:43,480 Nothing. 678 00:49:43,480 --> 00:49:45,729 Gives you 0 because they are in parallel of the angle, 679 00:49:45,729 --> 00:49:48,290 and they're in the same direction. 680 00:49:48,290 --> 00:49:53,230 So you only get a nonzero piece out of this minus times minus 681 00:49:53,230 --> 00:50:15,030 gives you a plus, and you end up with-- and that's 682 00:50:15,030 --> 00:50:16,850 what we set out to find. 683 00:50:21,010 --> 00:50:23,055 I said this is where I was trying to get, 684 00:50:23,055 --> 00:50:26,090 and now we're there. 685 00:50:26,090 --> 00:50:27,940 Now importantly, there's a couple 686 00:50:27,940 --> 00:50:31,410 of special cases of this. 687 00:50:31,410 --> 00:50:35,570 This can be a nuisance term to have to deal with. 688 00:50:35,570 --> 00:50:38,080 Lots of times you'd like to be able to get rid of it 689 00:50:38,080 --> 00:50:40,520 and just be able to go back to that old reliable formula, 690 00:50:40,520 --> 00:50:44,510 torque is time rate of change of angular momentum. 691 00:50:44,510 --> 00:50:48,530 So there are two obvious-- maybe obvious conditions 692 00:50:48,530 --> 00:50:52,220 in which this will go away. 693 00:50:52,220 --> 00:50:53,840 What's the most obvious one? 694 00:50:57,340 --> 00:50:59,010 AUDIENCE: [INAUDIBLE] change to 0. 695 00:50:59,010 --> 00:51:00,745 PROFESSOR: Something, what you say was 0? 696 00:51:00,745 --> 00:51:02,080 AUDIENCE: One of the terms 697 00:51:02,080 --> 00:51:02,996 PROFESSOR: Well, yeah. 698 00:51:02,996 --> 00:51:06,370 This term presumably not. 699 00:51:06,370 --> 00:51:24,650 This guy, if this is 0, it's just back 700 00:51:24,650 --> 00:51:25,885 to our old familiar formula. 701 00:51:29,820 --> 00:51:31,610 That's one case, so case one. 702 00:51:34,850 --> 00:51:39,550 But now-- actually, this is a really important result 703 00:51:39,550 --> 00:51:49,370 because A can be anywhere as long as it's not moving. 704 00:51:49,370 --> 00:51:53,080 So this allows you to do things, talk 705 00:51:53,080 --> 00:51:56,200 about rotations about fixed axes that 706 00:51:56,200 --> 00:51:57,560 aren't at the center of mass. 707 00:52:00,640 --> 00:52:03,260 So if you have a fixed axis of rotation and something 708 00:52:03,260 --> 00:52:08,810 going around it, that's what this allows you to do. 709 00:52:08,810 --> 00:52:09,710 That's one case. 710 00:52:09,710 --> 00:52:12,100 So this is-- 711 00:52:12,100 --> 00:52:14,150 We'll soon get to rigid bodies. 712 00:52:14,150 --> 00:52:17,370 Rigid bodies obey exactly the same formulas. 713 00:52:17,370 --> 00:52:19,740 And you can have a rigid body, now, 714 00:52:19,740 --> 00:52:23,040 that is not rotating about its center, 715 00:52:23,040 --> 00:52:26,130 but rotating maybe about its end like this. 716 00:52:26,130 --> 00:52:29,130 That formula applies if the velocity 717 00:52:29,130 --> 00:52:31,640 of that axis about which it's rotating, about which you're 718 00:52:31,640 --> 00:52:34,650 computing, is not moving, then you can just 719 00:52:34,650 --> 00:52:38,290 use-- you don't have to deal with that messy term. 720 00:52:38,290 --> 00:52:43,490 So this is one case in which this term goes away. 721 00:52:43,490 --> 00:52:47,600 The other case is if this velocity is parallel 722 00:52:47,600 --> 00:52:49,760 to the direction of the momentum. 723 00:52:52,380 --> 00:52:57,240 And there's a really useful time that that happens. 724 00:52:57,240 --> 00:53:01,530 So also this formula is true. 725 00:53:01,530 --> 00:53:09,150 Case two is when v A/o is parallel 726 00:53:09,150 --> 00:53:12,370 to P B/o, the direction. 727 00:53:12,370 --> 00:53:14,370 You're going in the same direction. 728 00:53:14,370 --> 00:53:18,840 Now that happens when A-- it's guaranteed 729 00:53:18,840 --> 00:53:22,890 to be true if A is at the center of mass 730 00:53:22,890 --> 00:53:26,920 because the momentum is defined as the mass of the object 731 00:53:26,920 --> 00:53:32,970 times the velocity of its center of mass, even for rigid bodies. 732 00:53:32,970 --> 00:53:45,035 So this is true when A is-- and I'll call it 733 00:53:45,035 --> 00:53:48,298 G-- at the center of mass. 734 00:53:53,570 --> 00:54:03,840 So this gives us another really important generalization 735 00:54:03,840 --> 00:54:07,500 that we'll use-- that we make great use of in dynamics. 736 00:54:22,330 --> 00:54:27,440 And that says that the torque, with respect 737 00:54:27,440 --> 00:54:34,030 to the center of mass, is time rate of change 738 00:54:34,030 --> 00:54:42,040 of h with respect to G. Now, I'm not 739 00:54:42,040 --> 00:54:44,860 going to go-- I did the proof-- went through this little proof 740 00:54:44,860 --> 00:54:46,340 just for a particle. 741 00:54:46,340 --> 00:54:49,540 But by summing a bunch of particles 742 00:54:49,540 --> 00:54:51,730 and going through all the summations, as we 743 00:54:51,730 --> 00:54:54,870 did, to prove the center of mass formula, 744 00:54:54,870 --> 00:54:59,490 you can show-- this allows you to very quickly show 745 00:54:59,490 --> 00:55:02,705 that this formulation is also true for rigid bodies. 746 00:55:05,730 --> 00:55:21,050 So the way you say it for rigid bodies is 747 00:55:21,050 --> 00:55:24,040 that the sum of the torques, with respect 748 00:55:24,040 --> 00:55:33,930 to some point for a rigid body, is capital H dot with respect 749 00:55:33,930 --> 00:55:42,960 to A plus the velocity of A with respect to o cross P. 750 00:55:42,960 --> 00:55:46,890 And now I'm going to say G, its center of mass, center 751 00:55:46,890 --> 00:55:50,588 of gravity, with respect to o. 752 00:55:50,588 --> 00:55:53,940 So the same statement for a rigid body 753 00:55:53,940 --> 00:55:56,810 is that the torque with respect to some point A, which 754 00:55:56,810 --> 00:56:01,020 can be moving now-- even accelerating-- 755 00:56:01,020 --> 00:56:03,520 is the time rate of change of the angular 756 00:56:03,520 --> 00:56:07,240 momentum with respect to A of that rigid body plus v A/o 757 00:56:07,240 --> 00:56:10,220 cross P B/o. 758 00:56:10,220 --> 00:56:15,070 And again, when would this messy second term go to 0? 759 00:56:15,070 --> 00:56:21,100 When the velocity of A is 0, fixed axis rotation, or when 760 00:56:21,100 --> 00:56:24,450 this is parallel to that, which is true for the center of mass 761 00:56:24,450 --> 00:56:25,960 always. 762 00:56:25,960 --> 00:56:31,650 So the same two special cases apply for rigid bodies 763 00:56:31,650 --> 00:56:37,920 when velocity of A with respect to 0 equals 0 764 00:56:37,920 --> 00:56:52,580 or when the velocity of A with respect to o is parallel the P. 765 00:56:52,580 --> 00:56:56,895 And the most important case of that is always true. 766 00:57:12,850 --> 00:57:13,890 It's always the case. 767 00:57:13,890 --> 00:57:15,560 And in these you can say this. 768 00:57:15,560 --> 00:57:19,590 Then, the torque, with respect to A, 769 00:57:19,590 --> 00:57:25,840 is dH, with respect to A dt. 770 00:57:25,840 --> 00:57:27,017 No second terms. 771 00:57:31,310 --> 00:57:35,610 And it's those kinds of-- you've applied, 772 00:57:35,610 --> 00:57:39,270 generally in your physics like in 801, you've used formulas 773 00:57:39,270 --> 00:57:40,120 like this a lot. 774 00:57:42,730 --> 00:57:44,220 Done problems either with respect 775 00:57:44,220 --> 00:57:51,500 to the center of mass so objects were doing things like that. 776 00:57:51,500 --> 00:57:54,740 You'll do the torque formulas with respect 777 00:57:54,740 --> 00:57:55,760 to the center of mass. 778 00:57:55,760 --> 00:58:00,900 Or when you have things that are pinned to points and rotate 779 00:58:00,900 --> 00:58:05,230 about fixed axes, then use the other formulation 780 00:58:05,230 --> 00:58:10,730 where it is with respect to a non-moving point. 781 00:58:10,730 --> 00:58:12,230 So this is fixed axis rotation. 782 00:58:27,660 --> 00:58:30,370 So that's the dry derivation part of it. 783 00:58:30,370 --> 00:58:33,100 I'm going to see if I can find an example here. 784 00:58:42,430 --> 00:58:44,029 This is quite a bit like a number 785 00:58:44,029 --> 00:58:45,070 of the homework problems. 786 00:59:18,620 --> 00:59:20,210 So I've got a carnival ride. 787 00:59:24,210 --> 00:59:26,470 I got a bar. 788 00:59:26,470 --> 00:59:29,890 And it's got a seat out here. 789 00:59:29,890 --> 00:59:32,240 You're riding in it. 790 00:59:32,240 --> 00:59:33,660 So you're taking this ride. 791 00:59:38,510 --> 00:59:40,010 It's rotating. 792 00:59:40,010 --> 00:59:42,330 This is some fixed axis here. 793 00:59:47,666 --> 00:59:53,120 You have a fixed coordinate system, your o x, y, z. 794 00:59:53,120 --> 00:59:55,690 And then you'd probably have some rotating coordinate system 795 00:59:55,690 --> 01:00:03,210 with the point A fixed here at the axis of rotation. 796 01:00:03,210 --> 01:00:09,580 And this is going-- what's unusual about this ride, 797 01:00:09,580 --> 01:00:11,860 not only can it go around and round, 798 01:00:11,860 --> 01:00:14,220 but the arm can go in and out. 799 01:00:17,230 --> 01:00:19,790 So it might take a path inwards, like you 800 01:00:19,790 --> 01:00:22,660 pull the arm in as you're going around. 801 01:00:22,660 --> 01:00:25,050 What are the forces that you feel in the ride? 802 01:00:28,040 --> 01:00:31,426 So this is my point B out here. 803 01:00:31,426 --> 01:00:32,550 It's where the person's at. 804 01:00:32,550 --> 01:00:36,165 You have some mass, m. 805 01:00:38,770 --> 01:00:49,010 And let's let, for now, the first case-- theta 806 01:00:49,010 --> 01:00:59,510 dot, let that be constant, the constant angular rate here. 807 01:00:59,510 --> 01:01:04,460 Here's theta and r dot. 808 01:01:13,220 --> 01:01:17,020 Now, you already know quite a bit about things like this. 809 01:01:17,020 --> 01:01:21,510 If your riding in that bucket, what forces 810 01:01:21,510 --> 01:01:23,920 do you think you would feel? 811 01:01:23,920 --> 01:01:26,600 Or I should say it more carefully. 812 01:01:26,600 --> 01:01:29,920 There will be accelerations that you feel in that bucket. 813 01:01:29,920 --> 01:01:31,700 You'll feel like forces on you. 814 01:01:31,700 --> 01:01:33,575 But what are the accelerations that you 815 01:01:33,575 --> 01:01:36,100 will feel-- you expect to be if you're riding in it? 816 01:01:39,812 --> 01:01:41,210 AUDIENCE: [INAUDIBLE]. 817 01:01:41,210 --> 01:01:42,460 PROFESSOR: I hear one here. 818 01:01:42,460 --> 01:01:43,690 AUDIENCE: Centripetal acceleration. 819 01:01:43,690 --> 01:01:45,960 PROFESSOR: So there will be centripetal acceleration, 820 01:01:45,960 --> 01:01:46,880 right. 821 01:01:46,880 --> 01:01:48,150 Everybody agree with that? 822 01:01:48,150 --> 01:01:49,070 Anything else? 823 01:01:49,070 --> 01:01:49,974 AUDIENCE: Coriolis acceleration. 824 01:01:49,974 --> 01:01:51,807 PROFESSOR: There's going to be some Coriolis 825 01:01:51,807 --> 01:01:54,762 because r dot is not 0. 826 01:01:54,762 --> 01:01:57,710 It's changing in position, but when 827 01:01:57,710 --> 01:01:59,740 the length of the arm of something 828 01:01:59,740 --> 01:02:03,054 rotating at constant speed changes, what momentum changes? 829 01:02:09,090 --> 01:02:09,840 AUDIENCE: Angular. 830 01:02:09,840 --> 01:02:11,048 PROFESSOR: Angular, for sure. 831 01:02:11,048 --> 01:02:12,980 How about linear momentum? 832 01:02:12,980 --> 01:02:16,260 Yeah, it's changing too because r cross P is angular momentum. 833 01:02:16,260 --> 01:02:17,810 So both linear and angular momentum 834 01:02:17,810 --> 01:02:22,320 are changing as this radius gets longer or shorter. 835 01:02:22,320 --> 01:02:25,500 And if that angular momentum changes, 836 01:02:25,500 --> 01:02:27,506 it takes torque to drive it. 837 01:02:27,506 --> 01:02:29,880 And so we ought to be able to use the formulas that we've 838 01:02:29,880 --> 01:02:32,280 just derived to calculate something 839 01:02:32,280 --> 01:02:35,430 about the torques required to make this happen 840 01:02:35,430 --> 01:02:36,750 and the forces on the rider. 841 01:02:44,580 --> 01:02:47,710 So we'll treat this one as a particle. 842 01:02:47,710 --> 01:02:52,420 So H, the angular momentum of the rider out here, 843 01:02:52,420 --> 01:03:07,480 at B with respect to o, is going to be r B/o cross P. 844 01:03:07,480 --> 01:03:10,780 And in this problem, I'll use polar coordinates. 845 01:03:10,780 --> 01:03:11,550 Pretty easy. 846 01:03:11,550 --> 01:03:13,470 This is a planar motion problem. 847 01:03:13,470 --> 01:03:17,040 It's confined to the x,y plane and rotation z, 848 01:03:17,040 --> 01:03:19,640 so polar coordinates are pretty convenient. 849 01:03:19,640 --> 01:03:26,020 So this should look like radius r r hat cross-- 850 01:03:26,020 --> 01:03:29,880 and the linear momentum of this is 851 01:03:29,880 --> 01:03:36,190 the mass-- times r dot r hat. 852 01:03:36,190 --> 01:03:41,090 That's the extension rate, but it also 853 01:03:41,090 --> 01:03:49,720 has velocity in this theta hat direction, r theta hat. 854 01:03:49,720 --> 01:03:56,450 So this is P was respected o, m, v. And this 855 01:03:56,450 --> 01:03:59,810 is the radius-- this is the r crossed into it. 856 01:03:59,810 --> 01:04:02,920 Now r hat cross r hat gives you 0, 857 01:04:02,920 --> 01:04:05,690 so you only get a single term out of this. 858 01:04:05,690 --> 01:04:10,970 And you get an r cross-- r hat cross, theta hat, positive k. 859 01:04:10,970 --> 01:04:25,750 So this looks like plus m r squared, theta dot, k hat. 860 01:04:25,750 --> 01:04:27,820 That's my H B/o. 861 01:04:34,910 --> 01:04:37,880 Now, I would-- that's my first piece that I wanted to get. 862 01:04:37,880 --> 01:04:38,790 That's the a part. 863 01:04:38,790 --> 01:04:41,200 That's find the angular momentum. 864 01:04:41,200 --> 01:04:43,688 So this is a. 865 01:04:43,688 --> 01:04:45,150 b, I want to know the torque. 866 01:04:54,950 --> 01:04:59,010 Well, which formula can we use? 867 01:04:59,010 --> 01:05:01,900 Are we allowed to use-- do we have to account 868 01:05:01,900 --> 01:05:04,202 for that second term? 869 01:05:04,202 --> 01:05:04,827 AUDIENCE: No. 870 01:05:04,827 --> 01:05:05,410 PROFESSOR: No. 871 01:05:05,410 --> 01:05:05,959 Why? 872 01:05:05,959 --> 01:05:06,750 AUDIENCE: Velocity. 873 01:05:06,750 --> 01:05:07,180 PROFESSOR: Yeah. 874 01:05:07,180 --> 01:05:08,971 The velocity of the point about which we're 875 01:05:08,971 --> 01:05:12,550 computing, the angular momentum, and therefore, the torques, 876 01:05:12,550 --> 01:05:13,380 is not moving. 877 01:05:13,380 --> 01:05:15,650 So you only have to deal with the first term. 878 01:05:15,650 --> 01:05:20,520 So the torque required to move that particle at B 879 01:05:20,520 --> 01:05:24,040 with respect to o is just a time rate of change 880 01:05:24,040 --> 01:05:28,870 d by dt of h B with respect to o. 881 01:05:28,870 --> 01:05:44,030 And that's d by dt of m r squared theta dot k hat. 882 01:05:44,030 --> 01:05:46,977 So what are the constants in this expression 883 01:05:46,977 --> 01:05:49,060 so we don't have to worry about their derivatives? 884 01:05:49,060 --> 01:05:51,160 Does k hat change length? 885 01:05:51,160 --> 01:05:51,660 No. 886 01:05:51,660 --> 01:05:53,010 Change direction? 887 01:05:53,010 --> 01:05:54,360 No. 888 01:05:54,360 --> 01:05:59,170 Theta dot, does it change in this problem? 889 01:05:59,170 --> 01:06:00,100 No, we fixed it. 890 01:06:00,100 --> 01:06:02,620 We arbitrarily started off saying that we'll just 891 01:06:02,620 --> 01:06:04,810 let theta dot be constant. 892 01:06:04,810 --> 01:06:06,390 r, though, is changing. 893 01:06:06,390 --> 01:06:08,420 So the time derivative of this particular one, 894 01:06:08,420 --> 01:06:10,910 we only have to deal with the r. 895 01:06:10,910 --> 01:06:13,175 Actually, I'll forget that for a second 896 01:06:13,175 --> 01:06:14,550 because I want to get both terms, 897 01:06:14,550 --> 01:06:16,120 and then we'll let it be 0. 898 01:06:16,120 --> 01:06:19,000 So this time derivative then, when you work it 899 01:06:19,000 --> 01:06:47,080 out-- the derivative of the r term 900 01:06:47,080 --> 01:06:56,730 gives you 2mr r dot theta dot k hat. 901 01:06:56,730 --> 01:06:58,764 And I'll just go ahead and forget for a minute 902 01:06:58,764 --> 01:06:59,930 that I said that's constant. 903 01:06:59,930 --> 01:07:03,870 Let's get the other term that might be there. 904 01:07:03,870 --> 01:07:16,785 That term gives us m r squared theta double dot k hat. 905 01:07:19,950 --> 01:07:24,580 This term comes from what we'll find is the Coriolis one. 906 01:07:24,580 --> 01:07:28,585 And this is from what we call the Eulerian acceleration. 907 01:07:31,445 --> 01:07:35,305 The torque-- let me get this up a little higher. 908 01:07:39,640 --> 01:07:48,270 We should be able to write as some r cross f. 909 01:07:48,270 --> 01:07:52,930 And so the r cross f terms, this will 910 01:07:52,930 --> 01:07:56,790 be from the Coriolis force. 911 01:07:56,790 --> 01:08:02,460 And this would be from that Eulerian force. 912 01:08:02,460 --> 01:08:06,700 So this will look like r r hat cross-- 913 01:08:06,700 --> 01:08:10,620 I'm just factoring this back out into its cross products-- 914 01:08:10,620 --> 01:08:17,819 r r hat cross 2m r dot theta dot, 915 01:08:17,819 --> 01:08:26,939 in the theta hat direction, plus r theta double dot, 916 01:08:26,939 --> 01:08:30,120 in the theta hat direction. 917 01:08:30,120 --> 01:08:32,689 So there's two terms in this torque expression. 918 01:08:32,689 --> 01:08:37,970 They come from r cross, two force terms. 919 01:08:37,970 --> 01:08:41,380 The first force term is what we know to be the Coriolis force. 920 01:08:41,380 --> 01:08:44,816 And the second force term-- I am missing an m here. 921 01:08:48,560 --> 01:08:50,880 r m theta double dot theta hat, that's 922 01:08:50,880 --> 01:08:57,140 the force that it takes to-- if the thing were accelerating, 923 01:08:57,140 --> 01:09:01,370 just to accelerate that ball, theta double dot takes a force. 924 01:09:01,370 --> 01:09:04,420 You do these cross products. r cross theta hat give you k. 925 01:09:04,420 --> 01:09:06,380 R cross theta hat give you k. 926 01:09:06,380 --> 01:09:09,200 It all comes out in the right directions. 927 01:09:09,200 --> 01:09:11,380 And now to do the problem we had, 928 01:09:11,380 --> 01:09:13,229 we said, oh yeah, let this one be 0. 929 01:09:13,229 --> 01:09:14,590 So we'll let this term go to 0. 930 01:09:17,220 --> 01:09:18,590 We're left with a single term. 931 01:10:14,430 --> 01:10:16,070 And if we plugged in some numbers-- 932 01:10:16,070 --> 01:10:17,670 let's just see how this works out. 933 01:10:17,670 --> 01:10:19,610 And this we know is just r cross the Coriolis. 934 01:10:27,570 --> 01:10:33,310 Well, let's let m-- if you let m be 100 kilograms 935 01:10:33,310 --> 01:10:44,580 and r be 5 meters and r dot, 0.4 meters 936 01:10:44,580 --> 01:10:48,380 per second-- they're all perfectly reasonable 937 01:10:48,380 --> 01:10:50,320 dimensions. 938 01:10:50,320 --> 01:10:59,020 And theta dot equals 3 radians per second. 939 01:10:59,020 --> 01:11:02,664 So 2 pi radians per second means it goes around once a second. 940 01:11:02,664 --> 01:11:04,830 So this is a little less than half a rev per second. 941 01:11:04,830 --> 01:11:05,442 Yeah? 942 01:11:05,442 --> 01:11:07,233 AUDIENCE: Is this still [INAUDIBLE] though, 943 01:11:07,233 --> 01:11:08,420 or does it-- 944 01:11:08,420 --> 01:11:10,310 PROFESSOR: Probably. 945 01:11:10,310 --> 01:11:11,350 Yeah. 946 01:11:11,350 --> 01:11:12,650 Good catch! 947 01:11:12,650 --> 01:11:15,870 Got to have on r in it because that r comes from here, right? 948 01:11:15,870 --> 01:11:18,180 So I had the Coriolis force written down, 949 01:11:18,180 --> 01:11:21,660 but not the r you have to multiply it by. 950 01:11:21,660 --> 01:11:24,020 So now if you plug in all of these numbers, 951 01:11:24,020 --> 01:11:26,510 let's see if it's a ride you could survive. 952 01:11:34,220 --> 01:11:38,069 I actually computed the Coriolis force first. 953 01:11:38,069 --> 01:11:40,360 I think it's just interesting to get a physical feeling 954 01:11:40,360 --> 01:11:43,570 for how-- whether or not you can feel these forces. 955 01:11:43,570 --> 01:11:45,910 So the Coriolis force is just everything 956 01:11:45,910 --> 01:11:49,860 but the r, 2m r dot theta dot. 957 01:11:49,860 --> 01:11:56,540 So if you calculate that, you get 240 newtons. 958 01:11:56,540 --> 01:12:00,262 And r times that, the torque. 959 01:12:03,650 --> 01:12:16,520 5 times that, about 1,200 Newton meters. 960 01:12:16,520 --> 01:12:21,960 And the acceleration, how do we get acceleration? 961 01:12:21,960 --> 01:12:26,650 Well, f cor, Coriolis force, is some mass 962 01:12:26,650 --> 01:12:27,930 times an acceleration. 963 01:12:27,930 --> 01:12:32,320 So we can solve, from that 240 newtons, 964 01:12:32,320 --> 01:12:34,040 the acceleration of the system. 965 01:12:34,040 --> 01:12:45,500 Our acceleration of B with respect to o, 240 newtons 966 01:12:45,500 --> 01:12:56,890 divided by 100 kilograms, 2.4 meters per second squared. 967 01:12:56,890 --> 01:13:00,134 And what's that in g's. 968 01:13:00,134 --> 01:13:01,070 AUDIENCE: About 1/4. 969 01:13:01,070 --> 01:13:03,082 PROFESSOR: Yeah, about a quarter of a g. 970 01:13:06,250 --> 01:13:08,230 And again, would you feel that? 971 01:13:08,230 --> 01:13:12,080 So now you riding in the bucket. 972 01:13:12,080 --> 01:13:13,730 Let's say it's just spinning around. 973 01:13:13,730 --> 01:13:15,760 The arm is not going in and out at all. 974 01:13:15,760 --> 01:13:16,790 So you're riding in it. 975 01:13:16,790 --> 01:13:19,840 What force would you feel? 976 01:13:19,840 --> 01:13:24,110 Would you feel any forces pushing you into your seat? 977 01:13:24,110 --> 01:13:26,150 Constant speed, r dot's 0. 978 01:13:36,810 --> 01:13:38,680 Somebody out there, would you feel any force 979 01:13:38,680 --> 01:13:41,120 if you're going around and around in this thing sitting 980 01:13:41,120 --> 01:13:41,680 in a seat? 981 01:13:41,680 --> 01:13:42,010 AUDIENCE: Yes. 982 01:13:42,010 --> 01:13:42,593 PROFESSOR: OK. 983 01:13:42,593 --> 01:13:44,000 What's it come from? 984 01:13:44,000 --> 01:13:44,310 AUDIENCE: Centripetal acceleration. 985 01:13:44,310 --> 01:13:45,813 PROFESSOR: Centripetal acceleration. 986 01:13:48,820 --> 01:13:51,264 And that is in what direction? 987 01:13:51,264 --> 01:13:52,180 AUDIENCE: It's inward. 988 01:13:52,180 --> 01:13:53,638 PROFESSOR: Accelerations is inward. 989 01:13:53,638 --> 01:13:55,560 What would you actually feel? 990 01:13:55,560 --> 01:13:59,350 You'd feel-- if this seat could swing out 991 01:13:59,350 --> 01:14:02,090 so your facing in, you'd feel like you're being thrown out 992 01:14:02,090 --> 01:14:04,800 in your seat, right? 993 01:14:04,800 --> 01:14:07,180 Because you have to have a force on you 994 01:14:07,180 --> 01:14:09,460 to make that acceleration happen. 995 01:14:09,460 --> 01:14:10,665 Acceleration is inward. 996 01:14:10,665 --> 01:14:13,360 The force had better be pushing you inward. 997 01:14:13,360 --> 01:14:16,460 So it's pushing on your back if you're look in, going around. 998 01:14:16,460 --> 01:14:22,750 So that's the-- I've forgotten the term for it. 999 01:14:22,750 --> 01:14:26,020 But put the astronauts in a centrifuge, 1000 01:14:26,020 --> 01:14:28,710 spin around to see if they can take high g's. 1001 01:14:28,710 --> 01:14:33,080 That's the inward high g acceleration 1002 01:14:33,080 --> 01:14:35,200 due to the centripetal acceleration, 1003 01:14:35,200 --> 01:14:36,800 makes you go in a circle. 1004 01:14:36,800 --> 01:14:40,520 But now not only are you going to feel that, but now you start 1005 01:14:40,520 --> 01:14:42,430 changing the length of the arm. 1006 01:14:42,430 --> 01:14:46,010 And if you change r dot and make it positive 1007 01:14:46,010 --> 01:14:47,820 so the arm is getting longer, and we made 1008 01:14:47,820 --> 01:14:49,340 this 0.4 meters per second. 1009 01:14:49,340 --> 01:14:51,350 So it's moving out about like that. 1010 01:14:51,350 --> 01:14:53,190 It's not real fast. 1011 01:14:53,190 --> 01:14:56,660 But it's going to create a quarter of a g acceleration 1012 01:14:56,660 --> 01:14:57,160 on you. 1013 01:14:57,160 --> 01:15:00,325 And if r dot's positive, which direction is that acceleration? 1014 01:15:03,500 --> 01:15:04,190 I was careful. 1015 01:15:04,190 --> 01:15:05,670 We walked all the way through this. 1016 01:15:05,670 --> 01:15:09,480 The acceleration is positive, but in what direction? 1017 01:15:09,480 --> 01:15:11,292 It's the Coriolis. 1018 01:15:11,292 --> 01:15:12,750 AUDIENCE: Theta hat. 1019 01:15:12,750 --> 01:15:14,760 PROFESSOR: Theta hat, positive theta hat, 1020 01:15:14,760 --> 01:15:16,800 so it's in the direction of increasing theta. 1021 01:15:16,800 --> 01:15:20,460 So this one is perpendicular to the arm. 1022 01:15:20,460 --> 01:15:24,190 And this is the-- one of my ping pong 1023 01:15:24,190 --> 01:15:28,510 balls in here, the force that actually causes it to speed up 1024 01:15:28,510 --> 01:15:30,940 is partly Coriolis and partly or Eulerian. 1025 01:15:30,940 --> 01:15:34,270 If I can make this go at constant speed, the force that 1026 01:15:34,270 --> 01:15:38,050 drives that-- the force that actually speeds that up 1027 01:15:38,050 --> 01:15:41,090 as they're going out is the Coriolis force, the normal one. 1028 01:15:41,090 --> 01:15:42,730 So if you were on this ride, you'd 1029 01:15:42,730 --> 01:15:45,760 be feel about a quarter of a g perpendicular to the arm. 1030 01:15:45,760 --> 01:15:47,780 You'd be filling another force inward 1031 01:15:47,780 --> 01:15:49,837 that's the centripetal-- caused by 1032 01:15:49,837 --> 01:15:51,045 the centripetal acceleration. 1033 01:15:51,045 --> 01:15:52,510 So you'd be feeling both of them. 1034 01:15:52,510 --> 01:15:54,176 How big is the centripetal acceleration? 1035 01:15:56,810 --> 01:16:02,274 Like r omega squared, right? 1036 01:16:02,274 --> 01:16:03,030 AUDIENCE: Right. 1037 01:16:03,030 --> 01:16:04,380 PROFESSOR: r is 5. 1038 01:16:04,380 --> 01:16:06,065 Omega is 3. 1039 01:16:06,065 --> 01:16:18,007 3 Squared is 9, times 5-- 9 times 5, 45, divided by 10, 1040 01:16:18,007 --> 01:16:18,590 4 and 1/2 g's. 1041 01:16:22,740 --> 01:16:24,815 You wouldn't notice the Coriolis very much. 1042 01:16:24,815 --> 01:16:29,470 It would be a tough ride on that side of things. 1043 01:16:32,980 --> 01:16:37,592 Now, we've done that-- you've seen a simple application, 1044 01:16:37,592 --> 01:16:39,050 pretty straightforward application, 1045 01:16:39,050 --> 01:16:44,350 of using time derivative of angular momentum 1046 01:16:44,350 --> 01:16:46,640 to calculate torques. 1047 01:16:46,640 --> 01:16:49,710 So on homework, again, there's a couple problems 1048 01:16:49,710 --> 01:16:52,400 very similar to the one I just did, things going around, 1049 01:16:52,400 --> 01:16:54,610 circus rides, that kind of stuff. 1050 01:16:54,610 --> 01:16:58,850 So have a good weekend. 1051 01:16:58,850 --> 01:17:01,460 See you on Tuesday.