WEBVTT 00:00:00.080 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.810 Commons license. 00:00:03.810 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.297 at ocw.mit.edu. 00:00:22.070 --> 00:00:24.590 PROFESSOR: The technical topic for today, 00:00:24.590 --> 00:00:26.210 we start-- I rushed, right at the end, 00:00:26.210 --> 00:00:28.380 a little bit about tangent and normal unit vectors. 00:00:28.380 --> 00:00:31.120 I'm going to just recap that quickly. 00:00:31.120 --> 00:00:35.070 And then we're going to go on really a review, which 00:00:35.070 --> 00:00:37.790 by a review, this is the sort of stuff that, for the most part, 00:00:37.790 --> 00:00:41.760 I'm sure you've seen in 801 Physics and other Physics 00:00:41.760 --> 00:00:42.550 you've had before. 00:00:42.550 --> 00:00:45.650 And that's impulse-- linear momentum, and impulse. 00:00:45.650 --> 00:00:47.580 So that'll be a quick review. 00:00:47.580 --> 00:00:51.280 And then the third subject is one that's much deeper, 00:00:51.280 --> 00:00:53.160 and this is angular momentum. 00:00:53.160 --> 00:00:55.530 And angular momentum with respect 00:00:55.530 --> 00:00:57.210 to moving points, which you probably 00:00:57.210 --> 00:00:59.290 haven't encountered before. 00:00:59.290 --> 00:01:02.130 So those are the three topics for the day. 00:01:02.130 --> 00:01:03.160 Let's get started. 00:01:06.070 --> 00:01:12.000 So last time, the piece that I rushed 00:01:12.000 --> 00:01:19.105 a bit is this notion of tangent and normal coordinates. 00:01:22.400 --> 00:01:24.180 So I gave this-- had this example. 00:01:24.180 --> 00:01:32.856 You're driving down the road, you're drunk or whatever. 00:01:35.770 --> 00:01:39.100 And when you're at this point, this point, and this point, 00:01:39.100 --> 00:01:41.580 I'd like to know what the accelerations are. 00:01:41.580 --> 00:01:43.840 So I'll call this 1, 2, 3. 00:01:43.840 --> 00:01:50.225 And this curve, y, is of the form some y of some f of x. 00:01:52.730 --> 00:01:59.600 And in this case, it's A sine kx. 00:01:59.600 --> 00:02:02.700 And k is what's known as wave number. 00:02:02.700 --> 00:02:07.170 This is 2 pi over the wave length, 2 pi over lambda. 00:02:07.170 --> 00:02:10.830 And the wave length then is, for example, from here to here. 00:02:17.520 --> 00:02:20.000 Now, the velocity-- I'm going to pick a point here. 00:02:20.000 --> 00:02:23.270 The velocity at any point, we know, 00:02:23.270 --> 00:02:26.267 is just the tangent to the path. 00:02:26.267 --> 00:02:28.100 So this is the path, and this is horizontal. 00:02:28.100 --> 00:02:32.100 So you're driving-- driving down the road like this. 00:02:32.100 --> 00:02:34.700 That's what we're trying to do here. 00:02:34.700 --> 00:02:36.390 The gravity's down into the board. 00:02:36.390 --> 00:02:38.300 It doesn't really come into the problem. 00:02:41.260 --> 00:02:45.930 So the velocity, at any point-- at anytime-- 00:02:45.930 --> 00:02:50.860 it's a vector-- we can describe as a magnitude. 00:02:50.860 --> 00:02:54.370 And in a unit vector, we'll call ut which 00:02:54.370 --> 00:02:56.610 is the tangent unit vector. 00:02:56.610 --> 00:02:59.750 And at any instant in time, it's just aligned with the tangent 00:02:59.750 --> 00:03:01.980 to the curve. 00:03:01.980 --> 00:03:08.730 And it's perpendicular partner is a normal unit vector, which 00:03:08.730 --> 00:03:11.450 it points inward on the curve. 00:03:11.450 --> 00:03:14.480 And this will be un. 00:03:14.480 --> 00:03:16.480 Now, we're interested in accelerations, 00:03:16.480 --> 00:03:18.890 so we'll need to take a derivative of this. 00:03:27.630 --> 00:03:29.960 So I was going to say, taken aside. 00:03:29.960 --> 00:03:30.792 But I won't. 00:03:30.792 --> 00:03:32.250 So the acceleration vector, we have 00:03:32.250 --> 00:03:33.541 to take the derivative of this. 00:03:33.541 --> 00:03:43.500 Well, it's a v dot ut plus a v ut dot. 00:03:43.500 --> 00:03:44.530 It's a unit vector. 00:03:44.530 --> 00:03:46.080 And we've encountered this problem 00:03:46.080 --> 00:03:49.780 before because if this rotates, the unit vector 00:03:49.780 --> 00:03:52.885 has a non-zero derivative because of this rotation. 00:03:57.020 --> 00:04:03.920 And this derivative of the unit vector 00:04:03.920 --> 00:04:13.380 is given by theta dot u m hat Now, what's theta dot? 00:04:13.380 --> 00:04:17.490 Well, on this curve, at any instant in time, 00:04:17.490 --> 00:04:21.410 here's u-- can't draw arrows today. 00:04:21.410 --> 00:04:23.200 Here's your ut. 00:04:23.200 --> 00:04:27.970 At any instant when you're traveling along on a curve, 00:04:27.970 --> 00:04:32.690 you are going around the circle of some radius we'll call rho. 00:04:32.690 --> 00:04:34.720 So there's some radius of curvature 00:04:34.720 --> 00:04:37.070 at any instant in time. 00:04:37.070 --> 00:04:44.780 And that radius, as you drive along in a little time delta t, 00:04:44.780 --> 00:04:51.990 you go forward an amount, rho theta dot, 00:04:51.990 --> 00:04:59.820 is the velocity at which you're traveling tangent to the curve. 00:04:59.820 --> 00:05:03.420 There's an angle in here, so in here, there's some delta. 00:05:03.420 --> 00:05:09.320 You advance some delta theta, so our omega is a velocity, right? 00:05:09.320 --> 00:05:12.530 So rho theta dot its velocity, and that gives you 00:05:12.530 --> 00:05:20.234 this velocity magnitude, v. And we've 00:05:20.234 --> 00:05:21.650 gone through-- I'm not going to go 00:05:21.650 --> 00:05:24.191 through that little derivative-- that little argument before, 00:05:24.191 --> 00:05:28.260 but the change in direction of this unit vector, 00:05:28.260 --> 00:05:35.560 ut, when you go forward a little bit is actually inward, un, 00:05:35.560 --> 00:05:42.080 by an amount rho theta dot-- by an amount theta dot one times 00:05:42.080 --> 00:05:42.874 theta dot. 00:05:42.874 --> 00:05:44.290 So we've done- I'm not going to do 00:05:44.290 --> 00:05:46.081 that piece of the derivative we did before. 00:05:46.081 --> 00:05:49.340 But here is this derivative of the unit vector. 00:05:49.340 --> 00:05:51.780 It's theta dot un. 00:05:51.780 --> 00:05:54.070 So if we substitute those back in here, 00:05:54.070 --> 00:05:57.625 we can get an expression for our acceleration. 00:06:02.600 --> 00:06:11.760 This is just the acceleration along the path plus v theta dot 00:06:11.760 --> 00:06:14.420 un. 00:06:14.420 --> 00:06:17.550 And then we need to take into account 00:06:17.550 --> 00:06:22.280 the fact that we know that v, the magnitude of velocity-- 00:06:22.280 --> 00:06:25.650 the speed in other words-- is rho theta dot. 00:06:28.300 --> 00:06:34.800 And that means theta dot is v over rho. 00:06:34.800 --> 00:06:37.670 So if we plug that in up here, then 00:06:37.670 --> 00:06:42.600 we get a final expression for our acceleration, 00:06:42.600 --> 00:06:52.610 v dot ut plus v squared over rho un. 00:06:52.610 --> 00:06:56.030 And that v squared over rho-- this 00:06:56.030 --> 00:06:59.850 is a centripetal acceleration term. 00:06:59.850 --> 00:07:01.640 You're going around a curve. 00:07:01.640 --> 00:07:04.050 There is an acceleration inward. 00:07:04.050 --> 00:07:07.280 It's just like as we did from polar coordinates. 00:07:07.280 --> 00:07:09.422 It comes from rotating things. 00:07:09.422 --> 00:07:11.130 Well, as soon as you go around the curve, 00:07:11.130 --> 00:07:14.030 you're going to generate an acceleration that 00:07:14.030 --> 00:07:16.340 is of the same kind, but now we-- 00:07:16.340 --> 00:07:19.720 because you tend to know speed, it's 00:07:19.720 --> 00:07:21.730 easier to express it this way when you're doing 00:07:21.730 --> 00:07:23.760 these tangent normal problems. 00:07:23.760 --> 00:07:27.120 And this is just the acceleration along a path, 00:07:27.120 --> 00:07:30.190 the usual hit the gas petal and speed up. 00:07:30.190 --> 00:07:31.115 That's this term. 00:07:34.240 --> 00:07:37.190 So the piece that I didn't have time 00:07:37.190 --> 00:07:47.760 to put up last time, which is in the book as 00:07:47.760 --> 00:07:50.010 is this little derivative-- the derivation of that 00:07:50.010 --> 00:07:53.310 is one page in one of the early chapters of the book. 00:07:53.310 --> 00:07:55.700 How do you get rho? 00:07:55.700 --> 00:07:58.900 And rho, when you have y as a function of x, 00:07:58.900 --> 00:08:03.470 there is just this formula from mathematics-- from calculus-- 00:08:03.470 --> 00:08:12.700 that says rho is dy dx 1 plus dy dx quantity squared to the 3/2 00:08:12.700 --> 00:08:22.710 power all over the magnitude of d2y dx squared. 00:08:22.710 --> 00:08:25.200 And you just calculate these quantities and plug them in. 00:08:25.200 --> 00:08:42.850 So for y equals a sine kx, then dy dx is Ak cosine kx 00:08:42.850 --> 00:08:56.720 and d2y dx squared is minus a k squared sine kx. 00:08:56.720 --> 00:08:59.580 So we now have these two quantities. 00:08:59.580 --> 00:09:01.550 We pick a value of x which we want 00:09:01.550 --> 00:09:04.720 to know the answer, like 0.1. 00:09:04.720 --> 00:09:10.510 So at 0.1, kx is pi over 2 because sine is a maximum. 00:09:17.330 --> 00:09:21.290 X is lambda over 4. 00:09:21.290 --> 00:09:22.970 kx is pi over 2. 00:09:26.370 --> 00:09:30.030 Sine kx is 1. 00:09:30.030 --> 00:09:34.400 Cosine kx is 0. 00:09:34.400 --> 00:09:36.280 And so we can calculate rho. 00:09:39.450 --> 00:09:47.440 1 plus and dy dx the derivative, is cosine. 00:09:47.440 --> 00:09:48.170 That's 0. 00:09:48.170 --> 00:09:55.560 So this term in here is 0. 00:09:55.560 --> 00:09:56.780 And this is to the 3/2. 00:09:56.780 --> 00:09:58.470 That's pretty easy to calculate. 00:09:58.470 --> 00:10:03.480 This term down here, d2y dx squared, well, sine of kx is 1. 00:10:03.480 --> 00:10:08.120 So that's minus Ak squared, but this is an absolute value sign. 00:10:12.650 --> 00:10:19.700 So this just turns out to be 1 over Ak squared. 00:10:19.700 --> 00:10:22.970 And you plug-in some numbers. 00:10:22.970 --> 00:10:31.036 I'm going to let my lambda be 150 meters. 00:10:31.036 --> 00:10:34.340 My amplitude here that I'm swerving back and forth, 00:10:34.340 --> 00:10:35.650 let's make that 5 meters. 00:10:38.790 --> 00:10:44.340 We need a-- that's all we need to get rho. 00:10:44.340 --> 00:10:50.110 So rho, in this case, works out to be 114 meters. 00:10:50.110 --> 00:10:53.360 So the radius of curvature, this road that I'm driving down, 00:10:53.360 --> 00:10:58.900 at that point right at the peak in the curve is 114 meters. 00:11:05.710 --> 00:11:08.890 So the acceleration that I'm looking for, 00:11:08.890 --> 00:11:13.600 a, well, it's v dot ut. 00:11:13.600 --> 00:11:15.330 And if I'm not accelerating-- I'm not 00:11:15.330 --> 00:11:17.010 hitting the gas petal at constant speed, 00:11:17.010 --> 00:11:19.500 I'll let that would be 0. 00:11:19.500 --> 00:11:21.870 And the other one, then, is v squared 00:11:21.870 --> 00:11:26.120 over rho of the un direction. 00:11:26.120 --> 00:11:32.140 And let's let v equal-- whatever I have here in my example-- 00:11:32.140 --> 00:11:38.600 20 meters per second, which is about 40 knots. 00:11:38.600 --> 00:11:42.060 And a knot is 15% more than a mile per hour, 00:11:42.060 --> 00:11:44.985 so it's somewhere about 45 miles an hour, so 00:11:44.985 --> 00:11:46.100 a typical road speed. 00:11:46.100 --> 00:11:48.650 Driving down the road, 20 meters per second, 00:11:48.650 --> 00:11:49.810 now you can plug-in here. 00:11:49.810 --> 00:11:51.360 You can plug-in here, and you'll find 00:11:51.360 --> 00:11:54.525 that the acceleration that I worked out here 00:11:54.525 --> 00:12:00.120 is 3.51 meters per second squared 00:12:00.120 --> 00:12:01.670 and in what direction is it? 00:12:07.470 --> 00:12:09.636 What direction is that acceleration? 00:12:12.384 --> 00:12:13.300 AUDIENCE: The normal. 00:12:13.300 --> 00:12:14.670 PROFESSOR: It's in the normal direction. 00:12:14.670 --> 00:12:17.200 Is it to the inside of the curve or the outside of the curve? 00:12:17.200 --> 00:12:17.690 AUDIENCE: Inside. 00:12:17.690 --> 00:12:18.440 PROFESSOR: Inside. 00:12:18.440 --> 00:12:20.266 It's always to the inside of the curve. 00:12:20.266 --> 00:12:23.800 So these tangent normal coordinates 00:12:23.800 --> 00:12:26.190 are really simple coordinates. 00:12:26.190 --> 00:12:28.620 They're meant for a particular kind of simple problem 00:12:28.620 --> 00:12:30.650 when you know the path. 00:12:30.650 --> 00:12:34.350 And it's just defined, the normal un is positive, 00:12:34.350 --> 00:12:36.620 always inward to the center of the curve, 00:12:36.620 --> 00:12:38.910 pointing toward the center of rotation 00:12:38.910 --> 00:12:41.450 where that radius of-- where your radius of curve 00:12:41.450 --> 00:12:44.640 is, always pointing at the origin 00:12:44.640 --> 00:12:48.920 of that radius of your curve. 00:12:48.920 --> 00:12:52.820 So g is order of 10 meters per second squared, 00:12:52.820 --> 00:12:56.265 so that's about a third of a g. 00:12:56.265 --> 00:13:00.690 Is 1/3 of a g enough to notice? 00:13:00.690 --> 00:13:01.737 Yeah, absolutely! 00:13:01.737 --> 00:13:03.820 You'll get-- when you push to the side of the car, 00:13:03.820 --> 00:13:04.896 you do that . 00:13:08.030 --> 00:13:10.440 So let's move-- any questions about that? 00:13:10.440 --> 00:13:11.640 I'm going to move on next. 00:13:11.640 --> 00:13:14.435 If not, I'm going to move on to linear impulse and momentum. 00:13:23.140 --> 00:13:28.860 And this will be a quick review because this 00:13:28.860 --> 00:13:35.104 consists of the kind of physics that you've done lots of times. 00:13:35.104 --> 00:13:37.020 We're going to hit it quickly and then move on 00:13:37.020 --> 00:13:37.853 to angular momentum. 00:13:41.390 --> 00:13:43.880 So we know for a particle-- this is what Newton told us. 00:13:43.880 --> 00:13:47.830 We've got a particle here and some mass, m. 00:13:47.830 --> 00:13:51.220 We know for a particle-- and it has some external forces on it. 00:13:54.480 --> 00:13:59.410 We know for a particle that the sum of these external forces-- 00:13:59.410 --> 00:14:03.710 it's a vector sum-- is equal to the mass 00:14:03.710 --> 00:14:07.280 times the acceleration for the particle, Newton's Second Law. 00:14:13.010 --> 00:14:17.880 And that's the mass times the derivative of the velocity 00:14:17.880 --> 00:14:19.000 with respect to time. 00:14:19.000 --> 00:14:21.830 That's where we get acceleration. 00:14:21.830 --> 00:14:24.810 So this formula we can just rearrange the dt's 00:14:24.810 --> 00:14:25.400 a little bit. 00:14:25.400 --> 00:14:35.360 So the summation of the forces, dt, is m dv. 00:14:35.360 --> 00:14:36.650 And I want to integrate these. 00:14:39.410 --> 00:14:44.670 So if I integrate this over time, from t1 to t2, 00:14:44.670 --> 00:14:49.150 this equality says we're going to find some change in velocity 00:14:49.150 --> 00:14:53.650 from v1 to v2. 00:14:53.650 --> 00:14:55.820 And this is our impulse-- this is 00:14:55.820 --> 00:14:59.530 the beginnings of our impulse-momentum relationship. 00:14:59.530 --> 00:15:03.620 We tend to call the integration of forces over time, impulse. 00:15:03.620 --> 00:15:07.280 And if you have a nonzero impulse, 00:15:07.280 --> 00:15:10.190 that leads to a change in the linear momentum. 00:15:13.660 --> 00:15:19.520 So when you carry it out so writing 00:15:19.520 --> 00:15:22.370 this a little bit different way, you can do summation 00:15:22.370 --> 00:15:25.940 because these are vectors, and we have rules about vectors. 00:15:25.940 --> 00:15:29.930 You can bring the summation outside of the integral. 00:15:29.930 --> 00:15:33.930 So that this says that the summation of the integral 00:15:33.930 --> 00:15:41.340 from t1 to t2 of these external forces 00:15:41.340 --> 00:15:46.853 is just mv2 minus mv1 because the integral 00:15:46.853 --> 00:15:51.390 on the right-hand side is really simple. 00:15:51.390 --> 00:15:56.280 And we'll finally state this in a way 00:15:56.280 --> 00:16:00.490 that we most commonly use it, moving this term to the left. 00:16:00.490 --> 00:16:04.950 So you start off with an initial momentum, mv1. 00:16:04.950 --> 00:16:11.750 You add to it the summation of the impulses that 00:16:11.750 --> 00:16:18.820 occur over time, t1, t2 of the external forces, 00:16:18.820 --> 00:16:22.670 and you get the final momentum mv2. 00:16:22.670 --> 00:16:27.544 And this is the way you usually use the formula. 00:16:27.544 --> 00:16:29.460 So what happens if there's no external forces? 00:16:32.680 --> 00:16:36.700 That's where we get the law of conservation of momentum. 00:16:36.700 --> 00:16:39.940 If there are no external forces on the particle, 00:16:39.940 --> 00:16:44.220 the momentum doesn't change and mv1 has got to be equal to mv2. 00:17:02.260 --> 00:17:07.390 So just a really trivial example drawing for something 00:17:07.390 --> 00:17:08.380 that we've done before. 00:17:08.380 --> 00:17:12.069 You got the block sliding down the hill. 00:17:12.069 --> 00:17:15.444 Draw your free body diagram. 00:17:15.444 --> 00:17:22.120 You got friction, got gravity, got a normal force. 00:17:27.010 --> 00:17:29.930 And we'll set ourselves up with a coordinate system aligned 00:17:29.930 --> 00:17:31.590 with the motion we're interested in. 00:17:31.590 --> 00:17:34.730 So this is an inertial system, x,y. 00:17:37.810 --> 00:17:39.380 And we've done this problem before. 00:17:39.380 --> 00:17:41.600 So we said the summation-- we found out 00:17:41.600 --> 00:17:46.160 that the summation of the forces in the x direction-- and this 00:17:46.160 --> 00:17:49.460 is a good moment to take an aside for a second. 00:17:49.460 --> 00:17:53.160 This was a vector expression. 00:17:53.160 --> 00:17:56.456 But you can break it-- you can implement it-- 00:17:56.456 --> 00:18:00.140 you can break it down into its individual vector components. 00:18:00.140 --> 00:18:02.220 So that equation gives you-- for particles-- 00:18:02.220 --> 00:18:06.220 gives you three sub-equations one in the x, one in the y, 00:18:06.220 --> 00:18:06.870 one in the z. 00:18:06.870 --> 00:18:09.030 And you can use them each independently. 00:18:09.030 --> 00:18:13.020 So in this case, we're going to use-- only need the x component 00:18:13.020 --> 00:18:14.320 in order to do the problem. 00:18:14.320 --> 00:18:17.400 So the sum of the forces in the x direction, 00:18:17.400 --> 00:18:20.860 in this problem which we had done before, 00:18:20.860 --> 00:18:28.260 is mg sine theta, that component of gravity pulling it 00:18:28.260 --> 00:18:35.360 down the hill, minus mu mg cosine theta. 00:18:35.360 --> 00:18:37.270 And that's the friction. 00:18:37.270 --> 00:18:41.490 And none of these are functions of time. 00:18:41.490 --> 00:18:44.681 So that whole thing is just some constant. 00:18:44.681 --> 00:18:47.980 I'll just call it k. 00:18:47.980 --> 00:18:51.149 So the total forces on the system in the x direction 00:18:51.149 --> 00:18:52.065 is just some constant. 00:18:52.065 --> 00:18:54.360 And that makes this integral up here pretty trivial 00:18:54.360 --> 00:18:56.140 to implement. 00:18:56.140 --> 00:19:03.530 So now we can say that mv1-- I'll make it v1x to emphasize 00:19:03.530 --> 00:19:07.020 that integral from 0, t1 to t2.. 00:19:07.020 --> 00:19:10.470 But I'm going to let t1 be 0 because it 00:19:10.470 --> 00:19:11.700 makes the problems easier. 00:19:11.700 --> 00:19:21.320 So from 0 to t of k dt equals mv2-- 00:19:21.320 --> 00:19:25.580 I can't write this morning-- in the x direction. 00:19:25.580 --> 00:19:34.390 And at v1, x is 0, starts off at 0 time and 0 velocity 00:19:34.390 --> 00:19:36.490 when you let it go to start with. 00:19:36.490 --> 00:19:38.350 Then, this term will go away. 00:19:38.350 --> 00:19:42.310 And we just implement this integral. 00:19:42.310 --> 00:19:46.960 And the integral of k dt is? 00:19:46.960 --> 00:19:47.940 kt, right? 00:19:47.940 --> 00:19:55.270 So this says then that kt evaluated from 0 to some time 00:19:55.270 --> 00:20:01.810 that we want to know the answer is mv2 in the x direction. 00:20:01.810 --> 00:20:05.930 And so let's let t equal 3 seconds. 00:20:08.470 --> 00:20:18.618 You find out that v 2x is 3k over m. 00:20:22.610 --> 00:20:30.760 And in this problem, then that looks like 3g sine theta 00:20:30.760 --> 00:20:35.670 minus mu cosine theta. 00:20:35.670 --> 00:20:42.870 So it's really an almost trivial example. 00:20:42.870 --> 00:20:44.990 It's not a hard example, but it emphasizes 00:20:44.990 --> 00:20:47.750 all of the key points in the problem. 00:20:47.750 --> 00:20:49.600 Start off with a vector equation. 00:20:49.600 --> 00:20:53.520 You can apply it in any one of the three vector component 00:20:53.520 --> 00:20:54.830 directions. 00:20:54.830 --> 00:20:56.870 We integrate the forces-- the sum 00:20:56.870 --> 00:21:00.160 of the forces on the object in that direction over time. 00:21:00.160 --> 00:21:02.860 And you apply the impulse-momentum formula, 00:21:02.860 --> 00:21:04.590 and you get the answer. 00:21:04.590 --> 00:21:05.320 You can do that. 00:21:05.320 --> 00:21:07.290 So that basic step-by-step process 00:21:07.290 --> 00:21:11.670 is how basically most impulse-momentum problems 00:21:11.670 --> 00:21:12.670 are done. 00:21:12.670 --> 00:21:14.390 And if there's no forces, then you 00:21:14.390 --> 00:21:15.750 have conservation of momentum. 00:21:19.800 --> 00:21:23.030 So let's do-- this was for particles. 00:21:37.390 --> 00:21:41.780 Just a quick reminder of we need to ask 00:21:41.780 --> 00:21:45.260 ourselves does this apply to groups of particles, systems 00:21:45.260 --> 00:21:46.360 of particles. 00:21:46.360 --> 00:21:48.590 Well, remember, that we said if you've got a bunch 00:21:48.590 --> 00:21:54.210 of particles, a system-- all of these are 00:21:54.210 --> 00:21:59.750 the mi''s-- we've already figured out that the total mass 00:21:59.750 --> 00:22:05.070 of the system times the velocity of the center of mass-- 00:22:05.070 --> 00:22:14.990 so g is my center of mass with respect to my inertial frame. 00:22:14.990 --> 00:22:19.330 So this is the momentum of the system that was just 00:22:19.330 --> 00:22:23.960 a summation of the individual mivi's with respect 00:22:23.960 --> 00:22:30.130 to-- And furthermore, we took the derivative of this, 00:22:30.130 --> 00:22:32.220 so this is the momentum of the system. 00:22:32.220 --> 00:22:34.240 The time derivative of the momentum 00:22:34.240 --> 00:22:38.810 should be the external forces so that we were able to-- 00:22:38.810 --> 00:22:43.810 by taking that derivative, you get mt, the total mass, 00:22:43.810 --> 00:22:47.950 times vg with respect to o dot. 00:22:47.950 --> 00:22:55.160 And that had better give you v equal to the summations 00:22:55.160 --> 00:23:00.466 of the external forces on all the particles 00:23:00.466 --> 00:23:02.340 because you can do it one particle at a time. 00:23:02.340 --> 00:23:05.890 And they all add-- each of these particles has forces. 00:23:05.890 --> 00:23:07.370 You sum them all up. 00:23:07.370 --> 00:23:12.920 You get these forces, and this allows 00:23:12.920 --> 00:23:17.290 you to say that the sum of all of the external forces 00:23:17.290 --> 00:23:19.540 on the system is equal to the total mass 00:23:19.540 --> 00:23:21.260 of the system times the acceleration 00:23:21.260 --> 00:23:23.860 of the center of mass. 00:23:23.860 --> 00:23:27.420 So this formula now can-- it looks 00:23:27.420 --> 00:23:31.380 very similar to where we started with this little derivation 00:23:31.380 --> 00:23:32.810 for a particle. 00:23:32.810 --> 00:23:36.350 You can now say that the summation-- this all 00:23:36.350 --> 00:23:41.830 leads to the summation of the integral 00:23:41.830 --> 00:23:52.570 from t1 to t2 of all of these courses over time-- 00:23:52.570 --> 00:23:58.030 gives you the change in the total linear momentum 00:23:58.030 --> 00:24:05.910 of the system, vg with respect to o1, 2 rather, 00:24:05.910 --> 00:24:12.160 minus vg with respect to o2, the second one. 00:24:12.160 --> 00:24:14.540 So this is exactly the same kind of formulation. 00:24:14.540 --> 00:24:20.660 The change in the linear momentum of the system 00:24:20.660 --> 00:24:24.730 is equal to the integral of the forces on the system over time, 00:24:24.730 --> 00:24:26.570 so the impulse to the system. 00:24:26.570 --> 00:24:30.410 So this allows you to do problems that you might not 00:24:30.410 --> 00:24:33.700 have thought about before. 00:24:33.700 --> 00:24:36.550 So I think about something like this. 00:24:36.550 --> 00:24:45.345 I've got an old Revolutionary War earlier period canon. 00:24:50.280 --> 00:24:51.910 I'm trying to draw it 00:24:51.910 --> 00:24:54.740 So here's my cannon barrel. 00:24:54.740 --> 00:25:03.870 And back in those days-- I'll do a little better job here. 00:25:03.870 --> 00:25:05.610 So we got this cannon sitting here. 00:25:05.610 --> 00:25:07.870 And in the barrel, you've got a bunch 00:25:07.870 --> 00:25:13.800 of what's known as-- in the old days, known as grapeshot. 00:25:13.800 --> 00:25:17.620 So sometimes, for anti personnel stuff, 00:25:17.620 --> 00:25:20.000 they would just throw a whole bunch of metal and junk 00:25:20.000 --> 00:25:20.530 into it. 00:25:20.530 --> 00:25:23.090 But just imagine a bunch of iron balls 00:25:23.090 --> 00:25:26.274 or lead ball rocks or whatever you want to put in there. 00:25:26.274 --> 00:25:28.940 And you put a whole mess of them in here, and you have a charge. 00:25:28.940 --> 00:25:31.807 You set it off, and it shoots them out the end of the gun. 00:25:31.807 --> 00:25:33.265 I want to know what's the reaction. 00:25:46.190 --> 00:25:48.210 What's the reaction force on the gun? 00:25:51.409 --> 00:25:53.200 When that gun-- when that charge goes off-- 00:25:53.200 --> 00:25:55.970 if you go down to the Constitution, which 00:25:55.970 --> 00:26:00.710 is the oldest commissioned warship in the US Navy 00:26:00.710 --> 00:26:02.850 still afloat-- I don't know if any of you 00:26:02.850 --> 00:26:05.130 have seen it down here in the dock in Chelsea. 00:26:05.130 --> 00:26:07.860 It was built around 1799. 00:26:07.860 --> 00:26:11.160 But when they shot one of those guns, 00:26:11.160 --> 00:26:15.600 the gun would roll back several feet and get dragged to a stop 00:26:15.600 --> 00:26:19.220 by a bunch of restraining lines and then dragged back up 00:26:19.220 --> 00:26:19.940 for another load. 00:26:19.940 --> 00:26:22.920 So the reaction force, the force pushing back-- 00:26:22.920 --> 00:26:25.640 these things were usually sitting on wheels-- 00:26:25.640 --> 00:26:26.642 could be pretty large. 00:26:26.642 --> 00:26:27.850 Let's take an estimate of it. 00:26:27.850 --> 00:26:29.845 So this is going to be a relatively small gun. 00:26:34.180 --> 00:26:40.270 The total mass of the shot here times g 00:26:40.270 --> 00:26:44.300 is 10 pounds, so 10 pounds a shot. 00:26:44.300 --> 00:26:45.840 A gallon of milk weighs 8 pounds, 00:26:45.840 --> 00:26:49.650 so this isn't a very big load. 00:26:49.650 --> 00:26:53.700 And let's say that this barrel here, from here to here, 00:26:53.700 --> 00:26:58.840 the acceleration length is 7 feet. 00:26:58.840 --> 00:27:02.220 So that shot is accelerated out of the barrel 00:27:02.220 --> 00:27:05.060 over a distance of 7 feet. 00:27:05.060 --> 00:27:16.270 And it has a muzzle velocity, an exit velocity, 00:27:16.270 --> 00:27:19.750 of 700 feet per second. 00:27:22.740 --> 00:27:26.140 Most guns, subsonic, supersonic, projectiles 00:27:26.140 --> 00:27:28.530 when they come out of guns, have you thought about that? 00:27:28.530 --> 00:27:32.240 I'm just asking-- seeing if you have a feeling for speed. 00:27:32.240 --> 00:27:36.362 And how close to the speed of sound is that? 00:27:36.362 --> 00:27:37.278 AUDIENCE: [INAUDIBLE]. 00:27:37.278 --> 00:27:38.736 PROFESSOR: So the speed of sound is 00:27:38.736 --> 00:27:43.100 about 1,100 feet per second, 340 meters per second. 00:27:43.100 --> 00:27:46.030 So this is order of mach 0.6 or something like that. 00:27:52.170 --> 00:27:56.540 Something like 0.6, not 0.060, 0.6. 00:27:56.540 --> 00:27:58.610 So about 60% of the speed of sound. 00:27:58.610 --> 00:28:00.490 That's slow actually as guns go. 00:28:03.800 --> 00:28:12.510 So the initial velocity-- the shot's just sitting theree-- 00:28:12.510 --> 00:28:13.640 is 0. 00:28:13.640 --> 00:28:21.350 The final velocity is 700 feet per second. 00:28:21.350 --> 00:28:23.935 The average velocity, which I'm going to need for a second 00:28:23.935 --> 00:28:25.810 because I'm just making some estimates here-- 00:28:25.810 --> 00:28:28.940 the average velocity, average of 0 plus 700 00:28:28.940 --> 00:28:30.690 is 350 feet per second. 00:28:33.650 --> 00:28:36.392 The reason I need this is I need to estimate the delta t. 00:28:36.392 --> 00:28:38.725 How long does it take to get the shot out of the barrel? 00:28:42.120 --> 00:28:45.120 So distance equals rate times time, right? 00:28:45.120 --> 00:28:56.260 So the v average times delta t equals 7 feet. 00:28:56.260 --> 00:28:59.050 At 350 feet per second, you find out 00:28:59.050 --> 00:29:09.050 delta t is about 0.02 seconds. 00:29:09.050 --> 00:29:12.950 So it gets out the barrel pretty quickly, 0.02 seconds, 00:29:12.950 --> 00:29:15.110 20 milliseconds. 00:29:15.110 --> 00:29:17.870 Now, that powder, when it goes off, 00:29:17.870 --> 00:29:21.420 it's putting a lot of forces on the shot. 00:29:21.420 --> 00:29:24.900 So if the total force on the shot is 5,000 pounds, 00:29:24.900 --> 00:29:27.030 what's the reaction force on the cannon? 00:29:35.040 --> 00:29:38.520 So we're going to apply a principle here 00:29:38.520 --> 00:29:40.890 that we talked about earlier. 00:29:40.890 --> 00:29:43.990 Newton had three laws. 00:29:43.990 --> 00:29:44.980 This is a group. 00:29:44.980 --> 00:29:49.970 This is just massive balls in there. 00:29:49.970 --> 00:29:53.190 The hot exploding gases are pushing them out the barrel. 00:29:53.190 --> 00:29:57.920 What must be the push of the same gas on the canon that's 00:29:57.920 --> 00:29:58.630 containing it? 00:29:58.630 --> 00:29:59.710 AUDIENCE: Equal and opposite. 00:29:59.710 --> 00:30:01.470 PROFESSOR: Equal and opposite, Newton's Third Law. 00:30:01.470 --> 00:30:01.970 Right. 00:30:01.970 --> 00:30:08.290 So the reaction force that we're looking for is minus the force 00:30:08.290 --> 00:30:10.290 that it takes to get the shot out of the barrel. 00:30:13.260 --> 00:30:19.890 So the force on the balls, on the shot, 00:30:19.890 --> 00:30:22.440 minus the reaction force. 00:30:22.440 --> 00:30:24.040 And that's Newton's Third Law. 00:30:31.400 --> 00:30:33.840 So we're almost there. 00:30:33.840 --> 00:30:35.760 We're just applying now this concept 00:30:35.760 --> 00:30:36.910 of impulse and momentum. 00:30:40.990 --> 00:30:44.860 So now we can say the integral-- and this is a summation. 00:30:44.860 --> 00:30:47.380 You got a bunch of balls in there. 00:30:47.380 --> 00:30:49.170 They all got forces on them, but we're 00:30:49.170 --> 00:30:51.660 treating the group of balls as a system, 00:30:51.660 --> 00:30:55.370 so we can use this notion of the total mass 00:30:55.370 --> 00:31:00.250 times the velocity of the center of gravity to do this problem. 00:31:00.250 --> 00:31:11.280 So this is the force on all of these little balls integrated 00:31:11.280 --> 00:31:11.970 over time. 00:31:14.560 --> 00:31:19.970 It's going to be the mass total of the balls times the change 00:31:19.970 --> 00:31:24.000 in the velocity of the center of gravity, the final velocity 00:31:24.000 --> 00:31:25.860 minus the initial velocity. 00:31:28.460 --> 00:31:29.225 This is 0. 00:31:32.800 --> 00:31:34.010 We know this. 00:31:34.010 --> 00:31:36.900 We know this. 00:31:36.900 --> 00:31:41.570 So this mt is the 10 pounds over g. 00:31:41.570 --> 00:31:44.470 We know this is 700 feet per second. 00:31:44.470 --> 00:31:46.980 We know that all of this mounts up 00:31:46.980 --> 00:31:54.700 to the force external on the group delta t. 00:31:58.700 --> 00:32:01.620 And this is the weight of the ball, w/g. 00:32:04.330 --> 00:32:10.430 And this is 700 feet per second. 00:32:10.430 --> 00:32:15.680 That's 10 pounds divided by gravity, 32.12. 00:32:15.680 --> 00:32:18.210 And we need to divide by delta t. 00:32:18.210 --> 00:32:21.990 So the external force on the balls 00:32:21.990 --> 00:32:32.710 is w/g delta t times 700 feet per second. 00:32:32.710 --> 00:32:45.170 And if you work out those numbers, you get 10,870 pounds. 00:32:45.170 --> 00:32:56.180 Now, this is an average force because we-- 00:32:56.180 --> 00:32:59.280 that delta t is the length of time 00:32:59.280 --> 00:33:04.130 it took, assuming we had some-- we assumed an average velocity 00:33:04.130 --> 00:33:05.250 going down the barrel. 00:33:05.250 --> 00:33:07.820 So the peak force is probably higher than this, 00:33:07.820 --> 00:33:10.340 and the pressure in the barrel probably isn't constant. 00:33:10.340 --> 00:33:12.610 This actually isn't a bad estimate. 00:33:12.610 --> 00:33:15.610 If you know the muzzle velocity, which you can figure out 00:33:15.610 --> 00:33:18.436 probably from the distance it goes and things like that, 00:33:18.436 --> 00:33:19.560 you can make this estimate. 00:33:19.560 --> 00:33:24.870 So 10,000 pounds of reaction force is quite a lot. 00:33:24.870 --> 00:33:27.020 This is a dinky gun. 00:33:27.020 --> 00:33:29.955 10 pounds a shot is not much. 00:33:29.955 --> 00:33:32.850 This is the force to get the balls out the barrel. 00:33:32.850 --> 00:33:35.100 If it's positive, what's the reaction force? 00:33:39.630 --> 00:33:42.340 Minus 10,870. 00:33:42.340 --> 00:33:45.080 And all through this, I've been doing this in a single vector 00:33:45.080 --> 00:33:46.920 direction, so I would have set up 00:33:46.920 --> 00:33:49.460 my coordinate system like here's o. 00:33:49.460 --> 00:33:52.780 This is the x direction, y direction. 00:33:52.780 --> 00:33:54.700 Obviously, in the positive x direction 00:33:54.700 --> 00:33:57.305 is what I've lined up my positive velocities 00:33:57.305 --> 00:33:59.670 and positive forces to be. 00:33:59.670 --> 00:34:01.241 So this is a really simple way. 00:34:01.241 --> 00:34:02.740 One of the kind of ways in which you 00:34:02.740 --> 00:34:05.610 use this notion of impulse and momentum 00:34:05.610 --> 00:34:07.290 that you can use to make estimates too. 00:34:12.830 --> 00:34:16.921 Any of you ever fired a shot gun. 00:34:16.921 --> 00:34:17.420 Not many. 00:34:17.420 --> 00:34:18.909 There's-- yeah. 00:34:18.909 --> 00:34:20.131 Any kick? 00:34:20.131 --> 00:34:20.630 Right. 00:34:20.630 --> 00:34:23.240 So what gauge shotgun did you fire? 00:34:23.240 --> 00:34:23.909 You don't know. 00:34:23.909 --> 00:34:24.409 OK. 00:34:24.409 --> 00:34:25.786 A 12-gauge shot gun. 00:34:25.786 --> 00:34:29.734 12-gauge shot gun, the cartridge is just about 3/4 00:34:29.734 --> 00:34:31.900 of an inch diameter and probably has a little amount 00:34:31.900 --> 00:34:32.750 of powder in there. 00:34:32.750 --> 00:34:34.747 It amounts up to about that much stuff. 00:34:34.747 --> 00:34:36.580 But it can give you a bruise in the shoulder 00:34:36.580 --> 00:34:38.739 if you don't hold it right. 00:34:38.739 --> 00:34:40.570 So that's what this is about. 00:34:43.610 --> 00:34:46.860 So this quick review, you've done 00:34:46.860 --> 00:34:49.710 lots of conservation of momentum problems in your time. 00:34:49.710 --> 00:34:53.690 The homework set this time has two or three problems on it 00:34:53.690 --> 00:34:56.850 that are conservation of momentum, linear momentum. 00:34:56.850 --> 00:35:01.990 But now I want to move on to talking about angular momentum. 00:35:01.990 --> 00:35:05.480 And angular momentum, in a way in which you probably 00:35:05.480 --> 00:35:09.060 haven't done angular momentum problems before. 00:35:09.060 --> 00:35:13.290 So anything about last thing on this. 00:35:13.290 --> 00:35:16.040 So mostly this, I want you to read the chapter. 00:35:16.040 --> 00:35:19.926 I think it's chapter 15. 00:35:19.926 --> 00:35:22.540 The first few sections of it are on linear momentum, 00:35:22.540 --> 00:35:24.880 the last few sections on angular momentum. 00:35:24.880 --> 00:35:28.160 Go read them and just work the problems. 00:35:28.160 --> 00:35:30.630 That's where you'll get most of your refresher 00:35:30.630 --> 00:35:33.620 is doing the practice problems. 00:35:33.620 --> 00:35:34.440 Angular momentum. 00:36:02.462 --> 00:36:03.670 So we'll start with particle. 00:36:03.670 --> 00:36:06.560 We're very rapidly going to get to rigid bodies. 00:36:11.990 --> 00:36:15.250 Now many times, you've done angular momentum problems 00:36:15.250 --> 00:36:18.610 before, mostly rotations about fixed axes. 00:36:18.610 --> 00:36:24.300 So here's our inertial frame, fixed x,y and you have 00:36:24.300 --> 00:36:27.170 a particle out here. 00:36:27.170 --> 00:36:31.240 And it just has some mass, m. 00:36:31.240 --> 00:36:35.009 And it has some total force on it, the force. 00:36:35.009 --> 00:36:36.550 And this says the particle is located 00:36:36.550 --> 00:36:41.180 at B here, so the force at B, total force at B, 00:36:41.180 --> 00:36:44.370 just some vector. 00:36:44.370 --> 00:36:47.580 It also is traveling with some velocity at that instant 00:36:47.580 --> 00:36:50.720 in time, so it has momentum. 00:36:50.720 --> 00:36:58.480 Momentum is p of this particle at B with respect to o. 00:37:02.630 --> 00:37:05.640 And I'd like to-- the standard expression 00:37:05.640 --> 00:37:13.960 then for the angular momentum of this particle at B 00:37:13.960 --> 00:37:23.700 with respect to o-- we have a position vector here, 00:37:23.700 --> 00:37:29.190 r Bo, which we've used lots of times by now, 00:37:29.190 --> 00:37:35.610 is just the cross product of the position vector 00:37:35.610 --> 00:37:37.540 with the linear momentum. 00:37:43.520 --> 00:37:46.400 And that's the definition of angular momentum. 00:37:46.400 --> 00:37:48.630 I'm using a lowercase h here, so they're 00:37:48.630 --> 00:37:50.970 going to do that to indicate single particles. 00:37:50.970 --> 00:37:53.200 And I'll use a capital H later on when we're 00:37:53.200 --> 00:37:55.010 referring to rigid bodies. 00:37:55.010 --> 00:37:58.000 So the angular momentum of this particle, 00:37:58.000 --> 00:38:01.234 with respect to this point, is given by that. 00:38:05.170 --> 00:38:07.710 And where you've used this before, 00:38:07.710 --> 00:38:11.630 then, it's the derivative of hB with respect 00:38:11.630 --> 00:38:17.830 to o dt is-- what's that? 00:38:17.830 --> 00:38:19.400 Remember, where does this give you? 00:38:19.400 --> 00:38:22.340 The time rate of change of the angular momentum is the? 00:38:22.340 --> 00:38:23.007 AUDIENCE: Torque 00:38:23.007 --> 00:38:23.756 PROFESSOR: Torque. 00:38:23.756 --> 00:38:24.330 Right. 00:38:24.330 --> 00:38:25.240 And it's the torque. 00:38:25.240 --> 00:38:29.150 It's a sum of the torques applied 00:38:29.150 --> 00:38:33.710 on that object with respect to the coordinate system 00:38:33.710 --> 00:38:36.540 in which you're computing the angular momentum. 00:38:36.540 --> 00:38:38.990 So that's-- we've used this formula many times. 00:38:38.990 --> 00:38:41.360 And in planar motion where you only 00:38:41.360 --> 00:38:46.410 have one axis of rotation using the z-axis of rotation, 00:38:46.410 --> 00:38:50.830 you usually would write this as I theta double dot. 00:38:50.830 --> 00:38:52.850 In simplest form, if it's a rigid body, 00:38:52.850 --> 00:38:55.260 it has a mass moment of inertia times 00:38:55.260 --> 00:38:58.190 theta double dot is equal to the sum of the external torque. 00:38:58.190 --> 00:39:00.337 So that's where you've met this before. 00:39:00.337 --> 00:39:02.170 But for the moment, this is just a particle. 00:39:02.170 --> 00:39:05.480 Let's stick with a particle. 00:39:05.480 --> 00:39:09.080 So the piece that's new here-- probably new 00:39:09.080 --> 00:39:18.440 for you-- is that what if I want to know the angular momentum 00:39:18.440 --> 00:39:21.630 with respect to another point? 00:39:21.630 --> 00:39:32.830 So here's a point A. I'd like to compute hB with respect to A. 00:39:32.830 --> 00:39:45.120 Well, that's rB with respect to A cross PB with respect to o. 00:39:45.120 --> 00:39:47.460 And this is really easy to forget. 00:39:47.460 --> 00:39:50.530 The momentum is always calculated with respect 00:39:50.530 --> 00:39:53.670 to your inertial frame. 00:39:53.670 --> 00:39:56.470 And that's why I keep in this with respect to 00:39:56.470 --> 00:39:59.710 and telling you what the frame is is pretty important. 00:39:59.710 --> 00:40:02.310 But we're out here at some arbitrary point, 00:40:02.310 --> 00:40:06.660 computing the cross product of the position vector 00:40:06.660 --> 00:40:10.870 from this arbitrary point to this moving mass. 00:40:10.870 --> 00:40:13.070 And we're defining-- it's just a definition 00:40:13.070 --> 00:40:16.910 defining the angular momentum with respect to this point A 00:40:16.910 --> 00:40:21.200 as r B/A cross PB with respect to o. 00:40:31.570 --> 00:40:42.680 And what I want to get to is now is the torque on this system, 00:40:42.680 --> 00:40:49.810 around this particle, B with respect to A 00:40:49.810 --> 00:40:57.260 is the time rate of change of h B/A with respect to t. 00:40:57.260 --> 00:41:00.250 But now it's a little more complicated. 00:41:00.250 --> 00:41:02.770 Plus, the velocity of A with respect 00:41:02.770 --> 00:41:12.900 to o-- these are all vectors-- cross PB with respect to o. 00:41:12.900 --> 00:41:13.700 Messy term. 00:41:17.352 --> 00:41:18.810 I was trying to think of an example 00:41:18.810 --> 00:41:23.630 where you might want to do this. 00:41:31.810 --> 00:41:38.320 So imagine that you've got an arm which can rotate. 00:41:38.320 --> 00:41:40.710 It might be on a robot or something like that. 00:41:40.710 --> 00:41:49.510 And it has attached to it another arm with a mass on it. 00:41:49.510 --> 00:41:54.850 And you've got a motor here, which can make this rotate. 00:41:54.850 --> 00:41:57.290 And you're trying to design-- the motor has 00:41:57.290 --> 00:42:00.240 to be able to put out a certain amount of torque. 00:42:00.240 --> 00:42:02.150 So this is o. 00:42:02.150 --> 00:42:05.930 This is B. This is A. And you actually-- 00:42:05.930 --> 00:42:09.880 I want to know the torque required in this motor 00:42:09.880 --> 00:42:12.990 to drive this thing around. 00:42:12.990 --> 00:42:15.380 But the motor is here, and it only 00:42:15.380 --> 00:42:19.420 cares about what it feels, so the torque at this point 00:42:19.420 --> 00:42:21.240 to drive this thing. 00:42:21.240 --> 00:42:24.550 But this whole system is now in motion. 00:42:24.550 --> 00:42:25.980 You'd have to use this formula. 00:42:28.900 --> 00:42:30.380 So there are practical times when 00:42:30.380 --> 00:42:33.360 you'd like to be able to calculate something like this. 00:42:42.870 --> 00:42:44.930 So there's a-- we're going to show 00:42:44.930 --> 00:42:52.530 you a very brief derivation of this 00:42:52.530 --> 00:42:54.620 just so you can get a feeling for where this comes 00:42:54.620 --> 00:42:56.450 from because there's a couple of outcomes 00:42:56.450 --> 00:42:57.890 that are very important to it. 00:43:10.020 --> 00:43:18.480 So the sum of the forces there at-- those forces at B, 00:43:18.480 --> 00:43:24.360 give you the time rate of change of the momentum 00:43:24.360 --> 00:43:26.080 at B with respect to o. 00:43:26.080 --> 00:43:29.930 This is the momentum vector of our particle. 00:43:29.930 --> 00:43:33.882 And this fB, this is the total external forces 00:43:33.882 --> 00:43:34.840 acting on the particle. 00:43:39.460 --> 00:43:42.715 And we know that's m. 00:43:42.715 --> 00:43:55.525 It's a single particle times the velocity of That 00:43:55.525 --> 00:43:56.830 certainly could be right. 00:44:03.920 --> 00:44:08.280 That's our familiar formula for F equals ma. 00:44:08.280 --> 00:44:11.970 So the time derivative of a linear momentum gives you this. 00:44:11.970 --> 00:44:17.690 And the torque of B with respect to A 00:44:17.690 --> 00:44:25.350 is r B/A cross-- I'm going let this just be a total vector. 00:44:25.350 --> 00:44:28.460 I'll call it fB. 00:44:28.460 --> 00:44:30.120 It's a vector. 00:44:30.120 --> 00:44:32.160 The torque with respect to A is just 00:44:32.160 --> 00:44:34.590 r B/A cross this total external force. 00:44:40.940 --> 00:44:53.283 r B/A cross time derivative of P B/o. 00:44:56.220 --> 00:44:58.680 Because the forces give us r if it's 00:44:58.680 --> 00:45:01.730 from the time rate of change of the linear momentum 00:45:01.730 --> 00:45:04.356 of that particle. 00:45:04.356 --> 00:45:06.630 So I can say it like this. 00:45:06.630 --> 00:45:12.640 But now there is just a little vector identity 00:45:12.640 --> 00:45:17.340 for products of vectors that I'm going to take advantage. 00:45:17.340 --> 00:45:31.620 And I'll call this Q. So Q is of the form-- 00:45:31.620 --> 00:45:35.290 and I'm going to say this is a quantity of vector A, 00:45:35.290 --> 00:45:38.600 and this is a quantity of vector-- time 00:45:38.600 --> 00:45:42.950 derivative of a vector B. It's A cross dB dt. 00:45:42.950 --> 00:45:47.080 So there's a little identity that you can use. 00:45:47.080 --> 00:45:55.330 It says A cross dB dt. 00:45:55.330 --> 00:46:01.510 You can alternatively write that as the time derivative 00:46:01.510 --> 00:46:14.499 of A cross B minus time derivative of A cross B. 00:46:14.499 --> 00:46:16.040 We're going to take advantage of this 00:46:16.040 --> 00:46:23.325 and just re-construct this formula using this expression. 00:46:30.540 --> 00:46:37.860 So that says torque of B with respect to A 00:46:37.860 --> 00:46:46.695 is the time derivative of r B/A cross P 00:46:46.695 --> 00:47:02.240 B/o minus the derivative of r B/A cross P B/o. 00:47:02.240 --> 00:47:05.360 So we've just made this substitution down here in terms 00:47:05.360 --> 00:47:09.375 of r B/A and P B/o all vectors. 00:47:12.540 --> 00:47:19.520 We know that r B/A, from all the previous work we've done, 00:47:19.520 --> 00:47:27.620 is just r B/o minus r A/o so the derivative 00:47:27.620 --> 00:47:43.900 of r B/A with respect to time is v B/o minus v A/o 00:47:43.900 --> 00:47:45.210 We're almost there. 00:47:45.210 --> 00:47:46.457 We're almost there. 00:47:46.457 --> 00:47:47.790 I'm going to need another board. 00:48:08.210 --> 00:48:15.700 This quantity here, this is just h B/A. 00:48:15.700 --> 00:48:19.510 This is the angular momentum of the particle with respect to A. 00:48:19.510 --> 00:48:22.145 It's just r cross B, if you recall. 00:48:52.360 --> 00:48:56.840 So then we can rewrite this expression for the torque 00:48:56.840 --> 00:49:02.800 as the time derivative of h of B with respect 00:49:02.800 --> 00:49:08.190 to A. That's because of what I pointed out there. 00:49:08.190 --> 00:49:19.217 Now, this is minus v B/o minus v A/o cross P B/o. 00:49:28.210 --> 00:49:31.670 So P of B with respect to o is just 00:49:31.670 --> 00:49:38.740 m velocity of B with respect to o. 00:49:38.740 --> 00:49:42.980 So v B/o cross P B/o gives you what? 00:49:42.980 --> 00:49:43.480 Nothing. 00:49:43.480 --> 00:49:45.729 Gives you 0 because they are in parallel of the angle, 00:49:45.729 --> 00:49:48.290 and they're in the same direction. 00:49:48.290 --> 00:49:53.230 So you only get a nonzero piece out of this minus times minus 00:49:53.230 --> 00:50:15.030 gives you a plus, and you end up with-- and that's 00:50:15.030 --> 00:50:16.850 what we set out to find. 00:50:21.010 --> 00:50:23.055 I said this is where I was trying to get, 00:50:23.055 --> 00:50:26.090 and now we're there. 00:50:26.090 --> 00:50:27.940 Now importantly, there's a couple 00:50:27.940 --> 00:50:31.410 of special cases of this. 00:50:31.410 --> 00:50:35.570 This can be a nuisance term to have to deal with. 00:50:35.570 --> 00:50:38.080 Lots of times you'd like to be able to get rid of it 00:50:38.080 --> 00:50:40.520 and just be able to go back to that old reliable formula, 00:50:40.520 --> 00:50:44.510 torque is time rate of change of angular momentum. 00:50:44.510 --> 00:50:48.530 So there are two obvious-- maybe obvious conditions 00:50:48.530 --> 00:50:52.220 in which this will go away. 00:50:52.220 --> 00:50:53.840 What's the most obvious one? 00:50:57.340 --> 00:50:59.010 AUDIENCE: [INAUDIBLE] change to 0. 00:50:59.010 --> 00:51:00.745 PROFESSOR: Something, what you say was 0? 00:51:00.745 --> 00:51:02.080 AUDIENCE: One of the terms 00:51:02.080 --> 00:51:02.996 PROFESSOR: Well, yeah. 00:51:02.996 --> 00:51:06.370 This term presumably not. 00:51:06.370 --> 00:51:24.650 This guy, if this is 0, it's just back 00:51:24.650 --> 00:51:25.885 to our old familiar formula. 00:51:29.820 --> 00:51:31.610 That's one case, so case one. 00:51:34.850 --> 00:51:39.550 But now-- actually, this is a really important result 00:51:39.550 --> 00:51:49.370 because A can be anywhere as long as it's not moving. 00:51:49.370 --> 00:51:53.080 So this allows you to do things, talk 00:51:53.080 --> 00:51:56.200 about rotations about fixed axes that 00:51:56.200 --> 00:51:57.560 aren't at the center of mass. 00:52:00.640 --> 00:52:03.260 So if you have a fixed axis of rotation and something 00:52:03.260 --> 00:52:08.810 going around it, that's what this allows you to do. 00:52:08.810 --> 00:52:09.710 That's one case. 00:52:09.710 --> 00:52:12.100 So this is-- 00:52:12.100 --> 00:52:14.150 We'll soon get to rigid bodies. 00:52:14.150 --> 00:52:17.370 Rigid bodies obey exactly the same formulas. 00:52:17.370 --> 00:52:19.740 And you can have a rigid body, now, 00:52:19.740 --> 00:52:23.040 that is not rotating about its center, 00:52:23.040 --> 00:52:26.130 but rotating maybe about its end like this. 00:52:26.130 --> 00:52:29.130 That formula applies if the velocity 00:52:29.130 --> 00:52:31.640 of that axis about which it's rotating, about which you're 00:52:31.640 --> 00:52:34.650 computing, is not moving, then you can just 00:52:34.650 --> 00:52:38.290 use-- you don't have to deal with that messy term. 00:52:38.290 --> 00:52:43.490 So this is one case in which this term goes away. 00:52:43.490 --> 00:52:47.600 The other case is if this velocity is parallel 00:52:47.600 --> 00:52:49.760 to the direction of the momentum. 00:52:52.380 --> 00:52:57.240 And there's a really useful time that that happens. 00:52:57.240 --> 00:53:01.530 So also this formula is true. 00:53:01.530 --> 00:53:09.150 Case two is when v A/o is parallel 00:53:09.150 --> 00:53:12.370 to P B/o, the direction. 00:53:12.370 --> 00:53:14.370 You're going in the same direction. 00:53:14.370 --> 00:53:18.840 Now that happens when A-- it's guaranteed 00:53:18.840 --> 00:53:22.890 to be true if A is at the center of mass 00:53:22.890 --> 00:53:26.920 because the momentum is defined as the mass of the object 00:53:26.920 --> 00:53:32.970 times the velocity of its center of mass, even for rigid bodies. 00:53:32.970 --> 00:53:45.035 So this is true when A is-- and I'll call it 00:53:45.035 --> 00:53:48.298 G-- at the center of mass. 00:53:53.570 --> 00:54:03.840 So this gives us another really important generalization 00:54:03.840 --> 00:54:07.500 that we'll use-- that we make great use of in dynamics. 00:54:22.330 --> 00:54:27.440 And that says that the torque, with respect 00:54:27.440 --> 00:54:34.030 to the center of mass, is time rate of change 00:54:34.030 --> 00:54:42.040 of h with respect to G. Now, I'm not 00:54:42.040 --> 00:54:44.860 going to go-- I did the proof-- went through this little proof 00:54:44.860 --> 00:54:46.340 just for a particle. 00:54:46.340 --> 00:54:49.540 But by summing a bunch of particles 00:54:49.540 --> 00:54:51.730 and going through all the summations, as we 00:54:51.730 --> 00:54:54.870 did, to prove the center of mass formula, 00:54:54.870 --> 00:54:59.490 you can show-- this allows you to very quickly show 00:54:59.490 --> 00:55:02.705 that this formulation is also true for rigid bodies. 00:55:05.730 --> 00:55:21.050 So the way you say it for rigid bodies is 00:55:21.050 --> 00:55:24.040 that the sum of the torques, with respect 00:55:24.040 --> 00:55:33.930 to some point for a rigid body, is capital H dot with respect 00:55:33.930 --> 00:55:42.960 to A plus the velocity of A with respect to o cross P. 00:55:42.960 --> 00:55:46.890 And now I'm going to say G, its center of mass, center 00:55:46.890 --> 00:55:50.588 of gravity, with respect to o. 00:55:50.588 --> 00:55:53.940 So the same statement for a rigid body 00:55:53.940 --> 00:55:56.810 is that the torque with respect to some point A, which 00:55:56.810 --> 00:56:01.020 can be moving now-- even accelerating-- 00:56:01.020 --> 00:56:03.520 is the time rate of change of the angular 00:56:03.520 --> 00:56:07.240 momentum with respect to A of that rigid body plus v A/o 00:56:07.240 --> 00:56:10.220 cross P B/o. 00:56:10.220 --> 00:56:15.070 And again, when would this messy second term go to 0? 00:56:15.070 --> 00:56:21.100 When the velocity of A is 0, fixed axis rotation, or when 00:56:21.100 --> 00:56:24.450 this is parallel to that, which is true for the center of mass 00:56:24.450 --> 00:56:25.960 always. 00:56:25.960 --> 00:56:31.650 So the same two special cases apply for rigid bodies 00:56:31.650 --> 00:56:37.920 when velocity of A with respect to 0 equals 0 00:56:37.920 --> 00:56:52.580 or when the velocity of A with respect to o is parallel the P. 00:56:52.580 --> 00:56:56.895 And the most important case of that is always true. 00:57:12.850 --> 00:57:13.890 It's always the case. 00:57:13.890 --> 00:57:15.560 And in these you can say this. 00:57:15.560 --> 00:57:19.590 Then, the torque, with respect to A, 00:57:19.590 --> 00:57:25.840 is dH, with respect to A dt. 00:57:25.840 --> 00:57:27.017 No second terms. 00:57:31.310 --> 00:57:35.610 And it's those kinds of-- you've applied, 00:57:35.610 --> 00:57:39.270 generally in your physics like in 801, you've used formulas 00:57:39.270 --> 00:57:40.120 like this a lot. 00:57:42.730 --> 00:57:44.220 Done problems either with respect 00:57:44.220 --> 00:57:51.500 to the center of mass so objects were doing things like that. 00:57:51.500 --> 00:57:54.740 You'll do the torque formulas with respect 00:57:54.740 --> 00:57:55.760 to the center of mass. 00:57:55.760 --> 00:58:00.900 Or when you have things that are pinned to points and rotate 00:58:00.900 --> 00:58:05.230 about fixed axes, then use the other formulation 00:58:05.230 --> 00:58:10.730 where it is with respect to a non-moving point. 00:58:10.730 --> 00:58:12.230 So this is fixed axis rotation. 00:58:27.660 --> 00:58:30.370 So that's the dry derivation part of it. 00:58:30.370 --> 00:58:33.100 I'm going to see if I can find an example here. 00:58:42.430 --> 00:58:44.029 This is quite a bit like a number 00:58:44.029 --> 00:58:45.070 of the homework problems. 00:59:18.620 --> 00:59:20.210 So I've got a carnival ride. 00:59:24.210 --> 00:59:26.470 I got a bar. 00:59:26.470 --> 00:59:29.890 And it's got a seat out here. 00:59:29.890 --> 00:59:32.240 You're riding in it. 00:59:32.240 --> 00:59:33.660 So you're taking this ride. 00:59:38.510 --> 00:59:40.010 It's rotating. 00:59:40.010 --> 00:59:42.330 This is some fixed axis here. 00:59:47.666 --> 00:59:53.120 You have a fixed coordinate system, your o x, y, z. 00:59:53.120 --> 00:59:55.690 And then you'd probably have some rotating coordinate system 00:59:55.690 --> 01:00:03.210 with the point A fixed here at the axis of rotation. 01:00:03.210 --> 01:00:09.580 And this is going-- what's unusual about this ride, 01:00:09.580 --> 01:00:11.860 not only can it go around and round, 01:00:11.860 --> 01:00:14.220 but the arm can go in and out. 01:00:17.230 --> 01:00:19.790 So it might take a path inwards, like you 01:00:19.790 --> 01:00:22.660 pull the arm in as you're going around. 01:00:22.660 --> 01:00:25.050 What are the forces that you feel in the ride? 01:00:28.040 --> 01:00:31.426 So this is my point B out here. 01:00:31.426 --> 01:00:32.550 It's where the person's at. 01:00:32.550 --> 01:00:36.165 You have some mass, m. 01:00:38.770 --> 01:00:49.010 And let's let, for now, the first case-- theta 01:00:49.010 --> 01:00:59.510 dot, let that be constant, the constant angular rate here. 01:00:59.510 --> 01:01:04.460 Here's theta and r dot. 01:01:13.220 --> 01:01:17.020 Now, you already know quite a bit about things like this. 01:01:17.020 --> 01:01:21.510 If your riding in that bucket, what forces 01:01:21.510 --> 01:01:23.920 do you think you would feel? 01:01:23.920 --> 01:01:26.600 Or I should say it more carefully. 01:01:26.600 --> 01:01:29.920 There will be accelerations that you feel in that bucket. 01:01:29.920 --> 01:01:31.700 You'll feel like forces on you. 01:01:31.700 --> 01:01:33.575 But what are the accelerations that you 01:01:33.575 --> 01:01:36.100 will feel-- you expect to be if you're riding in it? 01:01:39.812 --> 01:01:41.210 AUDIENCE: [INAUDIBLE]. 01:01:41.210 --> 01:01:42.460 PROFESSOR: I hear one here. 01:01:42.460 --> 01:01:43.690 AUDIENCE: Centripetal acceleration. 01:01:43.690 --> 01:01:45.960 PROFESSOR: So there will be centripetal acceleration, 01:01:45.960 --> 01:01:46.880 right. 01:01:46.880 --> 01:01:48.150 Everybody agree with that? 01:01:48.150 --> 01:01:49.070 Anything else? 01:01:49.070 --> 01:01:49.974 AUDIENCE: Coriolis acceleration. 01:01:49.974 --> 01:01:51.807 PROFESSOR: There's going to be some Coriolis 01:01:51.807 --> 01:01:54.762 because r dot is not 0. 01:01:54.762 --> 01:01:57.710 It's changing in position, but when 01:01:57.710 --> 01:01:59.740 the length of the arm of something 01:01:59.740 --> 01:02:03.054 rotating at constant speed changes, what momentum changes? 01:02:09.090 --> 01:02:09.840 AUDIENCE: Angular. 01:02:09.840 --> 01:02:11.048 PROFESSOR: Angular, for sure. 01:02:11.048 --> 01:02:12.980 How about linear momentum? 01:02:12.980 --> 01:02:16.260 Yeah, it's changing too because r cross P is angular momentum. 01:02:16.260 --> 01:02:17.810 So both linear and angular momentum 01:02:17.810 --> 01:02:22.320 are changing as this radius gets longer or shorter. 01:02:22.320 --> 01:02:25.500 And if that angular momentum changes, 01:02:25.500 --> 01:02:27.506 it takes torque to drive it. 01:02:27.506 --> 01:02:29.880 And so we ought to be able to use the formulas that we've 01:02:29.880 --> 01:02:32.280 just derived to calculate something 01:02:32.280 --> 01:02:35.430 about the torques required to make this happen 01:02:35.430 --> 01:02:36.750 and the forces on the rider. 01:02:44.580 --> 01:02:47.710 So we'll treat this one as a particle. 01:02:47.710 --> 01:02:52.420 So H, the angular momentum of the rider out here, 01:02:52.420 --> 01:03:07.480 at B with respect to o, is going to be r B/o cross P. 01:03:07.480 --> 01:03:10.780 And in this problem, I'll use polar coordinates. 01:03:10.780 --> 01:03:11.550 Pretty easy. 01:03:11.550 --> 01:03:13.470 This is a planar motion problem. 01:03:13.470 --> 01:03:17.040 It's confined to the x,y plane and rotation z, 01:03:17.040 --> 01:03:19.640 so polar coordinates are pretty convenient. 01:03:19.640 --> 01:03:26.020 So this should look like radius r r hat cross-- 01:03:26.020 --> 01:03:29.880 and the linear momentum of this is 01:03:29.880 --> 01:03:36.190 the mass-- times r dot r hat. 01:03:36.190 --> 01:03:41.090 That's the extension rate, but it also 01:03:41.090 --> 01:03:49.720 has velocity in this theta hat direction, r theta hat. 01:03:49.720 --> 01:03:56.450 So this is P was respected o, m, v. And this 01:03:56.450 --> 01:03:59.810 is the radius-- this is the r crossed into it. 01:03:59.810 --> 01:04:02.920 Now r hat cross r hat gives you 0, 01:04:02.920 --> 01:04:05.690 so you only get a single term out of this. 01:04:05.690 --> 01:04:10.970 And you get an r cross-- r hat cross, theta hat, positive k. 01:04:10.970 --> 01:04:25.750 So this looks like plus m r squared, theta dot, k hat. 01:04:25.750 --> 01:04:27.820 That's my H B/o. 01:04:34.910 --> 01:04:37.880 Now, I would-- that's my first piece that I wanted to get. 01:04:37.880 --> 01:04:38.790 That's the a part. 01:04:38.790 --> 01:04:41.200 That's find the angular momentum. 01:04:41.200 --> 01:04:43.688 So this is a. 01:04:43.688 --> 01:04:45.150 b, I want to know the torque. 01:04:54.950 --> 01:04:59.010 Well, which formula can we use? 01:04:59.010 --> 01:05:01.900 Are we allowed to use-- do we have to account 01:05:01.900 --> 01:05:04.202 for that second term? 01:05:04.202 --> 01:05:04.827 AUDIENCE: No. 01:05:04.827 --> 01:05:05.410 PROFESSOR: No. 01:05:05.410 --> 01:05:05.959 Why? 01:05:05.959 --> 01:05:06.750 AUDIENCE: Velocity. 01:05:06.750 --> 01:05:07.180 PROFESSOR: Yeah. 01:05:07.180 --> 01:05:08.971 The velocity of the point about which we're 01:05:08.971 --> 01:05:12.550 computing, the angular momentum, and therefore, the torques, 01:05:12.550 --> 01:05:13.380 is not moving. 01:05:13.380 --> 01:05:15.650 So you only have to deal with the first term. 01:05:15.650 --> 01:05:20.520 So the torque required to move that particle at B 01:05:20.520 --> 01:05:24.040 with respect to o is just a time rate of change 01:05:24.040 --> 01:05:28.870 d by dt of h B with respect to o. 01:05:28.870 --> 01:05:44.030 And that's d by dt of m r squared theta dot k hat. 01:05:44.030 --> 01:05:46.977 So what are the constants in this expression 01:05:46.977 --> 01:05:49.060 so we don't have to worry about their derivatives? 01:05:49.060 --> 01:05:51.160 Does k hat change length? 01:05:51.160 --> 01:05:51.660 No. 01:05:51.660 --> 01:05:53.010 Change direction? 01:05:53.010 --> 01:05:54.360 No. 01:05:54.360 --> 01:05:59.170 Theta dot, does it change in this problem? 01:05:59.170 --> 01:06:00.100 No, we fixed it. 01:06:00.100 --> 01:06:02.620 We arbitrarily started off saying that we'll just 01:06:02.620 --> 01:06:04.810 let theta dot be constant. 01:06:04.810 --> 01:06:06.390 r, though, is changing. 01:06:06.390 --> 01:06:08.420 So the time derivative of this particular one, 01:06:08.420 --> 01:06:10.910 we only have to deal with the r. 01:06:10.910 --> 01:06:13.175 Actually, I'll forget that for a second 01:06:13.175 --> 01:06:14.550 because I want to get both terms, 01:06:14.550 --> 01:06:16.120 and then we'll let it be 0. 01:06:16.120 --> 01:06:19.000 So this time derivative then, when you work it 01:06:19.000 --> 01:06:47.080 out-- the derivative of the r term 01:06:47.080 --> 01:06:56.730 gives you 2mr r dot theta dot k hat. 01:06:56.730 --> 01:06:58.764 And I'll just go ahead and forget for a minute 01:06:58.764 --> 01:06:59.930 that I said that's constant. 01:06:59.930 --> 01:07:03.870 Let's get the other term that might be there. 01:07:03.870 --> 01:07:16.785 That term gives us m r squared theta double dot k hat. 01:07:19.950 --> 01:07:24.580 This term comes from what we'll find is the Coriolis one. 01:07:24.580 --> 01:07:28.585 And this is from what we call the Eulerian acceleration. 01:07:31.445 --> 01:07:35.305 The torque-- let me get this up a little higher. 01:07:39.640 --> 01:07:48.270 We should be able to write as some r cross f. 01:07:48.270 --> 01:07:52.930 And so the r cross f terms, this will 01:07:52.930 --> 01:07:56.790 be from the Coriolis force. 01:07:56.790 --> 01:08:02.460 And this would be from that Eulerian force. 01:08:02.460 --> 01:08:06.700 So this will look like r r hat cross-- 01:08:06.700 --> 01:08:10.620 I'm just factoring this back out into its cross products-- 01:08:10.620 --> 01:08:17.819 r r hat cross 2m r dot theta dot, 01:08:17.819 --> 01:08:26.939 in the theta hat direction, plus r theta double dot, 01:08:26.939 --> 01:08:30.120 in the theta hat direction. 01:08:30.120 --> 01:08:32.689 So there's two terms in this torque expression. 01:08:32.689 --> 01:08:37.970 They come from r cross, two force terms. 01:08:37.970 --> 01:08:41.380 The first force term is what we know to be the Coriolis force. 01:08:41.380 --> 01:08:44.816 And the second force term-- I am missing an m here. 01:08:48.560 --> 01:08:50.880 r m theta double dot theta hat, that's 01:08:50.880 --> 01:08:57.140 the force that it takes to-- if the thing were accelerating, 01:08:57.140 --> 01:09:01.370 just to accelerate that ball, theta double dot takes a force. 01:09:01.370 --> 01:09:04.420 You do these cross products. r cross theta hat give you k. 01:09:04.420 --> 01:09:06.380 R cross theta hat give you k. 01:09:06.380 --> 01:09:09.200 It all comes out in the right directions. 01:09:09.200 --> 01:09:11.380 And now to do the problem we had, 01:09:11.380 --> 01:09:13.229 we said, oh yeah, let this one be 0. 01:09:13.229 --> 01:09:14.590 So we'll let this term go to 0. 01:09:17.220 --> 01:09:18.590 We're left with a single term. 01:10:14.430 --> 01:10:16.070 And if we plugged in some numbers-- 01:10:16.070 --> 01:10:17.670 let's just see how this works out. 01:10:17.670 --> 01:10:19.610 And this we know is just r cross the Coriolis. 01:10:27.570 --> 01:10:33.310 Well, let's let m-- if you let m be 100 kilograms 01:10:33.310 --> 01:10:44.580 and r be 5 meters and r dot, 0.4 meters 01:10:44.580 --> 01:10:48.380 per second-- they're all perfectly reasonable 01:10:48.380 --> 01:10:50.320 dimensions. 01:10:50.320 --> 01:10:59.020 And theta dot equals 3 radians per second. 01:10:59.020 --> 01:11:02.664 So 2 pi radians per second means it goes around once a second. 01:11:02.664 --> 01:11:04.830 So this is a little less than half a rev per second. 01:11:04.830 --> 01:11:05.442 Yeah? 01:11:05.442 --> 01:11:07.233 AUDIENCE: Is this still [INAUDIBLE] though, 01:11:07.233 --> 01:11:08.420 or does it-- 01:11:08.420 --> 01:11:10.310 PROFESSOR: Probably. 01:11:10.310 --> 01:11:11.350 Yeah. 01:11:11.350 --> 01:11:12.650 Good catch! 01:11:12.650 --> 01:11:15.870 Got to have on r in it because that r comes from here, right? 01:11:15.870 --> 01:11:18.180 So I had the Coriolis force written down, 01:11:18.180 --> 01:11:21.660 but not the r you have to multiply it by. 01:11:21.660 --> 01:11:24.020 So now if you plug in all of these numbers, 01:11:24.020 --> 01:11:26.510 let's see if it's a ride you could survive. 01:11:34.220 --> 01:11:38.069 I actually computed the Coriolis force first. 01:11:38.069 --> 01:11:40.360 I think it's just interesting to get a physical feeling 01:11:40.360 --> 01:11:43.570 for how-- whether or not you can feel these forces. 01:11:43.570 --> 01:11:45.910 So the Coriolis force is just everything 01:11:45.910 --> 01:11:49.860 but the r, 2m r dot theta dot. 01:11:49.860 --> 01:11:56.540 So if you calculate that, you get 240 newtons. 01:11:56.540 --> 01:12:00.262 And r times that, the torque. 01:12:03.650 --> 01:12:16.520 5 times that, about 1,200 Newton meters. 01:12:16.520 --> 01:12:21.960 And the acceleration, how do we get acceleration? 01:12:21.960 --> 01:12:26.650 Well, f cor, Coriolis force, is some mass 01:12:26.650 --> 01:12:27.930 times an acceleration. 01:12:27.930 --> 01:12:32.320 So we can solve, from that 240 newtons, 01:12:32.320 --> 01:12:34.040 the acceleration of the system. 01:12:34.040 --> 01:12:45.500 Our acceleration of B with respect to o, 240 newtons 01:12:45.500 --> 01:12:56.890 divided by 100 kilograms, 2.4 meters per second squared. 01:12:56.890 --> 01:13:00.134 And what's that in g's. 01:13:00.134 --> 01:13:01.070 AUDIENCE: About 1/4. 01:13:01.070 --> 01:13:03.082 PROFESSOR: Yeah, about a quarter of a g. 01:13:06.250 --> 01:13:08.230 And again, would you feel that? 01:13:08.230 --> 01:13:12.080 So now you riding in the bucket. 01:13:12.080 --> 01:13:13.730 Let's say it's just spinning around. 01:13:13.730 --> 01:13:15.760 The arm is not going in and out at all. 01:13:15.760 --> 01:13:16.790 So you're riding in it. 01:13:16.790 --> 01:13:19.840 What force would you feel? 01:13:19.840 --> 01:13:24.110 Would you feel any forces pushing you into your seat? 01:13:24.110 --> 01:13:26.150 Constant speed, r dot's 0. 01:13:36.810 --> 01:13:38.680 Somebody out there, would you feel any force 01:13:38.680 --> 01:13:41.120 if you're going around and around in this thing sitting 01:13:41.120 --> 01:13:41.680 in a seat? 01:13:41.680 --> 01:13:42.010 AUDIENCE: Yes. 01:13:42.010 --> 01:13:42.593 PROFESSOR: OK. 01:13:42.593 --> 01:13:44.000 What's it come from? 01:13:44.000 --> 01:13:44.310 AUDIENCE: Centripetal acceleration. 01:13:44.310 --> 01:13:45.813 PROFESSOR: Centripetal acceleration. 01:13:48.820 --> 01:13:51.264 And that is in what direction? 01:13:51.264 --> 01:13:52.180 AUDIENCE: It's inward. 01:13:52.180 --> 01:13:53.638 PROFESSOR: Accelerations is inward. 01:13:53.638 --> 01:13:55.560 What would you actually feel? 01:13:55.560 --> 01:13:59.350 You'd feel-- if this seat could swing out 01:13:59.350 --> 01:14:02.090 so your facing in, you'd feel like you're being thrown out 01:14:02.090 --> 01:14:04.800 in your seat, right? 01:14:04.800 --> 01:14:07.180 Because you have to have a force on you 01:14:07.180 --> 01:14:09.460 to make that acceleration happen. 01:14:09.460 --> 01:14:10.665 Acceleration is inward. 01:14:10.665 --> 01:14:13.360 The force had better be pushing you inward. 01:14:13.360 --> 01:14:16.460 So it's pushing on your back if you're look in, going around. 01:14:16.460 --> 01:14:22.750 So that's the-- I've forgotten the term for it. 01:14:22.750 --> 01:14:26.020 But put the astronauts in a centrifuge, 01:14:26.020 --> 01:14:28.710 spin around to see if they can take high g's. 01:14:28.710 --> 01:14:33.080 That's the inward high g acceleration 01:14:33.080 --> 01:14:35.200 due to the centripetal acceleration, 01:14:35.200 --> 01:14:36.800 makes you go in a circle. 01:14:36.800 --> 01:14:40.520 But now not only are you going to feel that, but now you start 01:14:40.520 --> 01:14:42.430 changing the length of the arm. 01:14:42.430 --> 01:14:46.010 And if you change r dot and make it positive 01:14:46.010 --> 01:14:47.820 so the arm is getting longer, and we made 01:14:47.820 --> 01:14:49.340 this 0.4 meters per second. 01:14:49.340 --> 01:14:51.350 So it's moving out about like that. 01:14:51.350 --> 01:14:53.190 It's not real fast. 01:14:53.190 --> 01:14:56.660 But it's going to create a quarter of a g acceleration 01:14:56.660 --> 01:14:57.160 on you. 01:14:57.160 --> 01:15:00.325 And if r dot's positive, which direction is that acceleration? 01:15:03.500 --> 01:15:04.190 I was careful. 01:15:04.190 --> 01:15:05.670 We walked all the way through this. 01:15:05.670 --> 01:15:09.480 The acceleration is positive, but in what direction? 01:15:09.480 --> 01:15:11.292 It's the Coriolis. 01:15:11.292 --> 01:15:12.750 AUDIENCE: Theta hat. 01:15:12.750 --> 01:15:14.760 PROFESSOR: Theta hat, positive theta hat, 01:15:14.760 --> 01:15:16.800 so it's in the direction of increasing theta. 01:15:16.800 --> 01:15:20.460 So this one is perpendicular to the arm. 01:15:20.460 --> 01:15:24.190 And this is the-- one of my ping pong 01:15:24.190 --> 01:15:28.510 balls in here, the force that actually causes it to speed up 01:15:28.510 --> 01:15:30.940 is partly Coriolis and partly or Eulerian. 01:15:30.940 --> 01:15:34.270 If I can make this go at constant speed, the force that 01:15:34.270 --> 01:15:38.050 drives that-- the force that actually speeds that up 01:15:38.050 --> 01:15:41.090 as they're going out is the Coriolis force, the normal one. 01:15:41.090 --> 01:15:42.730 So if you were on this ride, you'd 01:15:42.730 --> 01:15:45.760 be feel about a quarter of a g perpendicular to the arm. 01:15:45.760 --> 01:15:47.780 You'd be filling another force inward 01:15:47.780 --> 01:15:49.837 that's the centripetal-- caused by 01:15:49.837 --> 01:15:51.045 the centripetal acceleration. 01:15:51.045 --> 01:15:52.510 So you'd be feeling both of them. 01:15:52.510 --> 01:15:54.176 How big is the centripetal acceleration? 01:15:56.810 --> 01:16:02.274 Like r omega squared, right? 01:16:02.274 --> 01:16:03.030 AUDIENCE: Right. 01:16:03.030 --> 01:16:04.380 PROFESSOR: r is 5. 01:16:04.380 --> 01:16:06.065 Omega is 3. 01:16:06.065 --> 01:16:18.007 3 Squared is 9, times 5-- 9 times 5, 45, divided by 10, 01:16:18.007 --> 01:16:18.590 4 and 1/2 g's. 01:16:22.740 --> 01:16:24.815 You wouldn't notice the Coriolis very much. 01:16:24.815 --> 01:16:29.470 It would be a tough ride on that side of things. 01:16:32.980 --> 01:16:37.592 Now, we've done that-- you've seen a simple application, 01:16:37.592 --> 01:16:39.050 pretty straightforward application, 01:16:39.050 --> 01:16:44.350 of using time derivative of angular momentum 01:16:44.350 --> 01:16:46.640 to calculate torques. 01:16:46.640 --> 01:16:49.710 So on homework, again, there's a couple problems 01:16:49.710 --> 01:16:52.400 very similar to the one I just did, things going around, 01:16:52.400 --> 01:16:54.610 circus rides, that kind of stuff. 01:16:54.610 --> 01:16:58.850 So have a good weekend. 01:16:58.850 --> 01:17:01.460 See you on Tuesday.