0:00:00.415,0:00:02.103 - [Voiceover] When a major[br]league baseball player throws 0:00:02.103,0:00:05.637 a fast ball, that ball's[br]definitely got kinetic energy. 0:00:05.637,0:00:07.581 We know that cause if you get in the way, 0:00:07.581,0:00:09.429 it could do work on[br]you, that's gonna hurt. 0:00:09.429,0:00:10.583 You gotta watch out. 0:00:10.583,0:00:14.212 But here's my question: does[br]the fact that most pitches, 0:00:14.212,0:00:15.855 unless you're throwing a knuckle ball, 0:00:15.855,0:00:19.157 does the fact that most[br]pitches head toward home plate 0:00:19.157,0:00:21.805 with the baseball spinning[br]mean that that ball 0:00:21.805,0:00:24.197 has extra kinetic energy? 0:00:24.197,0:00:26.571 Well it does, and how[br]do we figure that out, 0:00:26.571,0:00:28.986 that's the goal for this video. 0:00:28.986,0:00:31.267 How do we determine what the rotational 0:00:31.267,0:00:33.647 kinetic energy is of an object? 0:00:33.647,0:00:35.770 Well if I was coming at[br]this for the first time, 0:00:35.770,0:00:37.803 my first guest I'd say okay, 0:00:37.803,0:00:40.561 I'd say I know what regular[br]kinetic energy looks like. 0:00:40.561,0:00:42.875 The formula for regular kinetic energy is 0:00:42.875,0:00:45.954 just one half m v squared. 0:00:45.954,0:00:48.567 So let's say alright, I want[br]rotational kinetic energy. 0:00:48.567,0:00:50.963 Let me just call that k rotational 0:00:50.963,0:00:52.497 and what is that gonna be? 0:00:52.497,0:00:54.964 Well I know for objects that are rotating, 0:00:54.964,0:00:58.764 the rotational equivalent of[br]mass is moment of inertia. 0:00:58.764,0:01:01.461 So I might guess alright instead of mass, 0:01:01.461,0:01:04.354 I'd have moment of inertia[br]cause in Newton's second law 0:01:04.354,0:01:06.813 for rotation I know that[br]instead of mass there's 0:01:06.813,0:01:09.091 moment of inertia so maybe I replace that. 0:01:09.091,0:01:12.188 And instead of speed[br]squared, maybe since I have 0:01:12.188,0:01:15.284 something rotating I'd[br]have angular speed squared. 0:01:15.284,0:01:16.898 It turns out this works. 0:01:16.898,0:01:20.014 You can often derive, it's[br]not really a derivation, 0:01:20.014,0:01:22.516 you're just kind of guessing[br]educatedly but you could 0:01:22.516,0:01:25.797 often get a formula for the[br]rotational analog of some 0:01:25.797,0:01:29.780 linear formula by just[br]substituting the rotational analog 0:01:29.780,0:01:32.415 for each of the variables,[br]so if I replaced mass with 0:01:32.415,0:01:35.247 rotational mass, I get[br]the moment of inertia. 0:01:35.247,0:01:37.721 If I replace speed with rotational speed, 0:01:37.721,0:01:40.743 I get the angular speed and[br]this is the correct formula. 0:01:40.743,0:01:43.303 So in this video we needed[br]to ride this cause that 0:01:43.303,0:01:44.913 is not really a derivation,[br]we didn't really 0:01:44.913,0:01:47.720 prove this, we just showed[br]that it's plausible. 0:01:47.720,0:01:50.111 How do we prove that[br]this is the rotational 0:01:50.111,0:01:52.991 kinetic energy of an[br]object that's rotating 0:01:52.991,0:01:54.349 like a baseball. 0:01:54.349,0:01:56.997 The first thing to recognize[br]is that this rotational 0:01:56.997,0:01:59.684 kinetic energy isn't really a new kind of 0:01:59.684,0:02:02.301 kinetic energy, it's[br]still just the same old 0:02:02.301,0:02:05.721 regular kinetic energy for[br]something that's rotating. 0:02:05.721,0:02:07.051 What I mean by that is this. 0:02:07.051,0:02:09.820 Imagine this baseball[br]is rotating in a circle. 0:02:09.820,0:02:13.320 Every point on the baseball[br]is moving with some speed, 0:02:13.320,0:02:15.462 so what I mean by that is[br]this, so this point at the top 0:02:15.462,0:02:18.613 here imagine the little[br]piece of leather right here, 0:02:18.613,0:02:20.367 it's gonna have some speed forward. 0:02:20.367,0:02:23.485 I'm gonna call this mass[br]M one, that little piece 0:02:23.485,0:02:27.188 of mass right now and I'll[br]call the speed of it V one. 0:02:27.188,0:02:29.744 Similarly, this point on[br]the leather right there, 0:02:29.744,0:02:32.288 I'm gonna call that M two,[br]it's gonna be moving down 0:02:32.288,0:02:35.713 cause it's a rotating circle,[br]so I'll call that V two 0:02:35.713,0:02:38.370 and points closer to the[br]axis are gonna be moving 0:02:38.370,0:02:41.027 with smaller speed so[br]this point right here, 0:02:41.027,0:02:43.779 we'll call it M three, moving[br]down with a speed V three, 0:02:43.779,0:02:46.771 that is not as big as V two or V one. 0:02:46.771,0:02:48.082 You can't see that very well, 0:02:48.082,0:02:51.587 I'll use a darker green[br]so this M three right here 0:02:51.587,0:02:54.921 closer to the axis, axis[br]being right at this point 0:02:54.921,0:02:58.781 in the center, closer to the[br]axis so it's speed is smaller 0:02:58.781,0:03:01.411 than points that are[br]farther away from this axis, 0:03:01.411,0:03:03.373 so you can see this is kinda complicated. 0:03:03.373,0:03:05.539 All points on this baseball[br]are gonna be moving with 0:03:05.539,0:03:08.126 different speeds so points[br]over here that are really 0:03:08.126,0:03:10.796 close to the axis, barely moving at all. 0:03:10.796,0:03:12.946 I'll call this M four and it would be 0:03:12.946,0:03:15.093 moving at speed V four. 0:03:15.093,0:03:17.676 What we mean by the[br]rotational kinetic energy is 0:03:17.676,0:03:19.892 really just all the regular[br]kinetic energy these 0:03:19.892,0:03:23.846 masses have about the center[br]of mass of the baseball. 0:03:23.846,0:03:26.681 So in other words, what[br]we mean by K rotational, 0:03:26.681,0:03:29.455 is you just add up all of these energies. 0:03:29.455,0:03:32.021 You have one half, this[br]little piece of leather 0:03:32.021,0:03:33.737 up here would have some kinetic energy 0:03:33.737,0:03:37.487 so you do one half M[br]one, V one squared plus. 0:03:38.415,0:03:41.050 And this M two has some kinetic energy, 0:03:41.050,0:03:43.152 don't worry that it points downward, 0:03:43.152,0:03:45.954 downward doesn't matter for[br]things that aren't vectors, 0:03:45.954,0:03:49.258 this V gets squared so[br]kinetic energy's not a vector 0:03:49.258,0:03:51.779 so it doesn't matter that[br]one velocity points down 0:03:51.779,0:03:54.443 cause this is just the[br]speed and similarly, 0:03:54.443,0:03:58.950 you'd add up one half M[br]three, V three squared, 0:03:58.950,0:04:00.517 but you might be like this is impossible, 0:04:00.517,0:04:02.926 there's infinitely many[br]points in this baseball, 0:04:02.926,0:04:05.388 how am I ever going to do this. 0:04:05.388,0:04:07.379 Well something magical is about to happen, 0:04:07.379,0:04:09.526 this is one of my favorite[br]little derivations, 0:04:09.526,0:04:12.133 short and sweet, watch what happens. 0:04:12.133,0:04:15.067 K E rotational is really just the sum, 0:04:15.067,0:04:17.661 if I add all these up[br]I can write is as a sum 0:04:17.661,0:04:21.494 of all the one half M V[br]squares of every point 0:04:22.457,0:04:25.416 on this baseball so imagine[br]breaking this baseball 0:04:25.416,0:04:27.756 up into very, very small pieces. 0:04:27.756,0:04:30.072 Don't do it physically but[br]just think about it mentally, 0:04:30.072,0:04:33.039 just visualize considering[br]very small pieces, 0:04:33.039,0:04:35.919 particles of this baseball[br]and how fast they're going. 0:04:35.919,0:04:38.938 What I'm saying is that[br]if you add all of that up, 0:04:38.938,0:04:41.359 you get the total[br]rotational kinetic energy, 0:04:41.359,0:04:42.967 this looks impossible to do. 0:04:42.967,0:04:44.552 But something magical is about to happen, 0:04:44.552,0:04:45.766 here's what we can do. 0:04:45.766,0:04:48.352 We can rewrite, see the problem here is V. 0:04:48.352,0:04:50.755 All these points have a different speed V, 0:04:50.755,0:04:52.711 but we can use a trick,[br]a trick that we love 0:04:52.711,0:04:55.125 to use in physics, instead[br]of writing this as V, 0:04:55.125,0:04:57.773 we're gonna write V as, so[br]remember that for things 0:04:57.773,0:05:01.564 that are rotating, V[br]is just R times omega. 0:05:01.564,0:05:04.133 The radius, how far from the axis you are, 0:05:04.133,0:05:06.885 times the angular velocity,[br]or the angular speed 0:05:06.885,0:05:09.358 gives you the regular speed. 0:05:09.358,0:05:12.145 This formula is really[br]handy, so we're gonna replace 0:05:12.145,0:05:16.185 V with R omega, and this[br]is gonna give us R omega 0:05:16.185,0:05:18.352 and you still have to[br]square it and at this point 0:05:18.352,0:05:19.993 you're probably thinking[br]like this is even worse, 0:05:19.993,0:05:21.079 what do we do this for. 0:05:21.079,0:05:24.023 Well watch, if we add this[br]is up I'll have one half M. 0:05:24.023,0:05:26.848 I'm gonna get an R squared[br]and an omega squared, 0:05:26.848,0:05:28.958 and the reason this is[br]better is that even though 0:05:28.958,0:05:32.626 every point on this baseball[br]has a different speed V, 0:05:32.626,0:05:35.491 they all have the same[br]angular speed omega, 0:05:35.491,0:05:38.315 that was what was good about[br]these angular quantities 0:05:38.315,0:05:41.618 is that they're the same for[br]every point on the baseball 0:05:41.618,0:05:43.870 no matter how far away[br]you are from the axis, 0:05:43.870,0:05:46.042 and since they're the[br]same for every point I can 0:05:46.042,0:05:48.634 bring that out of the[br]summation so I can rewrite 0:05:48.634,0:05:51.609 this summation and bring[br]everything that's constant 0:05:51.609,0:05:54.818 for all of the masses out[br]of the summation so I can 0:05:54.818,0:05:58.220 write this as one half times the summation 0:05:58.220,0:06:01.803 of M times R squared[br]and end that quantity, 0:06:02.782,0:06:06.597 end that summation and just[br]pull the omega squared out 0:06:06.597,0:06:08.565 because it's the same for each term. 0:06:08.565,0:06:11.444 I'm basically factoring[br]this out of all of these 0:06:11.444,0:06:13.857 terms in the summation, it's like up here, 0:06:13.857,0:06:15.548 all of these have a one half. 0:06:15.548,0:06:17.487 You could imagine factoring out a one half 0:06:17.487,0:06:18.985 and just writing this whole quantity as 0:06:18.985,0:06:22.135 one half times M one V one squared plus 0:06:22.135,0:06:24.167 M two V two squared and so on. 0:06:24.167,0:06:26.055 That's what I'm doing[br]down here for the one half 0:06:26.055,0:06:28.615 and for the omega squared,[br]so that's what was good 0:06:28.615,0:06:31.077 about replacing V with R omega. 0:06:31.077,0:06:32.540 The omega's the same for all of them, 0:06:32.540,0:06:33.816 you can bring that out. 0:06:33.816,0:06:35.514 You might still be[br]concerned, you might be like, 0:06:35.514,0:06:37.993 we're still stuck with[br]the M in here cause you've 0:06:37.993,0:06:39.990 got different Ms at different points. 0:06:39.990,0:06:42.160 We're stuck with all[br]these R squareds in here, 0:06:42.160,0:06:44.628 all these points at the[br]baseball are different Rs, 0:06:44.628,0:06:46.328 they're all different[br]points from the axis, 0:06:46.328,0:06:48.558 different distances from[br]the axis, we can't bring 0:06:48.558,0:06:51.449 those out so now what do we[br]do, well if you're clever 0:06:51.449,0:06:53.792 you recognize this term. 0:06:53.792,0:06:56.615 This summation term is[br]nothing but the total moment 0:06:56.615,0:06:59.296 of inertia of the object. 0:06:59.296,0:07:01.628 Remember that the moment[br]of inertia of an object, 0:07:01.628,0:07:04.394 we learned previously,[br]is just M R squared, 0:07:04.394,0:07:06.410 so the moment of inertia of a point mass 0:07:06.410,0:07:09.122 is M R squared and the moment of inertia 0:07:09.122,0:07:12.483 of a bunch of point[br]masses is the sum of all 0:07:12.483,0:07:15.402 the M R squareds and that's[br]what we've got right here, 0:07:15.402,0:07:19.514 this is just the moment of[br]inertia of this baseball 0:07:19.514,0:07:22.115 or whatever the object is,[br]it doesn't even have to be 0:07:22.115,0:07:24.287 of a particular shape, we're gonna add all 0:07:24.287,0:07:26.998 the M R squareds, that's[br]always going to be 0:07:26.998,0:07:28.637 the total moment of inertia. 0:07:28.637,0:07:30.925 So what we've found is[br]that the K rotational 0:07:30.925,0:07:34.066 is equal to one half times this quantity, 0:07:34.066,0:07:35.947 which is I, the moment of inertia, 0:07:35.947,0:07:38.284 times omega squared and that's the formula 0:07:38.284,0:07:40.004 we got up here just by guessing. 0:07:40.004,0:07:41.850 But it actually works[br]and this is why it works, 0:07:41.850,0:07:43.859 because you always get[br]this quantity down here, 0:07:43.859,0:07:46.204 which is one half I omega[br]squared, no matter what 0:07:46.204,0:07:47.676 the shape of the object is. 0:07:47.676,0:07:49.420 So what this is telling[br]you, what this quantity 0:07:49.420,0:07:52.346 gives us is the total[br]rotational kinetic energy 0:07:52.346,0:07:55.666 of all the points on that[br]mass about the center 0:07:55.666,0:07:58.591 of the mass but here's[br]what it doesn't give you. 0:07:58.591,0:08:01.036 This term right here does not include 0:08:01.036,0:08:03.451 the translational kinetic[br]energy so the fact that 0:08:03.451,0:08:06.292 this baseball was flying[br]through the air does not 0:08:06.292,0:08:08.142 get incorporated by this formula. 0:08:08.142,0:08:10.264 We didn't take into account the fact that 0:08:10.264,0:08:12.391 the baseball was moving through the air, 0:08:12.391,0:08:13.976 in other words, we[br]didn't take into account 0:08:13.976,0:08:16.791 that the actual center[br]of mass in this baseball 0:08:16.791,0:08:19.200 was translating through the air. 0:08:19.200,0:08:21.365 But we can do that easily[br]with this formula here. 0:08:21.365,0:08:24.279 This is the translational kinetic energy. 0:08:24.279,0:08:26.930 Sometimes instead of writing[br]regular kinetic energy, 0:08:26.930,0:08:29.841 now that we've got two we[br]should specify this is really 0:08:29.841,0:08:31.791 translational kinetic energy. 0:08:31.791,0:08:34.361 We've got a formula for[br]translational kinetic energy, 0:08:34.361,0:08:37.701 the energy something has due[br]to the fact that the center 0:08:37.701,0:08:40.522 of mass of that object is[br]moving and we have a formula 0:08:40.522,0:08:42.972 that takes into account the[br]fact that something can have 0:08:42.972,0:08:45.494 kinetic energy due to its rotation. 0:08:45.494,0:08:48.316 That's this K rotational,[br]so if an object's rotating, 0:08:48.316,0:08:50.483 it has rotational kinetic energy. 0:08:50.483,0:08:52.718 If an object is translating it has 0:08:52.718,0:08:54.500 translational kinetic energy, 0:08:54.500,0:08:56.515 i.e. if the center of mass is moving, 0:08:56.515,0:08:59.986 and if the object is[br]translating and it's rotating 0:08:59.986,0:09:02.435 then it would have those both[br]of these kinetic energies, 0:09:02.435,0:09:04.948 both at the same time and[br]this is the beautiful thing. 0:09:04.948,0:09:08.430 If an object is translating[br]and rotating and you want 0:09:08.430,0:09:11.390 to find the total kinetic[br]energy of the entire thing, 0:09:11.390,0:09:14.004 you can just add these two terms up. 0:09:14.004,0:09:17.147 If I just take the translational[br]one half M V squared, 0:09:17.147,0:09:20.573 and this would then be the[br]velocity of the center of mass. 0:09:20.573,0:09:22.157 So you have to be careful. 0:09:22.157,0:09:23.749 Let me make some room[br]here, so let me get rid 0:09:23.749,0:09:25.130 of all this stuff here. 0:09:25.130,0:09:28.741 If you take one half M,[br]times the speed of the center 0:09:28.741,0:09:31.655 of mass squared, you'll[br]get the total translational 0:09:31.655,0:09:33.239 kinetic energy of the baseball. 0:09:33.239,0:09:36.386 And if we add to that the[br]one half I omega squared, 0:09:36.386,0:09:39.184 so the omega about the[br]center of mass you'll get 0:09:39.184,0:09:43.688 the total kinetic energy, both[br]translational and rotational, 0:09:43.688,0:09:46.624 so this is great, we can[br]determine the total kinetic energy 0:09:46.624,0:09:49.889 altogether, rotational[br]motion, translational motion, 0:09:49.889,0:09:52.580 from just taking these two terms added up. 0:09:52.580,0:09:54.049 So what would an example of this be, 0:09:54.049,0:09:55.796 let's just get rid of all this. 0:09:55.796,0:09:59.180 Let's say this baseball,[br]someone pitched this thing, 0:09:59.180,0:10:02.582 and the radar gun shows that[br]this baseball was hurled 0:10:02.582,0:10:04.799 through the air at 40 meters per second. 0:10:04.799,0:10:07.452 So it's heading toward home[br]plate at 40 meters per second. 0:10:07.452,0:10:09.858 The center of mass of[br]this baseball is going 0:10:09.858,0:10:12.551 40 meters per second toward home plate. 0:10:12.551,0:10:15.094 Let's say it's also, someone[br]really threw the fastball. 0:10:15.094,0:10:18.107 This thing's rotating[br]with an angular velocity 0:10:18.107,0:10:20.190 of 50 radians per second. 0:10:22.264,0:10:24.376 We know the mass of a[br]baseball, I've looked it up. 0:10:24.376,0:10:28.781 The mass of a baseball[br]is about 0.145 kilograms 0:10:28.781,0:10:31.795 and the radius of the baseball,[br]so a radius of a baseball 0:10:31.795,0:10:35.388 is around seven centimeters,[br]so in terms of meters that 0:10:35.388,0:10:38.865 would be 0.07 meters, so[br]we can figure out what's 0:10:38.865,0:10:41.240 the total kinetic energy,[br]well there's gonna be 0:10:41.240,0:10:43.202 a rotational kinetic[br]energy and there's gonna be 0:10:43.202,0:10:45.048 a translational kinetic energy. 0:10:45.048,0:10:47.875 The translational kinetic[br]energy, gonna be one half 0:10:47.875,0:10:50.835 the mass of the baseball[br]times the center of mass speed 0:10:50.835,0:10:53.993 of the baseball squared which[br]is gonna give us one half. 0:10:53.993,0:10:57.626 The mass of the baseball was[br]0.145 and the center of mass 0:10:57.626,0:11:00.650 speed of the baseball is 40,[br]that's how fast the center 0:11:00.650,0:11:02.712 of mass of this baseball is traveling. 0:11:02.712,0:11:06.712 If we add all that up we[br]get 116 Jules of regular 0:11:06.712,0:11:08.894 translational kinetic energy. 0:11:08.894,0:11:11.246 How much rotational[br]kinetic energy is there, 0:11:11.246,0:11:13.281 so we're gonna have[br]rotational kinetic energy 0:11:13.281,0:11:16.088 due to the fact that the[br]baseball is also rotating. 0:11:16.088,0:11:19.587 How much, well we're gonna[br]use one half I omega squared. 0:11:19.587,0:11:22.484 I'm gonna have one half, what's[br]the I, well the baseball is 0:11:22.484,0:11:26.328 a sphere, if you look up the[br]moment of inertia of a sphere 0:11:26.328,0:11:29.665 cause I don't wanna have[br]to do summation of all 0:11:29.665,0:11:32.999 the M R squareds, if you[br]do that using calculus, 0:11:32.999,0:11:34.873 you get this formula. 0:11:34.873,0:11:36.995 That means in an algebra[br]based physics class 0:11:36.995,0:11:38.900 you just have to look this[br]up, it's either in your book 0:11:38.900,0:11:41.635 in a chart or a table or you[br]could always look it up online. 0:11:41.635,0:11:45.763 For a sphere the moment of[br]inertia is two fifths M R squared 0:11:45.763,0:11:48.619 in other words two fifths[br]the mass of a baseball 0:11:48.619,0:11:50.459 times the raise of the baseball squared. 0:11:50.459,0:11:53.627 That's just I, that's the[br]moment of inertia of a sphere. 0:11:53.627,0:11:56.235 So we're assuming this[br]baseball is a perfect sphere. 0:11:56.235,0:11:59.358 It's got uniform density,[br]that's not completely true. 0:11:59.358,0:12:00.954 But it's a pretty good approximation. 0:12:00.954,0:12:03.019 Then we multiply by this omega squared, 0:12:03.019,0:12:04.754 the angular speed squared. 0:12:04.754,0:12:07.137 So what do we get, we're[br]gonna get one half times 0:12:07.137,0:12:11.222 two fifths, the mass of[br]a baseball was 0.145. 0:12:11.222,0:12:13.284 The radius of the baseball[br]was about, what did we say, 0:12:13.284,0:12:18.027 .07 meters so that's .07[br]meters squared and then finally 0:12:18.027,0:12:20.494 we multiply by omega squared[br]and this would make it 0:12:20.494,0:12:23.238 50 radians per second and we square 0:12:23.238,0:12:25.821 it which adds up to 0.355 Jules 0:12:28.705,0:12:31.416 so hardly any of the[br]energy of this baseball 0:12:31.416,0:12:33.034 is in its rotation. 0:12:33.034,0:12:36.449 Almost all of the energy is[br]in the form of translational 0:12:36.449,0:12:38.521 energy, that kinda makes sense. 0:12:38.521,0:12:40.895 It's the fact that this[br]baseball is hurling toward 0:12:40.895,0:12:43.901 home plate that's gonna[br]make it hurt if it hits you 0:12:43.901,0:12:46.049 as opposed to the fact[br]that it was spinning when 0:12:46.049,0:12:48.545 it hits you, that doesn't[br]actually cause as much damage 0:12:48.545,0:12:50.705 as the fact that this[br]baseball's kinetic energy 0:12:50.705,0:12:54.466 is mostly in the form of[br]translational kinetic energy. 0:12:54.466,0:12:57.154 But if you wanted the total[br]kinetic energy of the baseball, 0:12:57.154,0:12:59.135 you would add both of these terms up. 0:12:59.135,0:13:02.641 K total would be the[br]translational kinetic energy 0:13:02.641,0:13:04.937 plus the rotational kinetic energy. 0:13:04.937,0:13:09.104 That means the total kinetic[br]energy which is the 116 Jules 0:13:10.046,0:13:12.546 plus 0.355 Jules which give us 0:13:14.425,0:13:15.592 116.355 Jules. 0:13:18.343,0:13:20.590 So recapping if an object is both rotating 0:13:20.590,0:13:23.156 and translating you can[br]find the translational 0:13:23.156,0:13:26.787 kinetic energy using one[br]half M the speed of the 0:13:26.787,0:13:29.564 center of mass of that[br]object squared and you can 0:13:29.564,0:13:32.071 find the rotational[br]kinetic energy by using 0:13:32.071,0:13:34.552 one half I, the moment of inertia. 0:13:34.552,0:13:36.161 We'll infer whatever shape it is, 0:13:36.161,0:13:38.640 if it's a point mass[br]going in a huge circle 0:13:38.640,0:13:41.035 you could use M R[br]squared, if it's a sphere 0:13:41.035,0:13:43.635 rotating about its center[br]you could use two fifths 0:13:43.635,0:13:46.209 M R squared, cylinders[br]are one half M R squared, 0:13:46.209,0:13:49.007 you can look these up[br]in tables to figure out 0:13:49.007,0:13:52.032 whatever the I is that[br]you need times the angular 0:13:52.032,0:13:56.319 speed squared of the object[br]about that center of mass. 0:13:56.319,0:13:58.423 And if you add these[br]two terms up you get the 0:13:58.423,0:14:01.423 total kinetic energy of that object.