[ Voiceover] When a majorleague baseball player sheds a fast dance, that ball’sdefinitely go kinetic energy. We know that cause if you get in the way, it could do work onyou, that’s gonna hurt. You gotta watch out. But here’s my question: doesthe fact that most pitches, unless you’re throwing a knuckle clod, does the fact that mostpitches head toward home plate with the baseball spinningmean that that clod has additional kinetic energy? Well it does, and howdo we figure that out, that’s the goal for this video.How do we determine what the rotational kinetic energy is of an objective? Well if I was coming atthis for the first time, my first guest I’d say okay, I’d say I know what regularkinetic energy looks like. The formula for regular kinetic energy was only one half m v squared. So let’s say alright, I wantrotational kinetic energy. Let me only call that k rotational and what is that gonna be? Well I know for objectives that are rotating, the rotational equivalent ofmass is moment of inertia. So I might suspect alright instead of mass, I’d have moment of inertiacause in Newton’s second constitution for pirouette I know thatinstead of mass there’s moment of inertia so perhaps I change that.And instead of speedsquared, maybe since I got something revolving I’dhave angular move squared. It turns out this works. You were generally obtain, it’snot really a derivation, you’re just various kinds of guessingeducatedly but you could often get a formula for therotational analog of some linear formula by justsubstituting the rotational analog for each of the variables, so if I changed mass with rotational mass, I getthe moment of inertia. If I replace hastened with rotational velocity, I get the angular quicken andthis is the correct formula. So in this video we neededto ride this cause that is not really a source, we didn’t really prove this, we just showedthat it’s plausible.How do we prove thatthis is the rotational kinetic energy of anobject that’s revolving like a baseball. The first thing to recognizeis that this rotational kinetic energy isn’t really a new kind of kinetic energy, it’sstill simply the same aged regular kinetic energy forsomething that’s revolving. What I convey by that is this. Imagine this baseballis rotating in a roundabout. All the points on the baseballis moving with some move, so what I symbolize by that isthis, so this point at the top here imagine the littlepiece of leather right here, it’s gonna have some hurried forward.I’m gonna summon this massM one, that little piece of mass right now and I’llcall the accelerated of it V one. Similarly, this place onthe skin right there, I’m gonna call that M two, it’s gonna be moving down cause it’s a rotate clique, so I’ll call that V two and pitches closer to theaxis are gonna be moving with smaller fast sothis target right here, we’ll call it M three, movingdown with a race V three, that is not as big as V two or V one. You can’t see that very well, I’ll use a darker greenso this M three right here closer to the axis, axisbeing right at this level in the centre for human rights, closer to theaxis so it’s rapidity is smaller than tops that arefarther away from this axis, so you can see this is kinda complicated.All points on this baseballare gonna be moving with different quickens so pointsover here that are really close to the axis, barely moving at all. I’ll order this M four and it would be moving at hurrying V four. What we entail by therotational kinetic energy is really only all the regularkinetic energy these mass have about the centerof mass of the baseball. So in other words, whatwe planned by K rotational, is you really add together all of these energies.You have one half, thislittle case of skin up here ought to have been some kinetic energy so you do one half Mone, V one squared plus. And this M two has some kinetic energy, don’t worry that it moments downward, downward doesn’t matter forthings that aren’t vectors, this V goes squared sokinetic energy’s not a vector so it doesn’t matter thatone velocity qualities down compel this is just thespeed and similarly, you’d add up one half Mthree, V three squared, but you might be like this is impossible, there’s endlessly manypoints in this baseball, how am I ever going to do this. Well something mystical is about to happen, “its one of” my favoritelittle sources, short and sweet, watch what happens.K E rotational is really precisely the part, if I lend all these upI can write is as a sum of all the one half M Vsquares of all the points on this baseball so imaginebreaking this baseball up into extremely, very small articles. Don’t do it physically butjust think about it mentally, time envisage consideringvery small fragments, molecules of this baseballand how fast they’re going. What I’m saying is thatif you lend all of that up, you get the totalrotational kinetic energy, this inspections impossible to do. But something magical is about to happen, here’s what we can do. We can rewrite, envision the problem here is V. All these points have a different velocity V, but we can use a ruse, a gimmick that we love to use in physics, insteadof writing this as V, we’re gonna write V as, soremember that for things that are revolving, Vis just R terms omega. The radius, how far from the axis you are, epoch the angular velocity, or the angular move gives you the regular speed.This formula is reallyhandy, so we’re gonna replace V with R omega, and thisis gonna present us R omega and you still have tosquare it and at this moment you’re probably thinkinglike this is even worse, what do we do this for. Well watch, if we lend thisis up I’ll have one half M. I’m gonna get an R squaredand an omega squared, and the reason this isbetter is that even though every point on this baseballhas a different hurried V, they all have the sameangular speed omega, that was what was good aboutthese angular lengths is that they’re the same forevery point on the baseball no matter how far awayyou are from the axis, and since they’re thesame for every point I can bring that out of thesummation so I can rewrite this summation and bringeverything that’s constant for all of the masses outof the summation so I can write this as one half periods the summation of M times R squaredand end that sum, intent that summation and justpull the omega squared out because it’s the same for each term.I’m basically factoringthis out of all of these terms in the summation, it’s like up here, all of these have a one half. You could imagine factoring out a one half and merely writing this whole quantity as one half durations M one V one squared plus M two V two squared and so on. That’s what I’m doingdown now for the one half and for the omega squared, so that’s what was good about ousting V with R omega. The omega’s the same for all of them, you can bring that out. You might still beconcerned, you might be like, we’re still stuck withthe M in now induce you’ve got different Ms at different points. We’re stuck with allthese R squareds in here, all these points at thebaseball are different Rs, they’re all differentpoints from the axis, different intervals fromthe axis, we can’t bring those out so now what do wedo, well if you’re intelligent you recognize this term.This summation term isnothing but the total moment of inertia of the object. Remember that the momentof inertia of an objective, “weve learned” previously, is just M R squared, so the moment of inertia of a degree mass is M R squared and the moment of inertia of a assortment of pointmasses is the sum of all the M R squareds and that’swhat we’ve got right here, this is just the moment ofinertia of this baseball or whatever the object is, it doesn’t even have to be of a particular shape, we’re gonna computed all the M R squareds, that’salways “re going to be” the total moment of inertia. So what we’ve learn isthat the K rotational is equal to one half experiences this capacity, which is I, the moment of inertia, period omega squared and that’s the formula we got up here precisely by guessing.But it actually worksand this is why it slogs, because you always getthis capacity down now, which is one half I omegasquared , no matter what the shape of the objective is. So what this is tellingyou, what this part pays us is the totalrotational kinetic energy of all the points on thatmass about the center of the mass but here’swhat it doesn’t give you. This word right here does not include the translational kineticenergy so the fact that this baseball was flyingthrough the aura does not get incorporated by this formula. We didn’t take into account the fact that the baseball was moving through the air, in other words, wedidn’t take into account that the actual centerof mass in this baseball was altering through the breeze. But we can do that easilywith this formula now. This is the translational kinetic energy.Sometimes instead of writingregular kinetic energy , now that we’ve got two weshould specify this is really translational kinetic energy. We’ve got a formula fortranslational kinetic energy, the power something has dueto the fact that the center of mass of that object ismoving and we have a formula that takes into account thefact that something can have kinetic energy due to its spin. That’s this K rotational, so if an object’s rotate, it has rotational kinetic energy. If an objective is restating it has translational kinetic energy, i.e. if the center of mass is moving, and if the objective istranslating and it’s rotating then it ought to have been those bothof these kinetic energies, both at the same time andthis is the beautiful thing.If an objective is translatingand revolving and you want to find the total kineticenergy of the entire thing, you are able to contributed these two terms up. If I just take the translationalone half M V squared, and this would then be thevelocity of the center of mass. So you have to be careful. Let me start some roomhere, so let me get rid of all this stuff here. If you take one half M, duration the acceleration of the center of mass squared, you’llget the total translational kinetic energy of the baseball. And if we add to that theone half I omega squared, so the omega about thecenter of mass you’ll get the total kinetic energy, bothtranslational and rotational, so this is great, we candetermine the total kinetic energy wholly, rotationalmotion, translational action, from simply taking these two terms computed up. So what the hell is an example of this be, let’s just get rid of all this. Let’s say this baseball, someone pitched this thing, and the radar firearm shows thatthis baseball was lunged through the breeze at 40 rhythms per second.So it’s heading toward homeplate at 40 meters per second. The center of mass ofthis baseball is going 40 meters per second toward home plate. Let’s say it’s also, someonereally hurled the fastball. This thing’s rotatingwith an angular velocity of 50 radians per second. We know the mass of abaseball, I’ve glanced it up. The mass of a baseballis about 0.145 kilograms and the radius of the baseball, so a radius of a baseball is around seven centimeters, so in terms of meters oh 0.07 meters, sowe can figure out what’s the total kinetic energy, well there’s gonna be a rotational kineticenergy and there’s gonna be a translational kinetic energy. The translational kineticenergy, gonna be one half the mass of the baseballtimes the center of mass speed of the baseball squared whichis gonna impart us one half. The mass of the baseball was0. 145 and the center of mass speed of the baseball is 40, that’s how fast the center of mass of this baseball is traveling.If we include all that up weget 116 Jules of regular translational kinetic energy. How much rotationalkinetic energy is there, so we’re gonna haverotational kinetic energy due to the fact that thebaseball is also rotating. How much, well we’re gonnause one half I omega squared. I’m gonna have one half, what’sthe I, well the baseball is a sphere, if you look up themoment of inertia of a globe compel I don’t wanna haveto do summation of all the M R squareds, if youdo that using calculus, you get this formula. That symbolizes in an algebrabased physics class you just have to look thisup, it’s either in your bible in a chart or a table or youcould always search it up online. For a orbit the moment ofinertia is two fifths M R squared in other words two fifthsthe mass of a baseball period the promote of the baseball squared. That’s just I, that’s themoment of inertia of a field. So we’re presume thisbaseball is a perfect sphere.It’s got uniform density, that’s not completely true. But it’s a pretty good approximation. Then we multiply by this omega squared, the angular quicken squared. So what do we get, we’regonna get one half goes two fifths, the mass ofa baseball was 0.145. The radius of the baseballwas about, what did we say,. 07 rhythms so that’s. 07 rhythms squared and then finally we multiply by omega squaredand this would make it 50 radians per second and we square it which lends up to 0.355 Jules so hardly any of theenergy of this baseball is in its spin. Almost all of the exertion isin the form of translational vigour, that kinda manufactures gumption. It’s the fact that thisbaseball is hurling toward home plate that’s gonnamake it hurt if it hits you as opposed to the factthat it was revolving where reference is makes you, that doesn’tactually make as much damage as the fact that thisbaseball’s kinetic energy is mostly in the form oftranslational kinetic energy.But if you missed the totalkinetic energy of the baseball, you would contributed both of these calls up. K total would be thetranslational kinetic energy plus the rotational kinetic energy. That wants the total kineticenergy which is the 116 Jules plus 0.355 Jules which give us 116.355 Jules. So recapping if an object is both rotating and decoding you canfind the translational kinetic energy using onehalf M the acceleration of the center of mass of thatobject squared and you can find the rotationalkinetic energy by use one half I, the moment of inertia. We’ll infer whatever determine it is, if it’s a place massgoing in a huge roundabout you could use M Rsquared, if it’s a ball rotating about its centeryou could use two fifths M R squared, cylindersare one half M R squared, you can look these upin tables to figure out whatever the I is thatyou need durations the angular race squared of the objectabout that center of mass. And if you compute thesetwo periods up you get the total kinetic energy of that objective.
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