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when a major league baseball player throws a fastball that balls definitely got kinetic energy we know that because if you get in the way you can do work on you that's going to hurt you got to watch help but here's my question does the fact that most pitches unless you're throw-in a knuckleball does the fact that most pitches head toward home plate with the baseball spinning mean that that ball has extra kinetic energy well it does and how do we figure that out that's the that's the goal for this video how do we determine what the rotational kinetic energy is of an object well if I was coming out this for the first time my first guess I'd say okay I say I know what regular kinetic energy looks like the formula for regular kinetic energy is just one-half MV squared so I'd say all right I want rotational kinetic energy let me just call that K rotational and what is that going to be well I know for objects that are rotating the rotational equivalent of mass is moment of inertia so I might guess all right instead of mass I'd have moment of inertia because in Newton's second law for rotation I know that instead of math there's a moment of inertia so maybe I replace that and instead of speed squared maybe since I have something rotating I'd have angular speed squared and it turns out this works so you can often derive not really a derivation you're just kind of guessing educated ly but you can often get the formula for the rotational analog of some linear formula by just substituting the rotational analog for each of the variables so if I replace mass with rotational mass I get the moment of inertia if I replace speed with rotational speed I get the angular speed and this is the correct formula so in this video we need to derive this because that it's not really in derivation we didn't really prove this we just showed that it's plausible how do we prove that this is the rotational kinetic energy of an object that's rotating like a baseball well the first thing to recognize is that this rotational kinetic energy isn't really a new kind of kinetic energy it's still just the same old regular kinetic energy for something that's rotating and what I mean by that is this so imagine this baseball that's rotating in a circle every point on that baseball is moving with some speed so what I mean by that is this so this point at the top here imagine this little piece of the leather right here it's going to have some speed forward I'm going to call this mass m1 that little piece of mass right there and I'll call the speed of it v1 similarly this point on the leather right there I'm going to call it m2 it's going to be moving down because this is rotating in a circle so I'll call that v2 and points closer to the axis are going to be moving with smaller speed so this point right here we'll call it m3 moving down with a speed V 3 that is not as big as v2 or v1 I you can't see that very well I'll use a darker green so this m3 right here closer to the axis axis being right at this point in the center closer to the axis so it's speed is smaller than points that are farther away from this axis so you can see this is kind of complicated all points on this baseball are going to be moving with different speeds so points over here that are really close to the axes barely moving at all I'll call this m4 and it would be moving at speed V for what we mean by the rotational kinetic energy is really just all the regular kinetic energy these masses have about the center of mass of the baseball so in other words what we mean by K rotational is you just add up all of these energies you'd have 1/2 this little piece of leather up here would have some kinetic energy so you do one-half m1 v1 squared plus and this m2 has some kinetic energy don't worry that it points downward downward doesn't matter for things that aren't vectors this V gets squared so kinetic energy is not a vector so it doesn't matter that one velocity points down because this is just speed and similarly you'd add up 1/2 m3 the three squared but you might be like this is impossible there's infinitely many points in this baseball how am I ever going to do this well something magical was about to happen this is one of my favorite little derivations it's short and sweet because watch what happens k irrotational is really just the sum if I add all these up I can write it as a sum of all the one-half MV squared zuv every point on this baseball so imagine breaking this baseball up into very very small pieces don't do it physically but just think about it mentally just visualize considering very small pieces particles of this baseball and how fast they're going what I'm saying is that if you add all of that up you get the total rotational kinetic energy this looks impossible to do but something magical is about to happen and here's what we can do we can rewrite see the problem here is V all these points have a different speed V but we can use a trick a trick that we love to use in physics instead of writing this as V we're going to write V as so remember that for things that are rotating V is just R times Omega the radius how far from the axis you are times the angular velocity or the angular speed gives you the regular speed this formula is really handy so we're going to replace V with R Omega and this is going to give us R Omega and you still have to square it and at this point you probably thinking like this is even worse what did we do this for we'll watch if we add the stuff I have 1/2 M I'm going to get an R squared and an Omega squared and the reason this is better is that even though every point on this baseball has a different speed V they all have the same angular speed Omega that was what was good about these angular quantities is that they're the same for every point on the baseball no matter how far away you are from the axis and since they're the same for every point I can bring that out of the summation so I can rewrite this summation and bring everything that's constant for all of the masses out of the summation so I can write this as 1/2 times the summation of M times R squared and end that quantity end of that summation and just pull the Omega squared out because it's the same for each term I'm basically factoring this out of all of these terms in the summation it's like up here all of these have a 1/2 you can imagine factoring out a 1/2 and just writing this whole quantity as 1/2 times m1 v1 squared plus m2 v2 squared and so on that's what I'm doing down here for the 1/2 and for the Omega squared so that's what was good about replacing V with R Omega D Omega is the same for all of them you can bring that out you might still be concerned you might be like we're still stuck with the M in here because you've got different M's at different and we're stuck with all these R Squared's in here all these points at the baseball are different ours they're all different points from the axis different distances from the axis we can't bring those out so now what do we do well if you're clever you recognize this term this summation term is nothing but the total moment of inertia of the object remember that the moment of inertia of an object we learned previously is just M R squared so the moment of inertia of a point mass is M R squared and the moment of inertia of a bunch of point masses is the sum of all the M R Squared's and that's what we've got right here this is just the moment of inertia of this baseball or whatever the object is it doesn't even have to be of a particular shape we're going to add all the M R Squared's that's always going to be the total moment of inertia so what we found is that the K rotational is equal to one-half times this quantity which is I the moment of inertia times Omega squared and that's the formula we got up here just by guessing but it actually works and this is why it works because you always get this quantity down here which is one-half I Omega squared no matter what the shape of the object is so what this is telling you what this quantity gives us is the total rotational kinetic energy of all the points on that mass about the center of mass but here's what it doesn't give you this term right here does not include the translational kinetic energy so the fact that this baseball was flying through the air does not get incorporated by this formula we didn't take into account the fact that this baseball was moving through the air in other words we didn't take into account that the actual center of mass of this baseball was translating through the air but we can do that easily with this formula here this is the translational kinetic energy so sometimes instead of writing regular kinetic energy now that we've got to we should specify this is really translational kinetic energy so we've got a formula for the translational kinetic energy the energy something has due to the fact that the center of mass of that object is moving and we have a formula that takes into account the fact that something can have kinetic energy due to its rotation that's this K rotational so if an object's rotating it has rotational kinetic energy if an object is translating it has translational kinetic energy ie if the center of mass is moving and if the object is translating and it's rotating then it would have both of these kinetic energies both at the same time and this is the beautiful thing if an object is translating and rotating and you want to find the total kinetic energy of the entire thing you can just add these two terms up if I just take the translational one-half MV squared and this would then be the velocity of the center of mass so you have to be careful this let me make some room here so let me get rid of all this stuff here if you take one-half M times the speed of the center of mass squared you'll get the total translational kinetic energy of the baseball and if we add to that the one-half I Omega squared so the Omega about the center of mass you'll get the total kinetic energy both translational and rotational so this is great we can determine the total kinetic energy all together rotational motion translational motion from just taking these two terms add it up so what would an example of this be let's just get rid of all this and let's say this baseball someone pitched this thing and the radar gun shows that this baseball was hurled through the air at 40 meters per second so it's heading toward home plate at 40 meters per second the center of mass of this baseball is going 40 meters per second toward home plate but let's say it's also someone really threw a fastball and this thing's rotating with an angular velocity of 50 radians per second and we know the mass of a baseball I looked it up the mass of a baseball is about 0.14 five kilograms and the radius of the baseball so a radius of a baseball is around seven centimeters so in terms of meters that would be zero point zero seven meters so we can figure out what's the total kinetic energy well there's going to be a rotational kinetic energy and there's going to be a translational kinetic energy the translational kinetic energy going to be one-half the mass of the baseball times the center of mass speed of the baseball squared which is going to give us 1/2 the mass of the baseball was zero point one four five and the center of mass speed of the baseball is forty that's how fast the center of mass of this baseball is traveling if we have that all up we get a hundred and sixteen Joule of regular translational kinetic energy but how much rotational kinetic energy is there so we're going to have rotational kinetic energy due to the fact that the baseball is also rotating how much will we're going to use 1/2 I Omega squared so I'm going to have 1/2 what's the I will the baseball is a sphere if you look up the moment of inertia of a sphere because I don't want to do I don't want to have to do some nation of all the M R Squared's if you do that using calculus you get this formula so that means in an algebra based physics class you just have to look this up together in your book in a chart or a table or you could always look it up online for a sphere the moment of inertia is two-fifths M R squared so in other words two-fifths the mass of the baseball times the radius of the baseball squared that's just I that's the moment of inertia of a sphere so we're assuming this baseball is a perfect sphere that's got uniform density that's not completely true but it's a pretty good approximation and then we multiply by this Omega squared the angular speed squared so what do we get we're going to get 1/2 times 2/5 the mass of the baseball was 0.14 5 the radius of the baseball was about would we say 0.0 7 meters so that's 0.0 7 meters squared and then finally multiplied by omega squared and this Omega was 50 radians per second and we square it which adds up to 0.35 5 joules so hardly any of the energy of this baseball is in its rotation almost all of the energy is in the form of translational energy that kind of makes sense it's the fact that this baseball is hurtling toward home plate that's going to make it hurt if it hits you as opposed to the fact that it was spinning when it hits you that doesn't actually cause as much damage as the fact that this baseball is kinetic energy is mostly in the form of translational kinetic energy but if you wanted the total kinetic energy of the baseball you would add both of these terms up k total would be the translational kinetic energy plus the rotational kinetic energy that means the total kinetic energy would just be a hundred and sixteen joules plus 0.35 five joules which gives us a hundred and sixteen point three five five joules so recapping if an object is both rotating and translating you can find the translational kinetic energy using 1/2 m the speed of the center of mass of that object squared and you can find the rotational kinetic energy by using 1/2 I the moment of inertia what if for whatever shape it is if it's a point mass going in a huge circle you could use M R squared if it's a sphere rotating about its center you could use 2/5 M R squared cylinders or 1/2 M R square and you can look these up in tables to figure out whatever the eye is that you need times the angular speed squared of the object about that center of mass and if you add these two terms up you get the total kinetic energy of that object