good day everybody just ignore us are coached physics here we're gonna look at some more aspects of a rotation and let's just get to it here so I got a question for you take this point mass here let's say it's a point mass highly localized and I spin it around like this I say this that point mass have kinetic energy yeah it's moving through space right it's moving it has the velocity it's the mass mass and velocity and that substantiates some sort of kinetic energy if I got hit with this it would hurt because it would be partying its kinetic energy to me and I don't really want that all right can hit something else with it knock something over do work all sorts of stuff right but the main point is yeah has kinetic energy so what we want to start looking at is well how do we have quantify kinetic energy of like extended objects or objects that are ultimately composed of multiple point masses I can take this entire meter stick here and I can start spinning around say hey does this meter stick have kinetic energy you say well yeah it's moving oh yeah that's great it is it's moving but different parts of its different mass elements if you will are moving at different rates through space this little orange mass element versus this little blue mass element are moving at vastly different rates through space as this rotates around as a rigid object everything's moving at the same angular velocity but not at the same tangential velocity not at the same rate through space and that's what kinetic Energy's really it's functional right how fast is something moving so I want to starts doing is looking at location energy and generally I just call it king Issa are so we know that we're talking about a rotation subscript R in there so consider a point mess cheating is a circular path at a constant angular rate [Music] axis of rotation will take some point mass I'll call it m1 it is some distance from the axis of rotation a distance R 1 and this point mass here is going to be initial but 1 will be rotating around this axis of rotation with a constant angular velocity Omega 1 stay well what is its kinetic energy it comes down to this right comes down to how fast is actually moving through space this max likewise has a tangential speed will really is kinetic energy is going to be one-half times the mass times speed squared tangential speed so that's that's fine that's just our old notion of kinetic energy when half MV squared let's write it in terms of the rotational quantity though let's write it in terms of Omega and I put a little subscript one there one day so we know and what do we have we have that the tangential speed of this mass m1 is equal to R 1 Omega 1 so we can write then that rotational kinetic energy of this mass m1 is then equal to one-half m1 times the quantity of r1 Omega 1 squared there we go we've got it's rotational kinetic energy written in terms of the radius at which it rotates and its angular velocity we'll use the constant angular velocity doesn't need to be constant if it's not constant what's the rotational kinetic energies changing with respect to time that's kind of perspective if anything at this point so the main point is that we can quantify it in terms of rotational quantities but what happens if we now mass two point masses that are originally connected now consider two point masses rigidly connected take some axis of rotation take some M one like this and attach to it we have some m2 this is one rigid system here we've got two different masses we've got this one exists at a distance of r1 and this distance will call r2 so they're both different distances from this axis of rotation we say well what is the total kinetic energy of this rotating object what is the total rotational kinetic energy well if we're looking for the total and it would be the sum of the individual rotational kinetic energies right we're just saying we got two objects rotating and just sum up what their individual contributions are so we have that ke sub rotation total going to be ke r1 plus ke are two rotational kinetic energy of mass number one plus rotational kinetic energy of mass number two which we can write out as one-half m1 v tangential one squared plus one-half m2 v2 and the masses and what else do we have with this we've got the be tangential one as equal to r1 Omega 1 and be tangential to it is equal to R 2 Omega 2 and there's a little connection that we need to make here if this is a rigid object every point on it rotates at the same angular rate everything goes to the same angle in the same amount of time so everything has the same angular velocity even though that we're saying there's 2 and Omega 1 and Omega 2 Omega 1 is equal to Omega 2 if the object is rigid so we don't really need to make a distinction between those omegas if we're saying that some rigid object but there is a distinction to be made between the actual or tangential speeds right we have the tangential one we have a in general to which objects moving faster through space the one that's farthest away from the axis of rotation because it has a much greater distance or a relatively greater distance than the most traveling the same amount of time as this object as it sweeps with the same angle in the same amount of time so this is where rotation really necessitates they take a close look at this it's not just how fast the objects moving because different points on the object are moving at different rates so go this will just go ahead and put in our v kendra to win or be tangential to here and there and we can get the total rotational kinetic energy of this system is going to be one-half m1 it's the quantity of r1 Omega squared don't need Omega 1 there because both rotate at the same angular rate plus one-half m2 quantity of our two Omega squared all right so the nice thing about this again is that everything is moving to the constant angular rate of the same angular rate not necessarily constant but they exhibit the same changes in angular position with respect to time Omega say for the entire rigid object but depending on how far away the mass element is from the axis of rotation really has something to say with how much kinetic energy that particular element has the further away from the axis of rotation the greater the contribution kinetic energy wise that that particular mass element has again we're just looking at two point masses but we can still do this right what happens if I say well m1 and m2 those are two different mass elements with different masses they are at r1 and r2 two different radial distance away from the axis of rotation I have the same Omega we rewrite this a little bit well I can factor out an Omega squared factor out a 1/2 this as 1/2 multiplied by the quantity of m1 r1 squared + m2 r2 squared multiplied by Omega squared the reason that we want to write it out like this is because again this quantity right in the middle here is what is dependent on the mass distribution where the individual point masses are relative and axis of rotation and what those individual point masses are how much mass what happens if we have three masses you stick on another one for five or a hundred or billion or aquapod rows never what are we gonna do if we got three different masses at three different radii we would end up having three different terms we have m1 r1 squared + m2 r2 squared plus m3 r3 squared more messes we'd add on them and for our 4 squared 5 and 5 are 5 squared we just keep on adding things up right we did another term in here for every mass that we add into the system it really doesn't matter if we're lining them all up or putting something else over here within our 4m 4 as long as it's the rigid object all we need to know is what is the mass element and how far away is it from the axis of rotation and add that into this particular section right here so what this allows us to do is actually write out rotational kinetic energy for a system of rigidly bound point masses rotating and pinky were great Omega as this this middle term just write this down for system of rigid Dooley down point masses let's hit it a rape I mean you are a mega total rotational kinetic energy is you can write it like this we've got ourselves key total this function equal to 1/2 divided by some sort of sum here write the sum of individual masses mass elements times their radial distance away from the axis of rotation squared let's write that as the sum from I equals 1 to N of M sub I R sub I squared times Omega squared this term right here M sub I R sub I squared is just 1 R 1 squared plus 2 R 2 squared plus + and some end R sub N squared tell us how many nasa elements you have that's how many terms you're gonna have in there we're just right in summation notation well we can define this as a certain quantity is defined what is the mass moment of inertia for system of point masses moment of inertia is equal to M sub I R sub I sweat where mSpy R sub I is what we just described with the individual mass elements are and what radial distance away from the axis of rotation they reside which nonetheless we can define this quantity moment of inertia and note that can be written as one-half I Omega squared which is very much like one-half MV squared but this is for rotational kinetic energy object rotating with an angular velocity Omega as some sort of kinetic energy with respect to the mass elements rotating around some axis of rotation substantiating some moment of inertia of the object this is really a measurement of the distribution of the mass will relative to the axis of rotation so moment of inertia is and this is necessitated by looking at both extended objects or objects with more than one localized mass under rotations Viru important quantity and very often it is confusing to people but it kind of get used to it and understand this is just really a matter of where are all the maths elements and where are they where are mass M is relative to the axis of rotation and what are their individual masses in terms of their part mentalization if you will so in the end you think about any mass right got this extended object this is just a bunch of point masses a bunch of protons and neutrons and electrons all highly localized at different positions and depending on where I rotate this those individual mass elements are different radial distances away from this axis of rotation I rotate it from the middle I've got a different mass distribution or relative to that axis of rotation so what you have to be careful of is again what is your axis of rotation where you're finding an object's moment of inertia it isn't something that just a matter of object it's a matter of what point in space are you measuring it with respect to and for us we're measuring it with respect to an axis of rotation so that is a nice quantity units that's how you miss a moment of inertia what do they have to be hope we go back and look it has to be mass times radius squared so we got kilogram meter squared there's that yeah SI units of it and system of point masses so that's so we wanted to look at be back so what if we don't have point masses or we can start to consider extended masses that are continuous master's divisions which is really not - because like I said everything's made of a bunch of highly localized poop times electrons neutrons but if we can consider things to be sort of continuous because there's so many of them packed in then we can start to develop other ways of determining what moments of inertia are for some common types of objects so in order to do that that involves a little bit of a calculus so we'll split that into another segment and just note that generally for the problems that we're looking at you just need to go look up what the objects moment of inertia is relative to its particular axis of rotation so common objects moments objects moments of inertia about their stop their sex so we start looking at something like we have a wheel here well the wheel has some sort of fixed axis that's in the center here and everything rotates about that now this could be a solid wheel like a solid cylinder or it could just be this little strip here in the end that we consider like a hoop or it could be something like a ball that's rolling a sphere growing about its center those are the general types of objects that you'll be dealing with and have very well-defined moments of inertia so I'll just tell you what those are I'm not going to derive them here show you what one of them is this easy to figure out without any calculus but there's something that you're going to use from time to time some of these problems I'm just going to know what they are so look it I for a solid cylinder or a solid cylinder moment of inertia is one-half and R squared is equal to the total mass and r is the radius this is about its center so we got ourselves a cylinder and we're looking at it rotating about its center see your wheels something like then solid cylinder we've also got a cylindrical ring ring that ends up being just M times R squared R again times the total mass R to the radius so finally it's a hoop that has the same radius and same mass as the solid cylinder has a greater momentum inertia that's myself why would that be you grab something really quick I've got two things right here these actually I'm right about the same mass as soon as you're happy face before and they have about the same radii as whether relative to their centers which they would roll about just allowed to roll this one has a greater moment of inertia why because all of its mass all of its individual mass elements are as far away as they can be from the central axis of rotation whereas this one you have the same amount of mass elements but there's a spread summer as far away as they can be some are really close again for point masses the moment of inertia depends on this heavily how far away the individual mass elements are away from the axis of rotation so if it turns out that moment of inertia for a solid cylinder is less than the moment of inertia for a comfortable ring so those are two that you'll deal with quite a bit we've also got solid sphere at the moment of inertia and a solid sphere is equal to two this M R squared and a spherical shell much like a a ball of it has a very thin shell two thirds are square these are all considered about the very center of the object itself so those are some kind of moments of inertia that you'll run into we can actually get this one pretty easily so for this loop here okay take some central axis and there's a circle we say that there's our axis of rotation and we have that all of the mass elements have the same radii so we can even consider this to be made of a bunch of little mass elements all packed together not even treat this as continuous treated as truly a thin layer of these mass elements all separated by so we've got and one two three four five six seven eight m8 and we go all the way around and how many mass elements we break this up into is how many we've got right so we look at the moment of inertia of this we have that it is going to be equal to the sum from I equals 1 to N of M sub I R sub I squared well R sub 1 is equal to R sub 2 is equal to R sub n every mass elements at the same radial distance away from the axis of rotation so this is just one value we're summing up over one kind of just gets pulled out so we could write this thing as is equal to R squared multiplied by the sum from I equals 1 to N of xfi so what happens when we sum up all the individual mass elements this is going to be equal then 2r squared multiplied by M 1 plus M 2 plus M 3 plus dot dot dot plus M sub n what happens when you add up all the individual mass elements that's the total mass right this is equal to M R squared so there you go there's this cylindrical hoop or a ring if you will this moment of inertia extracted out pretty easy just by directly looking at the fact that everything has the same radio distance even up for the spherical shell to a little bit more tricky but not that bad and the other ones generally you need to be able to for some calculus to add up the variation the radial distances but any rate there's some common ones that you'll utilize some read things carefully and use alright good stuff and everybody we want to look at a little problem now involving some rotational kinetic energy well also consider not just rotational but translational as well because our linear type of kinetic energy so consider this I take some object and I say spinning spinning around right has some rotational kinetic energy because of all the mass element at different locations are moving relative to the axis of rotation I take the moment of inertia of the object that's fine one-half I Omega squared I figure out what its rotational kinetic energy is well what happens if I have an object that's actually translating its rotation into linear motion that is both rotating and moving floored let's just think of the moving Ford part this object moves forward as a whole at one rate whether I'm dragging it across this or whether it's actually rolling the whole objects moving this total mass is moving up one velocity or all the mass elements are moving forward at one velocity as a whole and then we have all the individual mass elements rotating around having their different rotational velocities or rotational velocities but rotational kinetic energies so when we start talking about rolling objects the object has both translational and rotational kinetic energy so we just want to appreciate that object under translation can have two different types in the sense of kinetic energy it still manifestations of all moving parts really moving around let's just take a little look at a problem that involves some rotational and translational kinetic energy and we'll go from there so we're gonna take ourselves a solid sphere that has particular radius and we're gonna phone cost around and ask some questions about it so here's the problem you all right so we've got ourselves some Louisville ground here and we're gonna take ourselves a solid sphere as some radius and it is going to have some angular acceleration that said that that is three radians per second squared and then it was initially at rest and note that this is a solid sphere and what do we do what we need is moment of inertia if we're considering anything regarding rotational kinetic energy so we know that I is equal to we're looking on up for a solid sphere rotating about its center which a rowing object does will be 2/3 M R squared so let's say that the mass of this object is 2.50 kilograms and the radius of this object is 0.35 zero meters it's initially at rest undergoing an angular acceleration of 3 radians per second squared we just have some questions so the first question was what is the rotational kinetic energy at T equals four seconds it was 40 seconds - what Oh we'll just go ahead and figure out what we need to know we know that rotational kinetic energy is one-half I Omega squared I for this given object here rotating about its center is this two-thirds mr squared that's constant when asked constant this is what's changing with respect to time right that angular velocity based upon the angular acceleration so we've got the Omega final is equal to alpha t plus Omega initial we wanted to we could write this out and say that ke sub R as a function of time it's going to be one half times two-thirds M R squared multiplied by Omega squared where that is going to be alpha t plus Omega initial quantity squared so well that looks sort of big it's really not because this reduces down to so number one Oh Meg initial is equal to zero so we can get rid of this one half times two thirds I'll just call it one third and we are left with this this is now [Music] one third and R squared times falls into place and there we go so what do we want to do we want to evaluate this at t equals four seconds so this is going to go to one-third x 10.5 kilograms and multiply it by 0.35 meters quantity squared multiplied by the quantity of three-year radians per second squared times the time of four seconds it's that whole thing squared which gives us the rotational kinetic energy evaluated at t equals four seconds to be once 17.7 what joules 2.7 joules of a rotational kinetic energy after 4 seconds goes by alright so what about the translational kinetic energy what is kinetic energy of this object at T equals 4 seconds translational eyes so this well that is just going to be equal to 1/2 MV squared evaluated at T equals 4 seconds so we need to know how fast is this object moving forward at T equals 4 seconds this is translational velocity so we know that V is equal to R Omega through equations of a translation so we can go ahead and figure out what that is either in terms of Omega or V it doesn't really matter ourselves Omega because we already got our radius and we know what the angular acceleration is because I can finally go to the alpha t plus Omega initial so we can write this as being equal to one-half M times V squared where V is equal to R times Omega we've got R squared times Omega squared which is then break the sentence so make a final squared which is that equal to and this is really a capital R here radio 7th 1/2 M R squared times Omega final squared which is alpha so again alpha is just three readings per second squared 2 T is equal to the 4 minutes per second and we got ourselves five times five twenty two point zero five joules of translational kinetic energy so there we go we've got this what else can we ask about this seat what is the total kinetic energy at T equals four seconds well the total kinetic energy is just the sum of the rotational and translational so this is going to be equal to KT sub R evaluated at T equals four seconds plus ke T evaluated at G is equal to four seconds which is merely just our fourteen point seven joules plus our 2010 point zero five joules which is what going to give us 36 point seven five joules of total kinetic energy after four seconds this come by right could you just ask that in the very beginning quantified what these terms are there you got to be able to see that we can break it down into the individual pieces if you will of kinetic energy rotational and translational there's another question hey at what point in time will the total kinetic energy of this object equal to so we need to figure out when is the total kinetic energy equal to this so what do we have hopefully cut ourselves that 75 joules must be equal to the sum of the translational and rotational kinetic energies so we got a can't use of T plus a KT piece of our whoa we know that this is one-half MV squared plus one-half I Omega squared I am we can start opening some things up is equal to R Omega so we can go ahead and put that in there let's write this out 75 juice 1/2 M R R Omega quantity squared for the V squared plus one-half I Omega sort of just weave in I how it is just right now both factor that common factor beam Omega Omega squared does that mean holds the time dependence in that right we know this Omega is going to be alpha T initial is equal to zero but Omega as a function of time so there's the time right there we gotta be able to isolate that so what does expansion stuff out who can group some terms and kind of see what we can do with this so let us write this as 75 tools is equal to 1/2 M R squared multiplied by alpha T quantity squared R squared that Omega squared plus a 1/2 times 2/3 and R squared times Omega squared which is so one of the stuff is gonna collapse down right cuz we've gotta come in term of M and R squared and then alpha squared all the things that we know and then we can factor out the T squared lose everything on the other side nice nice final expression there so let's collect some terms here we can make this one third and then we've got a 1/2 M R squared plus a 1/3 M R squared comes out to the squared quantity times seventy five joules is equal to the quantity of one-half M R squared alpha squared plus one third M R squared alpha squared in factoring that alpha L times T squared that's what we want to isolate we want to figure out the time when that occurs that okay everything is the same one half of something plus one third of something that's something being M R squared alpha squared so what can we do here you can tell this to six five six mr squared alpha squared times T squared so 75 joules is equal to one third plus a 1/2 we got 3 sixths plus 2/6 which gives us a five six five six and R squared alpha squared T squared I want to solve this F of T do that I had a little problem with video back though so we have that again seventy five tools it's five six mr squared of the squared times T squared and go ahead and solve this a forty six times seventy five times five divided by five times two point five M R squared alpha squared square root of all that and we get our nice values five point seven one four seconds that is the time at which the total kinetic energy of the object is seventy five joules compared to T equals four seconds when it was approximately thirty six tools it's not much more time right so about one point seven seconds goes by and it increased its kinetic energy almost twofold almost doubled it really close what's with that I'm gonna remember that the kinetic energy itself is dependent on how fast this objects spinning around because how fast it moves forward depends on how fast it's spinning around and well that is that it depends on it squared but the rate of the change of the angular velocity is linear it's not one-to-one we don't get a linear relationship with respect to time and kinetic energy it varies as Omega squared ultimately so there you go there we go all right have a good one