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Current time:0:00Total duration:9:40

AP.PHYS:

CON‑5.E (EU)

, CON‑5.E.2 (EK)

, CON‑5.E.2.1 (LO)

so we saw in previous videos that a ball of mass M rotating in a circle of radius R at a speed V has what we call angular momentum and the symbol we use for angular momentum is a capital L and the amount of angular momentum that it would have would be the mass of the ball times the speed of the ball so that means this is basically just the magnitude of the momentum but then we multiply by the radius of the circle that's traveling in and that gives us the angular momentum of this ball going in a circle which is great and good to know but sometimes you don't have a ball going in the circle and you want to know the angular momentum so for instance instead of this case let's say you had this case where instead of a ball going around in a circle you've got a rod of mass m and radius R and the whole rod rotates around in a circle let's say the outside edge travels at a speed V just like the ball did so the question is will this rod also have an angular momentum that's equal to MV R and it won't you could probably commit yourself for that because for the ball all the mass was traveling at a speed V and all the mass was at the outside edge of the circle that it traces out in other words all this mass is traveling at a radius of our but for this rod some of the mass in fact only part of the mass only this outside edge of the mass is actually traveling at a radius R that's the part that travels at the full radius R the rest of these pieces of mass like this one in here traces out a circle it definitely traces out a circle but the circle it traces out is not equal to the radius R it's got a diminished R value so how do we determine the angular momentum of an object whose mass is distributed in a way that some of the mass is close to the axis and some of the masses far away from the axis that's what we're going to do in this video that's the goal and the approach physicists take to this is almost always the same we say well I've got the formula for the angular momentum of a single particle traveling at a single radius so let's just imagine our continuous object being composed of a bunch of single masses all traveling at a single radius so if I break this continuous mass up into individual pieces right so if I imagine being broken up into all these little pieces well then if I found the angular momentum of each piece and added it up I'd get the total angular momentum for the whole object so let's try this so the angular momentum of some piece of the object let's say that little piece of mass is going to be well the mass of that little piece it's I'm going to write em it's not the entire mass of this entire rod it'd just be the mass of that small piece times the speed of that piece times the radius that it is at so to make this clear let me just write this as like piece one so this would be m1 v1 and r1 and this would be the momentum angular momentum of that small piece and you can do this for mass two over here and you would get that the angular momentum of mass 2 would be M 2 V 2 and R 2 now keep in mind that these these are all going to be different so the speed out here at the outside edge is going to be fastest this speeds not going to be as great and this speed closer to the middle is even smaller because they're tracing out smaller circles in the same amount of time as these outside pieces trace out larger circles in the same amount of time so at this point you might be worried you might be like this is gonna be really hard we're gonna have to add all these up they've all got different speeds they're all at different radii how are we going to do this well you got to have faith and something magical is about to happen so let me show you what happens if we imagine adding all these up I only draw two you got to imagine those are an infinite amount of these right so that makes it seemed even harder but imagine breaking this up into an infinite amount of these little discrete masses and considering each individual angular momentum they'd be very small because this M one would be an infinitesimal very small piece of mass and let's add them all up and see what we get so if we add up all of the MV R's of every piece of mass on this rod that would be the total angular momentum of the rod so in other words this is really just M 1 V 1 R 1 plus M 2 V 2 R 2 and you'd so on you'd have a infinite amount of them right I can't write them all out because there's an infinite amount but just imagine that so what can we possibly do with this how do we clean this up when you're doing a physics problem you don't want to solve an infinite series by writing each term out infinitely we want a clever way to deal with this and there's a really clever way to deal with this watch this so if we write this as L equals the sum of M V our one problem we have is that each mass has a different V right if I can pull things out of this summation it would help me out because it would simplify things I can just factor them out but right now I can't factor out the r because these all are at different radii from the axis you always measure your radius from the axis here and they're all at different radii from the axis and they'll have different speeds but remember we like writing quantities in terms of angular variables because the angular variables are the same for every point on this mass so every point on this rotating rod has a different speed V but they all have the same angular speed Omega so that's a key that's often what we do and we're going to do that here I'm gonna write this as summation of M but instead of writing V I'm going to write this as R times Omega so remember that for something rotating in the circle the speed V is going to be equal to R times Omega and that's what I'm going to substitute down here so the speed of any point here is the radii of that point times the angular speed of this rod rotating in a circle and I still have to multiply by the last R here so this was V we substituted in what V was but we have to multiply by R and what do we get we get that L is going to be the summation of M R squared Omega and this is great D Omega is the same for every single mass in here every single mass travels at the same angular speed so we could factor this out of the summation imagine all these terms would have an Omega we can factor that out and I can just bring that outside of the summation so I'll write this as the summation of M R squared and to make this clear I'm going to put parentheses here it's that summation and then that whole thing times Omega because we're just factoring out Omega and you might not be impressed you know if you like all right big deal we've still got an infinite sum in here what the heck am I going to do with that you don't have to do anything with that this is where the magic happens look at what some you got you got the sum of all the M R Squared's remember what mr squared was M R squared was the moment of inertia of a point mass and if I add up all the M R Squared's I get the moment of inertia of the entire mass this entire object I get its total moment of inertia it's what we found was a really handy way to write the angular momentum of an object it's just the moment of inertia of an object I times the angular speed of that object so this is this is a great formula and it totally makes sense for this reason think about regular momentum right regular momentum P was just equal to mV well if you then told me determine the angular momentum and I didn't want to go through this derivation I might have just been like all right angular momentum shoot well I'm just going to replace mass with angular mass and the angular mass the angular inertia is just the moment of inertia and I'll just replace the speed with the angular speed and look I just get this formula so it makes sense because if you replace all the linear quantities with their angular counterpart you indeed just get the angular momentum of a rotating object so this is how you do it if you've got an object an extended object where the mass is distributed about the entire object if you just take the moment of inertia of that object and multiply by its angular speed you get its angular momentum so for instance if this rod has a mass of let's say 3 kilograms and that mass is evenly distributed let's say the radius of this object is 2 meters so that's the distance from the axis to the outside edge now let's say the angular speed of this object was let's say 10 radians per second we can figure out the angular momentum of this rod by saying that the angular momentum is going to equal the moment of inertia well the moment of inertia of a rod about one end is equal to one-third ml squared that's the moment of inertia of a rod about the end and I then multiply by the angular speed of the object so if I plug in numbers I get the angular momentum of this rod is going to be I'll use purple here one-third and then times three kilograms times the length of the object was two meters and we square that and then we multiply by the angular speed and that was ten radians per second which gives us an angular momentum of 40 kilogram meters squared per second so recapping if you've got a point mass where all the mass rotates at the same radius and you want to find the angular momentum the easiest way to get it is probably with the formula M V R however if you have a mass whose mass is distributed throughout the object so that different points on the object or a different radii the easiest way to get the angular momentum of that object is most likely with the formula I Omega where I is the moment of inertia of the object and Omega is the angular velocity of the object