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Current time:0:00Total duration:14:54

Video transcript

what's up everybody I want to show you something kind of crazy when I first heard about this it really bothered me so if you have that reaction it's a natural but I'll help you get over it and it'll hopefully make sense by the end of this the crazy thing is this if you've got an object say this ball this is a bouncy ball and it's going in a straight line it can have angular momentum and I'll say that again a ball traveling in a straight line can have angular momentum when I first heard this I was like what there's no way this ball can have angular momentum it's moving in a straight line it's not even rotating don't things have to have some sort of rotational motion in order to have angular momentum and it turns out they don't but the proper response it gets a little weirder before I show you how to make sense let me show you this it doesn't even have to have angular momentum it could be that this has no angular momentum and before you get really confused let me explain the proper response to someone saying does this ball have angular momentum is to respond with angular momentum about which axis so you have to specify the axis so the axis is the point about which you're going to consider the rotation so if I said does this ball have angular momentum if the ball is moving in a straight line does it have angular momentum relative to this axis that's a proper question but if I just ask you does this ball have any low momentum and don't specify the axis then it's not even a meaningful question so let's try to figure this out why why does this ball have angular momentum at all regardless of any axis right that's the confusing part it's not even rotating how does a ball moving in a straight line have angular momentum we know it has a regular momentum because objects with mass and velocity have momentum but it's not even rotating how can it have angular momentum first let me explain conceptually why that makes sense so imagine you have this bar here right and this bar say this is a bird's-eye-view we're looking down this bar is attached to an axis that can rotate and we're looking down this this is like a this is on a tabletop let's say this is all happening on a tabletop and we're looking down at it so imagine we throw this ball right this ball moving in a straight line hits the edge of this bar and this bar what is it going to do we know what it's going to do it's going to rotate it's going to rotate about its axis and now the question is this bar initially had no angular momentum because the bar was just sitting here at rest then it did have angular momentum after the ball hit it right because objects rotating around in a circle have angular momentum where did this bar get its angular momentum well the only thing that that bar interacted with was this ball so this ball must have come in and transferred some angular momentum to the bar because where else with the bar get the angular momentum from I mean if we believe in conservation of angular momentum that angular momentum has to come from somewhere it can't just pop out of nowhere so the only place that angular momentum that the bar got had to be from the ball because that was the only other thing in the problem so this ball had to come in with its own angular momentum even though it was travelling in a straight line which is kind of weird but that's the case that's physics and that's also why it makes sense why it depends on where the axis is because if I take this bar and I move this bar over to here instead right so now we consider our axis over at this point well imagine the ball hitting the bar at the axis point right it's just his right there boink it's not even going to cause that bar to rotate because it's hitting it at the axis so the location of the axis is going to determine how much angular momentum an object has that's moving in a straight line because if it hits this bar at the axis it's going to transfer no angular momentum but if it hits this bar far away from the axis it can transfer a lot of angular momentum because the bars going to rotate a lot because there was a large amount of torque applied since this force was applied at a distance that was far away from the axis now you still might not be all that impressed you might be like that just sounded like a bunch of physics witchcraft to me how do you like calculate this exactly how do you exactly define what we mean by the fact that this ball has angular momentum so let's define this exactly let's say this ball has a speed or a velocity V and the mass of the ball will say the mass of this ball is M and the distance from the axis to the ball let's just draw that on here so that's going to be from the axis to the ball we'll call that little R and now we can define precisely what we mean by the angular momentum of a point mass the angular momentum of a point mass L is the symbol for angular momentum is going to be M the mass of the ball so the mass of the object that has the angular momentum times V the speed of the ball and this is looking pretty familiar because M times V is just momentum so this is regular momentum but if you just left it like this that'd just be regular momentum we have to turn this into angular momentum and we do that by multiplying by r and r is defined to be the distance from the axis or your origin to the mass that you're considering and so this is the total distance R but you're not done yet you need one more term in here that's going to be sine of the angle sine of the angle between the velocity and the R vector so you're going to have sine of this angle right here the angle between the velocity and the R vector now I'm being a little sloppy so people pay an attention out there that already know this might be a little concerned technically R goes from the axis to the mass so this isn't the angle between V and R technically you have to imagine are being extended out this way and then that's the angle but because we're taking sine sine of either of these angles these angles are supplementary the sign of either of them are going to be the same number so you're safe by just taking any angle here between V and R and you'll get the right answer but this is complicated if I were you I'd be like oh man and the R sine theta I don't want to have to figure out what the angles are between things and you don't there's a trick so check this out if you just consider what is R sine theta even mean what is this right visually what does this represent on here well here's the total amount R here's theta think about it R times sine theta imagine making the triangle out of this I make a triangle that goes from there to there and then here to here so you've got this triangle here R times sine theta is just this right here so it's just this this length right here this OP sit side because alright if you if you didn't catch that that might be a little bit weird so sine of theta is always opposite over hypotenuse and our opposite side opposite to that angles just R so that's just R and then the hypotenuse excuse me is just this little R this pink R and we'll call that little R so if I multiply both sides by little R I'll get that R times sine theta and this theta here is this theta here so this is that theta right there our sine theta is just equal to this right here and what is this this is just this point of closest approach so when the ball makes it to this point right here when the ball gets to this point going this way at some speed V it's going to be R away so all you really need to do to find the angular momentum of an object of a point mass even if that point mass is going in a straight line is take the mass times V and then if you don't want have to worry about sine theta and all of that mess just multiplied by R which is the distance of closest approach to the axis so that's what this R is this R is the distance of closest approach to the axis and that's just this right here that's just this distance right here between the axis and the point where the ball will be closest to the axis so that's this distance right here turns out this R sine theta is always just equal to that so you can make life easy just imagine when this ball comes in at what point is it closest to the axis that would be this point and then how far is it when it is closest that gives you this R value you can take em VR that gives you the angular momentum of this point mass tells you the total amount of angular momentum that thing could transfer to something else if it lost all of its angular momentum that's how much angular momentum something like that rod could get so let's try an example let's do this example it's actually a classic a ball hitting a rod man I'm telling you physics teachers and professors they love this thing you should know how to do this let's get you prepared here so let's say this ball comes in it hits a rod right and so the ball is going to come in ball is going to hit a rod and let's put some numbers on this thing so we can actually solve this example let's say the ball had a mass of five kilograms on 8 m/s hits the end of the rod and the rod is 10 kilograms 4 meters long and listing let's assume this rod has uniform density so this rod has a nice mass distributed evenly throughout it and it can rotate around the end so when the ball gets in here strikes the end of the rod the rod is going to rotate around its axis and let's make another assumption let's assume when this ball does hit the rod the ball stops so after hitting the rod the ball is stopped and the rod moves on with all the angular momentum that the ball had that'll just make it a little easier we'll talk about what to do if that doesn't happen it's not that much harder let's just say that's the case initially here so I move the ball back over to here how do we solve this problem well we're going to try to use conservation of angular momentum we're going to say that even though there's an axis here exerting a force the force that that axis is going to exert on our system is going to exert zero torque because the R value right torque is equal to RF sine theta and if if the R is zero R is the distance from the axis to the force if R is zero to me no torque exerted by that axis and if there's no torque exerted externally there's no change in angular momentum of the system so this system of ball and rod is going to have no external torque on it that means the angular momentum has to stay the same this is a classic conservation of angular momentum problem so we're going to say that L initial the initial angular momentum has to equal the final angular momentum we just say for our entire system what had angular momentum initially well it was this mass so this mass had the angular momentum and how do we find that remember it's M times V times R and the R is that distance of closest approach so we're going to use this here for the whole four meters as this R yes you can consider this hypotenuse our end of the sine of the angle but that's harder than it needs to be you can find angular momentum with MV R that's going to equal the final angular momentum remember this ball stops so since this ball comes to rest and it's only the bar that has a the momentum afterward we only have to worry about the angular momentum of the bar on the final side and to find the angular momentum of an extended object a rigid object you can use I Omega and this would let us solve for what is the final angular velocity of this rod after the collision so that's what we want to figure out what is the final angular velocity of the rod after the collision now we can figure it out we know the mass of the ball and we know the speed of the ball initially we know the our line of closest approach that's 4 meters what's the moment of inertia here well it's just going to be one-third ml squared let me clean this up a little bit let me take let me take this I'll just copy that put that right down over here and we can say that the moment of inertia of a mass of a rod rotating around its end is always going to be one-third ml squared so one-third times the mass of the rod times the length of the rod squared which is going to be the same as this R here because this balls line of closest approach was just equal to the entire length of the rod since it's struck it at the very end and then times Omega so we can solve this for Omega now we can say that Omega I'm going to bring this down around here so I've got some room Omega final of the rod is just going to be what it's going to be mass of the ball times the initial speed of the ball times the line of closest approach and I'm going to divide by 1/3 the mass of the rod times the length of the rod I can just call that R is the same variable length of the rod squared and that's what I get so I can cancel off one of these R's and then I can plug in numbers if I wanted to actually get a number I could say that the final angular velocity of this rod it's going to be 5 kilograms that was the mass of the ball times 8 meters per second that was the initial speed of the ball and I'm gonna divide by one third of the mass of the rod was 10 kilograms and then the length of the rod which is this line of closest approach was 4 meters and if you solve all that you get 3 radians per second so that's how much angular speed or angular velocity this rod had after the ball hit it and transferred the ball's angular momentum into the rod giving the rod angular momentum causing it to spin around and rotate at 3 ian's per second now what would be different if instead of getting stopped the ball balance to backward let's say at two meters per second well now the final angular momentum wouldn't just be the angular momentum of the rod you'd have to include the angular momentum of the ball but here's the tricky part if the ball was coming in this way initially which would mean it has angular momentum essentially around this way and the ball went backward the other way now it has angular momentum around this way you would have to have a negative sign in here in other words when you include this angular momentum on the right hand side you'd have to treat it as well there's a couple ways you could do it you could just do plus if you wanted to and then you do the mass of the ball times this negative two meters per second times the same four meters as the line of closest approach you could put the negative year with a plus out here or you could put the negative out here with a plus in here you could do it either way but this term for the final angular momentum of the ball would have to have the opposite sign as this term for the initial angular momentum of the ball so recapping a ball can have angular momentum even if it's moving in a straight line and you could determine the angular momentum of that ball by using MV R sine theta where R is the distance from the axis to the point where the ball is and theta is the angle between R and the velocity or if you don't want to use R sine theta you can use MV capital R where this capital R represents the closest that ball will ever be to the axis as it's traveling along it straight line