David explains how a mass can have angular momentum even if it is traveling along a straight line. Then David shows how to solve the conservation of angular momentum problem where a ball hits a rod which can rotate. Created by David SantoPietro.
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- I still don't quite get how the ball has angular momentum. What I see happening is a ball with regular momentum hitting a rod and causing said rod to move with angular momentum around it's axis. I mean, if the rod wasn't connected to an axis it would just rotate about it's own center of mass, so if it got hit in the middle it should move in the same direction as the ball. Just consider the point of mass where the rod was hit. If that point mass were not part of the rod and therefore not connected to the axis, it would move in the same direction as the ball, with no angular momentum. Isn't David actually describing the transfer and conversion of regular momentum in the ball to angular momentum in the rod?
I'm sure it's me and I'm just missing something, so any clarification would be appreciated. :) Thanks(38 votes)
- I know this is an old comment but I think this may help future viewers. You have the right idea, but you have to be careful when you say "transfer and conversion of regular momentum in the ball to angular momentum in the rod." They are completely different quantities with different units. As you said, if the ball hit the pivot, then no angular momentum would be created. Therefore we know the angular momentum does not depend solely on linear momentum, but on linear momentum and some other quantity, in this case, the radius of closest approach. Therefore we can say relative to some origin (in this case the pivot), the ball has some angular momentum that is equal to the linear momentum times that radius. It is sort of artificial and can be confusing, but necessary in order to solve the problem. Think of it like a torque on a rod, the amount of torque depends on both the perpendicular force and the distance along the rod at which that force acts, not JUST the force, the same basic principle holds true for angular momentum.(13 votes)
- can we solve this problem by energy conservation(19 votes)
- But there is conservation of energy.
Where does the kinetic energy go if it is not all converted from the ball to the rod?(4 votes)
- Can anyone tell me why is the moment of inertia for the rod 1/3 MLsquared? I do not quite understand. Thanks in advance. (At11:52)(13 votes)
- Well that is derived using integral calculus. https://www.scienceabc.com/nature/universe/moment-of-inertia-calculate-rod.html(5 votes)
- I don't understand why we have to take the sin of theta. Could someone please explain? Thanks(7 votes)
- Because angular momentum is
L = m (r x v)
The (r x v) is the cross product between the radius and velocity vectors, which when you take the magnitude it reduces to
|L| = m * |r| * |v| * sin(Θ)(11 votes)
- How did the torque become zero?(7 votes)
- When did the torque become zero? I can help you out if you can tell me when this happened in the video.(4 votes)
- If the ball hits the rod at the axis, the rod wouldn't rotate. Now, where does the angular momentum of the ball go. Because according to the law of conservation of angular momentum, it should be going somewhere. NO?(4 votes)
- no, because if the ball is moving straight towards the axis, it has no angular momentum relative to that axis. See my comment to the top question. Think of it as a set of polar coordinates, if the ball is moving towards the axis then the angle is the same for each step of the ball's motion. Try drawing the axis as a dot ad the ball as a dot, and draw a line between the two at each step of its motion. you will see the ball is going towards the axis, every point will be on a straight line. Otherwise each time it will be at a different angle.(2 votes)
- What if the ball after hitting the rod did not stop and went on straight with a speed of +2m/s. Is this situation the same as described at13:25?(4 votes)
- Yes. You are going to add another momentum with +2 velocity becouse the +6 velocity is transfered to rod and +2 is stayed within the ball as momentum is always conserved .(4 votes)
- why is the moment of inertia for the rod 1/3 MLsquared? Isnt it supposed to be 1/12 MLsquared?(5 votes)
- But what happens to total momentum ?
Is it linear + angular momentum ??(2 votes)
- They are 2 entirely different concepts. You can not add apples with oranges . Linear momentum is measure of motion ( momentum) contained in the body , it does not depend upon what origin you are choosing . But angular momentum is always about some Origin , and its value keeps on changing depending upon the origin about which it is being calculated.(8 votes)
- So i did the problem if the ball bounces back with 2m/s, and i got the final angular velocity to be 3.75rad/s, which is greater than when ball stops. How is that make sense? When the ball stops all the momentum got transferred to the rod, but when ball bounces back, only part of it go transferred, shouldnt the final angular velocity of the rod be less?(4 votes)
- [Instructor] What's up everybody? I wanna show you something kind of crazy. When I first heard about this, it really bothered me. So if you have that reaction, it's natural, but I hope you get over it, and it'll hopefully makes sense by the end of this. The crazy thing is this. If you got an object, say this ball, say it's 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 it makes sense, let's 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 gonna 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 it just ask you, "Does this ball have angular momentum?" and don't specify the axis, then it's not even a meaningful question. So let's try to figure this out. 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 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 is on a table top. Let's say this is all happening on a table top, 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 gonna do? We know what it's gonna do, it's gonna rotate. It's gonna 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, 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've come in and transferred some angular momentum to the bar. Because where else would 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 at 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 traveling 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, So now we can set our axis over this point. Well, imagine the ball hitting the bar at the axis point. It just hits right there. Boink! It's not even gonna cause that bar to rotate because it's hitting it at the axis. So the location of the axis is gonna 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 gonna 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 bar is gonna 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 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, we'll 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 gonna 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. It's gonna 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 it just left you 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 defining to be the distance from the axis or your origin to the mass that you've considering. And so, this is the total distance r. But you're not done yet. You need one more term in here. That's can be sine of the angle, sine of the angle between the velocity and the r vector. So you're gonna 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 paying 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 r being extended out this way, and then that's the angle. But because we're taking sine, sine of either these angles, these angles are supplementary. The sine of either of them are gonna give you 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, "mvr sine theta. "I don't wanna 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'll make a triangle that goes from to there, then here to here. So you've got this triangle here. r times sine theta, it's just this right here, so it's just this. This length right here, this opposite side, because... All right, 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 angle is just R, so that's just R. And then the hypotenuse, excuse me, is just this little r, this pink r. Let me 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. r 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 gonna 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 wanna have to worry about sine theta and all of that mess, just multiply it 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. It turns out this r sine theta is always just equal to that, so you could make your 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 mvR. That gives you the angular momentum of this point mass. It 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 gonna come in. Ball is gonna 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. It was going eight meters per second, hits the end of the rod, and the rod is 10 kilograms, four meters long. 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 gonna 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 has stopped, and the rod moves on with all the angular momentum that the ball had. That will 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 or so. I'll move the ball back over to here. How do we solve this problem? Well, we're gonna try to use conservation of angular momentum. We're gonna say that even though there's an axis his exerting a force, the force that that axis is gonna exert on our system is gonna exert zero torque, because the r value. Torque is equal to r F sine theta. And if the r is zero, r is the distance from the axis to the force, if r is zero, there's gonna be 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 gonna have no external torque on it. That means the angular momentum has to stay the same. This is a classic conservation of angular momentum problems. So we're gonna say that L initial, the initial angular momentum, has to equal the final angular momentum. And we'll 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 gonna use this here for the whole four meters as this R. Yes, you can consider this hypotenuse R and a sine of the angle, but that's harder than it needs to be. You can find angular momentum, mvR, that's gonna equal the final angular momentum. Remember this ball stops. So since this ball comes to rest, and it's only the bar that has angular 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 wanna figure out. What is the final angular velocity of the rod after the collision? Now we can figure it out. We know the mass the of the ball, m. We know the speed of the ball initially. We know the R, line of closest approach. That's four meters. What's the moment of inertia here? Well, it's just gonna be 1/3 m L squared. Let me clean this up a little bit. Let me take this. I'll just copy that. Put that right down over here, and we could say that the moment of inertia of a mass of a rod, it's rotating around its end, is always gonna be 1/3 m L squared. So 1/3 times the mass of the rod, times the length of the rod squared, which is gonna be the same as this R here, because this ball's line of closest approach was jus equal to the entire length of the rod, since it 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 gonna bring this down around here, so we go some room. Omega final of the rod is just gonna be, what? It's gonna be mass of the ball times the initial speed of the ball, times the line of closet approach. And then I'm gonna divide by 1/3 the mass of the rod times the length of the rod. I can just call that R, it's the same variable, length of the rod squared, and that's what I get. So I can cancel off one of these Rs, 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 is gonna be five kilograms, that was the mass of the ball, times eight meters per second, that was the initial speed of the ball, and then I'm gonna divide by 1/3 of the mass of the rod was 10 kilograms, and then the length of the rod, which is this line of closest approach, was four meters. And if you solve all that, you get three radians per second. So that's how much angular speed or angular velocity this rod had after the ball hit it, and transfer the ball's angular momentum into the rod, giving the rod angular momentum, causing it to spin around and rotate at three radians per second. Now what would be different if instead of getting stopped, the ball bounced 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 is 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'd 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 of 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 here with the plus out here, or you could put the negative out here with the 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 m v 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 wanna use r sine theta, you can use m v capital R, where this capital R represents the closest that ball will ever be to the axis as it's traveling along its straight line.