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16. Angular Momentum

1

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Hey, guys. So in this video, we're going to start talking about the conservation of angular momentum. Let's check it out. So remember that when we talked about linear momentum that the most important part about linear momentum was the fact that it was conserved or that it is conserved. Right? Um, it's conserved in certain situations. We'll talk about that in just a bit. The same thing is gonna happen for angular momentum. Most any little mental problems are actually going to be about the conservation of angular momentum. So they're gonna be about conservation. They're gonna be conservation problems. Okay, so what I want to do here is do sort of a compare and contrast between linear momentum and its angular equivalent. Angular momentum. So linear momentum. Little p is mass. Times velocity, angular momentum. Big l is Iomega. Moments of inertia and angular speed. Linear momentum is conserved. If there no external forces. Yeah, and angular momentum is conservative there no external torques, right? And this should make sense. Um, angular. Momentum is the rotation equivalent of linear momentum. Torques is the rotation equivalent of forces. Now, even better description is it's actually not that there are no forces. Um, it's just that there are no external forces or, um, that there are that if there are external forces, they cancel, they at least cancel each other out. So even better definition is if there's some off external forces again, they could exist. Jason. They just have to add up to zero. Something here does. Some of the external torques has to be zero. This is the condition, the condition for conservation of linear momentum in the conservation of angular momentum. In the vast majority of physics problems, those quantities are conserved. Certainly all the problems we're gonna look into from now on for angular momentum will have conservation. Alright. One difference between these is that most problems for linear momentum involved two objects. Pretty much all of them involve two objects collide against each other. Okay, so the conservation equation will look like this p initial equals P final. Right? So it's saying that momentum doesn't change. This is of the system. So I can expand this equation and I have to object. Soapy initial becomes P initial one plus p initial two equals p initial one plus p initial too. So it's gonna be this very familiar equation. M one v one initial plus and to be to initial equals m one v one final plus m two. The two final Now angular momentum is a little bit different, that there's a lot of momentum angular momentum problems that involved just a single object, right? So the most classic, probably the most classic, angular momentum conservation of angular momentum. Question is when you have a nice skater. So let's say you have ah, girl ice skating. And she is spinning with her arms open and she closed her arms that she's gonna spin faster. This is a conservation of angular momentum. Question. We're going to solve this later, and it's just one object that's one body that's spinning. Okay, now conservation equation will be similar. I'm gonna have that. L initial equals l final right, because l doesn't change. That's the whole deal. And l is Iomega. So I'm going to say that I initial Omega initial is not going to change. It's a constant. Okay, but what I want to do here is I wanna expand this equation a little bit to show you something. So moment of inertia I for a point mass is something like M R Square for a shape. It's something like, Let's say, for a for a solid cylinder, you'd be half M r square for another object for like, a solid sphere would be to fifth M R Square. The point that I want to make here is that it's something M r squared something M r squared, right? What changes is that here you have half here have to fits here. There's, like a one that hides in there, right, That's implicit. We don't have to write. So I'm going to say that this takes the shape this I takes the shape of box, which is some fraction m r squared. And then I have omega. So I'm just expanding Iomega to show this, and I'm going to say that this is a constant. This is a constant constant. Okay, meaning this number doesn't change. So really, the kinds of problems you're gonna have. There's two basic types of problems in one type. The mass will change. I'm gonna put a little delta here on top of the mass will change, which will cause a change in omega on the other type of problem. Um, the R will change and cause the change in omega. So if the mass of a system changes, the system will slow down. Right, You might be able to see here. If this mass grows, the system will slow down. Or if the radius of the system the effective total radius of the system increases, then the mass, the velocity of rotation will go down as well. So the opposite case of what I just mentioned with the girl spinning is if she's spinning like this and then she opens her arms, she slows down. And that's because her total are right. You can see that these things are going away from the axis of rotation, so the are grows. Therefore the omega becomes smaller. Okay, So the two types of changes we're gonna have for one objects is that either, um either Emma or are will change. And those will cause a change in omega. Okay, change maker. Now, when we have to objects when we have two objects, we have problems where you're essentially adding or removing mass. So the classic example here there's a disc that's spinning. You add a little block to it. What happens while the disks Now I'm going to spend a little bit slower, and we can calculate that. Okay. When we had linear momentum, the two big groups of of big groups of problems we had were push away problems where two things would like. When you shoot a gun, the bullet goes this way. The gun goes this way or collision problems. So push away. Two things are going away from each other. Collision two things. Air coming into each other. Okay. And we also had We also had these types of problems we're adding or removing a mass, adding or removing a mass, um, in linear motion, which, if you think about it adding a mass is a collision, right? One mass joins the other, and removing a mass is really a push away. Problem is, if you jump out of a escape or something, All right, so anyway, that's it for that. Let's do I have, uh, an introductory example here talking about a bunch of different situations to see so we can discuss what happens in these situations, and we want way. Want to figure out whether the angular speed omega will increase or decrease? All right, so a nice skater. We just mentioned this nice skater spins and frictionless ice. What happens to her angular momentum if she closes or arms. If you close your arms, you spin faster. You might know this from class from just watching TV from doing it yourself, or we're gonna use the equation here. So what I'm gonna do is I'm gonna say I'm l is a constant L, which is Iomega is a constant mhm. I'm gonna expand Iomega into something m r squared Omega. And this is a constant. What's gonna write to see and look what's happening here is that by closing her arms by closing her arms, her are is decreasing. Therefore, her omega is going to increase. So the answer is that omega increases Omega will increase. Alright, be ah, large horizontal disk spins around itself. What happens to discs? Angular speed if you land on it. So there's a disk spinning around itself like this. You land on it right here. So this is you. You got added to the disc. What happens to the disc's speed? Well, I equals Iomega is constant. I'm gonna expand Iomega to be something m r squared. Okay, times omega. I M R squared times. Omega is a constant. What's happening here is there's mass being added to the system. Therefore, the system will slow down someone right here that Omega will decrease. All right, Uh, see, a knob checked is tied to a point via a string that spins horizontally around it. So here's an object, and it's tied to a point here. It's connected by a string, and it's gonna spin horizontally around the string. So the object is going like this because it's connected to a string. And what we want to know is what happens if you shorten the string. So again, I equals Iomega is a constant. See, Iomega, I'm gonna expand to be something, um, m r squared. It doesn't matter what this something is for these problems. We're just doing a quick analysis of what would happen if you remove if you shorten the string. Um, if you shorten the length of the string, you're shortening the radius of rotation of this object. Therefore, the Omega will increase omega increases. Okay, you can imagine that if you spend something a really long cable the second you pull the cable in, um, it's gonna stand Instead of being like this, it's gonna get faster like this. Okay? And in the last one, a star like the sun spins around itself. Cool. Um, and I wanna know what happens if it collapses and loses half of its mass and half of its radius. Okay, so you might know this. You may know this stars live for obviously billions of years. Eventually, they run out of star Fuel and they collapse. And what that means is that they're going to significantly shrink in sighs, um, in volume and in mass. Okay, so that's gonna happen to our son, like in 10 billion years. You're safe. Don't worry. So what happens if it collapses and loses half of its mass and half of its radio? So l equals Iomega? See, it's an object that spins, but it's not going to, uh, it's angular. Momentum is conserved even though this thing is blowing up. Right. So I have This is gonna be something m r squared omega. And that's a constant. So here we actually have precise numbers, half and half. So if this goes down by a factor of two, yet that and then this goes down by a factor of two. Notice that our ISS squared, Um, so I'm gonna actually square the factor of to get out of the way. So the net result of this going down by a factor of two is that it actually goes down by the whole thing, goes down by a factor of four. So I have this going down by a factor of two that's going down by a factor of four. I multiply those two and I have a I have this thing growing by a factor of eight. Okay, so two times four is AIDS. If the if these two variables here become eight times smaller, this variable has to become eight times greater so that the whole thing is a constant. So this Starwood then spin a time faster eight times faster than it was before it collapsed. Okay, so that's it for this one. Some introduction in terms of what to expect in these different kinds of problems. We're going to solve most of these later on. But that's it. Let me do you have any questions and let's get going

2

example

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Hey, guys. So let's check out this classic example of conservation of angular momentum. Okay? Remember, conservation of angular momentum are about objects that are spinning that will change either their m and then that will cause a change in omega. Or they will change the radius since of some sort. And that will cause a change in omega. Those are the two types that you have to look out for. So here we have a nice skater that has a moment of inertia of six. So I equals six when she spins with her arms open. So her eye is six. When she spins with arms will open. And four, if she closes her arm, I close equals form. It says here if she spins with six with rpm with their arms open. So rpm open is 20. What? Our pm you should have is a result of closing her arms. What will be our PM close? Okay, so you can think of open as in this show, because that's where we start. And you can think of clothes as final. Okay, so we're gonna use the conservation of angular momentum equation which is l I equals L f. In this case, we have one person. So it's just gonna be this is gonna be just Iomega for one person. I initial or make initial equals I final Omega final. Okay, the eyes were given, three eyes were given, so this is going to be initial is six, and then I have a mega. I'm gonna make a little space here. And then this is four and I'm gonna take a little space here. Now, notice that this is Omega and this is Omega. But I gave you one rpm and I asked you for another rpm. So this whole question is in terms of RPM, but the equations in terms of omega, like, usual, all of our equations are in terms of Omega. We always have to convert rpm into omega. But what I want to show you is that you can actually think of You can actually rewrite this equation here in terms of rpm, so let's do that real quick. Remember, um, Omega is two pi f or two pi over tea or two pi. I'm gonna plug in f here, and it's gonna be rpm over 60. So what I don't wanna do real quick cause I want to show you that there are three variations of this question. OK, so let's do that real quick. So this is like the official legit version. Number one. Here's version number two instead of omega. I'm gonna write two pf and look what happens. I initial two pi frequency initial equals I Final two pi frequency final notice that the two pies cancel and you end up with I f equals I f Okay, so this is another version of this equation. You can just basically replace Omega by frequency, okay? And they are both on the top. If you do this with period, this is what you get. This is version two I two pi period. Initial equals I Final Two Pi period final. Okay, these guys will cancel and you end up with I initial period initial because I final period final and you could do the same thing for our PM. Okay? And this is the last one. That's the one we're gonna use here. You weaken. Say I in issue now instead, off instead of two pf, we're gonna use two pi r p m over 60 So two pi our PM over 60 equals. I final trying to squeeze us in here. Out of the way two pi our PM over 60. Okay. And almost unfit. And look what I can do here. I can cancel the two pies in the in the 60 and you're left with I initial rpm initial equals I final rpm final. Okay, so this is the conservation equation. But you can think of it in these three alternative versions as well. This just makes it really easy for you to solve these questions, Uh, by basically briefly rewriting the equation. So one point I wanna make here is that a way to know how to make thes thes exchanges very quickly is Look at Omega. Omega is on the top. It's on the denominator. A numerator up here f is in the numerator appear they're both up top. That's why they both show up up top here. T is on the denominator. That's why T shows up at the bottom when you replace it and rpm is at the top. That's why rpm shows up at the top here. Okay, so in this question, were actually we don't have to convert the RPMs into omega and then back into our PM, we can just actually plug in the rpm, so I'm gonna plug in rpm initial rpm final. Okay? It would have been quick to just replace stuff, but I wanted to show you that we can do this. So rpm in issue is rpm initial is open, which is 1 20. And then this is four rpm final. So rpm final will be six times 1 divided by four. Okay. And the answer here is 180 our PM Now, the last point I wanna make. So this is the final answer. Boom. The last point I wanna make is notice that our I went from 6 to 4. It is it changed by a factor of 1.5. It went down by a factor of 1.5 and then So it went down by 1.5. Um, and then the rpm went from 1 20 to 1 80. It went up by a factor of 1.5. And that's because conservation momentum here of angel momentum is a linear relationship. There's no squares or whatever. Eso if one goes down by by 1.5. The other one has to go up by 1.5. Okay, Alright, so that's it for this one. This question is actually really easy. I just took a little longer because I wanted Thio do a little bit of analysis and I wanted to introduce you to these three alternative versions of the conservation equations so you can solve some of these questions faster. All right, so that's it for this one. Let me know if you need any help. If you have any questions and let's keep going.

3

Problem

Suppose a diver spins at 8 rad/s while falling with a moment of inertia about an axis through himself of 3 kg•m^{2}. What moment of inertia would the diver need to have to spin at 4 rad/s?

BONUS:How could he accomplish this?

A

4 kg•m_{2}

B

6 kg•m_{2}

C

11 kg•m_{2}

D

24 kg•m_{2}

4

example

6m

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Hey guys, So let's check out this example here. So a very common example of conservation of angular momentum questions is that of a star dying and that's because the star spins around itself and when the star dies or collapses it will change. Not only its mass, it will become lighter but it'll also reduce in size and radius and volume. Okay, so this is a good setup for a conservation of angular momentum because momentum will be conserved. So let's check it out. Um when they start exhaust all of its stellar energy, it dies. That's why it's sad, poor thing. Um at which point a gravitational collapse happens, causing its radius and mass to decrease substantially. So it's just telling you what happens. No actual information there. Our son spins around itself at its equator at the middle point right there every 24.5 days. The time that it takes for you to go around yourself, for you to complete a full revolution of any sort is period. So period of the sun is around itself 24. days. If our somewhere to collapse and shrink 90% mass and 90% in radius. In other words, our new mass, the new mass of the sun, I'm gonna call this m prime um is going to be, it's shrinking 90% in mass, meaning my new mass is 10% of the original mass. Okay, in the new radius is 10% of the original radius. I want to know how long would its new period of rotation take in days. In other words, if it's taken 24 a half days for the center spin around itself, how long would it take for the center spin around itself? Once these changes happen. In other words, what is my T. Final? Right? Think of this as as T. Sun initial. I want to know what is my T. Final. And what I wanna do here is um instead of writing L. Initial equals L. Final. Because we don't have actual numbers here, we just have percentages in terms of drop. This is really a proportional reasoning question. What I'm gonna do is I'm gonna write actually like this I'm gonna say L. Initial equals a constant. I'm gonna expand this. Okay I'm gonna expand this. The gonna be I initial omega initial is a constant eye. So a sun. The sun can be treated as a um as a solid sphere even though it's actually like a huge ball of gas. Um So so treating is a solid sphere is kind of kind of bad but it won't matter as I show in the second. So I have let's just do that for now half M. R squared. Um And then omega um I want not omega but I want period and remember omega is two pi over t. So I'm gonna rewrite this as two pi over tea and it's a constant because this is a proportional reasoning question. Um This number doesn't matter and this number doesn't matter. The only thing that matters are the variables that are changing. So even though it was kind of crappy to um to model the sun as a solid sphere, it doesn't matter because that fraction goes away anyway. So just just write em are square and you're good. It's sort of what I did earlier, what I had like box M r square right omega. So something like that, because the fraction doesn't matter. So here's what's happening um this guy here is decreasing by by 90%. So basically it's being multiplied by 0.1, right? And then this guy here is being multiplied. Imagine that you're putting a 0.1 in front of the M and a 0.1 in front of the are now the r is squared, so if you square 0.1 you get 0.1. So think of it as the left side of the equation here Is being multiplied by a combination of these two numbers, which is 0.001. Basically the left side of the equation becomes 1000 times smaller. Therefore the right side of the equation has to become 1000 times greater. Okay, so the right side of the equation has to become 1000 times greater. So this side here grows by 1000. Now the problem here is it's a little bit complicated because t is in the bottom. So if the whole thing grows by That means that T, which is in the denominator actually goes down by a factor of 1000. Okay if your fraction goes up, your denominator went down and that's how your fraction goes up, right? Imagine for example you have 100 divided by 10, that's 10 Um 100 divided by two. That's 50. Okay. Your entire thing went up because your denominator went down. So if this goes down by a lot, right side right here goes up by a lots. Uh and then the denominator goes down by a lot. So basically your new period is 1000 times smaller than your old period. So I can write T. Final is um one over T. Initial. So it's basically 24.5 days divided by 1000. And then what you wanna do is you want to convert this into ours? Okay Um or or some some measure of that sort. So we're gonna do here that one Day has 24 hours. This canceled his cancels and you multiply some stuff. Um you multiply some stuff and I actually have this in minutes. Okay I actually have this in minutes. So let me change that in two minutes. So it's gonna be um 24 hours times minutes. And when you do this you get that the answer is about 35 Minutes. I did minutes. Because otherwise you end up with like .55 hours and minutes makes more sense. It's easier to make sense out of it. Okay? So imagine this, this the sun takes 24 days to go around itself, and after it collapses and it shrinks significantly. It's gonna spend 1000 times faster, So it's gonna make a full revolution around itself in just 35 minutes. Okay, so that's it for this one. Let me know if you have any questions and let's keep going.

5

Problem

Two astronauts, both 80 kg, are connected in space by a light cable. When they are 10 m apart, they spin about their center of mass with 6 rad/s. Calculate the new angular speed they’ll have if they pull on the rope to reduce their distance to 5 m. You may treat them as point masses, and assume they continue to spin around their center of mass.

A

ω_{f} = 1.5 rad/s

B

ω_{f} = 4 rad/s

C

ω_{f} = 6 rad/s

D

ω_{f} = 24 rad/s

6

example

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Hey, guys. So here we have another classic problem in conservation of angular momentum, which is the problem where you have some sort of disc that spins. And then there's either a person or an object in that disk, and the personal object will either move closer to the disk or closer to the center of the disc or away from the center of the disc. And in doing this, you are going to change how fast the disk spins. Okay, so it's conservation of angular momentum. Let's check it out. So you're moving on a rotating disc, so we have a disc of Mass. 200. Mass equals 200. Radius is 4 m. Now. This is a disk. This means that the moment of inertia we're going to use for the disk is the moment of inertia of a solid cylinder, which is half M R Square. Um, it spends about a perpendicular access through its center. Um, if you have a disc, here's a disk perpendicular axis means that your axis of rotation imaginary line perpendicular means it makes 90 degrees with the face of the disk. So it means that the disk spends like this about an axis, and there's gonna be like a person here walking around. All right, um, so through its center, at two ratings per second. So the disk is initially rotating with an omega initial of two radiance per second on, then you have a person. So this is Big M. We're going to say that the person has massed little M equals 80 that falls into the disk with no horizontal speed. So here is the disk, and the person will just, like parachute into the disc land on top of the disk. You're probably imagining that this will cause the disk to rotate at a lower rate lower speed because you added mass to its It's heavier. Now that's actually what's going to happen. So I'm going to draw a top view of the disk, so let's make it a little rounder. Alright, so here's disk. The radius of the disk R is for the person is going to land somewhere over here at a distance of three. So remember, radius is big. Our distance from the center is little. Our little our is three, so the person lands there. Initially, there is no person, and then, after the person lands, we will have a person, which means our omega will change. And what we want to know is what is the disks? New angular speed. So Omega initial for the disk is to Omega Final for the disk is what we're looking for. So question Mark. Okay, So conservation of angular momentum. So I'm gonna write that angular momentum Initial equals angular momentum final. Uh, I'm gonna expand this. This is I initial Omega initial, and then this is I final Omega final. However, in the beginning, there's Onley a disk. So I'm gonna say that this is I of Big M in Omega of Big M But at the end, at the end, there are two things. There is big, um, and there's also little m So we're gonna do this. I f omega f off little m a little. Okay, so we added mass to the system. So the system now has two l's instead of just one, but the whole thing still has to be equal. Okay, so let's see, we're looking for Omega Final of the disc, which is this. Now let me point out to you that the final make of the disk is the same as the final omega of the person. Because if you land on a disk that's spinning, right, if the disks spinning you land on it, you're going to rotate with the disc, right? So you rotate with the disk, so you have the same omega. So these two guys here are actually the same, Which means you could do something like this instead of saying Omega final Big M and Omega Final little M you could just call this Omega final. And instead of having two variables, you have one variable, which is simpler so you can do a mega final in both of these, you can factor it out and then have I final M plus I final little m. Okay, then you have I initial Big m then Omega Initial Big M, which we have. So we're looking for this. We have this, so all we gotta do is calculate all the eyes. Okay, so let's do that. The I'm gonna do this off to the side over here and the moment of inertia of the disc. In the beginning, the disk is by itself. So you have half m r squared and I have all these numbers, So this is easy to calculate. Half M is 200 and our is four squared. And when you do all of this, when you do all of this, you get 1600. So this is gonna be 1600 right here. Okay, so this number is 1600. The initial speed of the disk is too equals Omega Final. And then these two numbers here now at the end, after you land on it. Well, the disk the disc still has the same moment of inertia, right? The mass of the disk didn't change. The radius of the disk didn't change. What change was the mass of the whole system. So this is still 1600. But the difference is now that I have something else and that's what we have to find. We have to find it. The final. The final moments of inertia off this person we're treating the person is a point mass. So we're gonna use the moment of inertia off a point mass, which is M R. Squared, where r is distance from the center. The person is 80. It is a distance not four, but three three square. So if you multiply all of this, you get that? This is 7. 20. This is 7. 20 and that's the number that goes right here. 7. 20. Cool. So if you solve here you have This is on the left. Uh, the stuff on the right side adds up to 23 to 0. And if you divide, you get in omega of 1.38 radiance per second. Now, this should make sense. You started off at two. You added mass. The disk became heavier. Therefore, the disk slows down a little bit, and now it has an Omega final off 1.38. Okay, so I went from 2 to 1 point 3/8. This is part A. I'm gonna make some space here and then solve part B. Part B is calculate the person's new tangential speed. Okay, If you are a point mass going around in a circular path, not only do you have in omega, but you have an equivalent tangential speed that is tied to your omega, and that's given by our Omega. So that's all we gotta do is plug this in. The person spins with the disk, so The person also has an omega of 1.38. Okay. And 1.38. Now, what are do you think I use? Well, R is the distance that the person has from the center you are. The radius of the disk is four. But you're spinning around the circular path of radius of three. So that you put it three here. All right. And if you multiply this answers 4.14. Very straightforward. That is your tangential velocity. As a result, part C part C. So you landed on the disk. Now you're going to walk towards the center of the disc, okay? And this is what we talked about in terms of moving inside of the disk. So here we're adding mass to a disk when you land and removing within the disk after you land. So it says the person starts walking towards the disc center, calculate the disc's speed once the person reaches it. So I want to draw here. Quick diagram. You got the person here. That's our equals three. Okay, Max, what's going do this, Um, and the person is initially at r equals three right there. So that's my are think of this as your our initial, okay? And then the person is going toe walk in this direction here until he gets to the center. So the person is going to have our final off zero. Okay, So you think the disk is gonna be spinning faster or slower? Faster or slower? Well, one of the ways you can think about this is similar to when you close your arms, you're gonna spin faster. The effective total radius of this system, um, is going to be smaller. They're more mass is closer to the axis of rotation. Therefore, it spins faster. Okay, so the disk will speed up. How do we do this? Well, conservation of angular momentum. Every time you have something that spins and something about that system changes like the radius of the earth mass, then we're gonna right. The conservation of angular momentum equation l initial equals l final. In the beginning, we have in the beginning here. We're talking about this to this. So in the beginning, we have disc. We have disc plus guy. When the guy is at R equals three. And at the end, we also have disc plus guy, but here. The guy is in a position of r equals zero. The guy sits on the axis of rotation. Okay, so let's expand this equation Here. I initial mega initial I Final Omega Final, and for both of these, we have guy and, um, we have guy and disk. So what you can do is you can write I as I disk. Plus I guy, it's the moment of inertia. The system. You can do it together like this Omega initial equals I disk plus I guy Omega Final. So this is obviously idisk initial I guy initial idisk Final I guy final. The moment of inertia of the disc doesn't really change because the disc didn't change its mass or its radius. What will change is the I put a little delta here for change. What will change is the final moment of inertia of the guy will be different than the initial. Okay, so here I have half M R square. Remember the moment of inertia of the guy since returning has a point. Mass is m little r squared. This is little our initial um, this is the number that's changed and this is a three in the beginning in the zero at the end, Omega initial. So the guy lands at this thing and hey automatically picks up a speed of 1.38. So that is the Omega initial off. The guy in the disk together. One point. I'm 38. It's not too, too. Was the speed of the disc before you landed on it? Okay, on then, here on the other side. You have same thing, half M r squared. That doesn't change. But this is going to be m our final square in this final here, our final is zero. Okay. And Omega Final is what we're looking for. We have all the numbers, so should be able to plug everything in and software made a final. So the rest here is just algebra. Most important step was getting, um, here. Okay, so let's do this. Half the mass of the disk is 200 actually. Already calculated this whole thing here before, This is just 1600 right? Yeah, this is 1600. Plus this thing here, which is Did we have this before? The mass of the guy is 80 distances, three squared. Feel like we have that before Yeah, that's 7 20 right? So we have 1600 plus 7. 20 and the whole thing is 1 38. We had these two guys from before from part a, the end of party. And then here we have the same 1600. But now check it out. This whole thing will be zero because the are is zero. He's sitting on the axis of rotation Ah, point mass. Sitting on your axis of rotation doesn't contribute towards your moment of inertia, right? It's as if it wasn't there. It's basically the same thing as if this guy just got, like, picked off from the disc. Alright, so I'm gonna put a plus. You're here just to be very clear and omega final, So Omega Final will be. Now, once you plug this in here, this entire left side will be 3200. I divide by 1600 I get the Omega Final is to radiance per second. Now, where have we seen to before? Two was the original speed that the disc had without the guy, which kind of goes with what I just mentioned that if you walk to the center, you have no contribution towards the moment of inertia. It's as if the guy got picked off. So you're back to the original part of the problem, which is disk without the person was spending with two. You add a person, you're slower. The person walks to the center, which means that the person's contribution contribution to moments of inertia disappears. So you're back to where you were before the person was on the disc. Cool. So too, is not a coincidence. It should make sense. Hope the whole question makes sense. Let me know if you have any questions and let's keep going.

Additional resources for Conservation of Angular Momentum

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- Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and calle...
- A thin uniform rod has a length of 0.500 m and is rotating in a circle on a frictionless table. The axis of ro...
- A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg...
- A Gyroscope on the Moon. A certain gyroscope precesses at a rate of 0.50 rad/s when used on earth. If it were ...
- CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless c...
- CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless c...
- The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be con...
- A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg...

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