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 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? It's conserved in certain situations. We'll talk about that in just a bit. The same thing is going to happen for angular momentum. Most angular momentum problems are going to be about the conservation of angular momentum. So they're going to be about conservation. They're going to 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 I omega. Moment of inertia and angular speed. Linear momentum is conserved. There are no external forces. And angular momentum is conserved. There are no external torques. Right? And this should make sense. Angular momentum is the rotation equivalent of linear momentum. Torques is the rotation equivalent of forces. Now, an even better description is it's actually not that there are no forces. It's just that there are no external forces, or that there are, that if there are external forces, they cancel, they at least cancel each other out. So an even better definition is if the sum of external forces again, they could exist just as they just have to add up to 0. Same thing here. The sum of the external torques has to be 0. This is the condition for conservation of linear momentum and the conservation of angular momentum. In the vast majority, of physics problems, those quantities are conserved. Certainly, all the problems we're going to look into from now on for angular momentum will have conservation. Alright. One difference between these is that most problems for linear momentum involve 2 objects. Pretty much all of them involve 2 objects colliding against each other. Okay. So the conservation equation will look like this: P initial = 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 2 objects. So P_initial becomes P_initial_1 + P_initial_2 = P_final_1 + P_final_2. So it's going to be this very familiar equation. M1*v1_initial + M2*v2_initial = M1*v1_final + M2*v2_final. Now, angular momentum is a bit different; there are a lot of angular momentum problems that involve just a single object. Right? So the most classic probably the most classic angular momentum, conservation of angular momentum question is when you have an ice skater. So let's say you have a girl ice skating, and she is spinning with her arms open and she closes her arms, and she's going to spin faster. This is a conservation of angular momentum question. We're going to solve this later. And it's just one object. It's one body that's spinning. Okay? Now, the conservation equation will be similar. I'm going to have that L_initial = L_final. Right? Because L doesn't change. That's the whole deal. And l is I omega. 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 want to expand this equation a little bit to show you something. So moment of inertia, I, for a point mass is something like m r^2. For a shape, it's something like, let's say, for a solid cylinder to be (1/2)m r^2. For another object, for like a solid sphere, it'd be (2/5)m r^2. The point that I want to make here is that it's something m r^2. Right? What changes is that here you have half, you have 2/5. 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 of the box, which is some fraction, m r^2. And then I have omega. So I'm just expanding I omega to show you this, and I'm going to say that this is a constant. This is constant. Constant. K? Meaning, this number doesn't change. So, really, the kinds of problems you're going to have, there are 2 basic types of problems. In one type, the mass will change. I'm going to put a little Delta here on top of M, the mass will change, which will cause a change in Omega. And the other type of problem, the r will change and cause the change in omega. So if the mass of the 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 r, right, you can see that these things are going away from the axis of rotation. So the r grows, therefore, the omega becomes smaller. Okay? So the two types of changes we're going to have for one object is that either, either m or r will change, and those will cause a change in omega. Okay? Change in omega. Now when we have 2 objects, when we have 2 objects, we have problems where you're essentially adding or removing mass. So the classic example here is there's a disc that's spinning. You add a little block to it. What happens? Well, the disc is now going to spin a little bit slower, and we can calculate that. Okay? When we had linear momentum, the 2 big groups of problems we had were push-away problems where 2 things would like when you shoot a gun, the bullet goes this way, the gun goes this way, or collision problems. So push away, 2 things are going away from each other. Collision, 2 things are coming into each other. Okay? And we also had, we also had these types of problems where you're adding or removing a mass, adding or removing a mass, 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, as if you jump out of a, of a, of escape or something. Alright? So anyway, that's it for that. Let's do I have 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, we want to figure out whether the angular speed, omega, will increase or decrease. Alright. So a nice skater, we just mentioned this nice skater spins on frictionless ice. What happens to her angular momentum if she closes her 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 going to use the equation here. So what I'm going to do is I'm going to say, L is a constant. L, which is Iω, is a constant. I'm going to expand Iω into something m r^2 Omega, and this is a constant. I'm just going to write c. And look, what's happening here is that by closing her arms, by closing her arms, her r is decreasing, therefore, her omega is going to increase. So the answer is that omega increases. Omega will increase. Alright. B, a large horizontal disc spins around itself. What happens to the disc's angular speed if you land on it? So there's a disc 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 I omega is constant. I'm going to expand Iω to be something m r^2 k times omega. I m r^2 times omega is a constant. What's happening here is there's mass being added to the system. Therefore, the system will slow down. So I'm going to write here that omega will decrease. Alright. C, an object 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 going to spin horizontally around the string. So an 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 I omega is a constant. C I omega, I'm going to expand to be something, m r^2. 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, 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 spin something on a really long cable, the second you pull the cable in, it's going to spin instead of spinning like this, it's going to get faster like this. K. And then the last one, a star like the sun spins around itself. Cool. And I want to 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 size, in in volume, and in mass. K? So that's going to happen to our sun, 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 radius? So I l equals Iω c. It's an object that spins, but it's not going to, its angular momentum is conserved even though this thing is blowing up. Right? So I have this is going to be something m r^2 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 2, get that, and then this goes down by a factor of 2, Notice that r is squared. So I'm going to actually square the factor of 2. Let me get out of the way. So the net result of this going down by a factor of 2 is that it actually goes down by the whole thing goes down by a factor of 4. So I have this going down by a factor of 2, this going down by a factor of 4. I multiply those 2, and I have, this thing growing by a factor of 8. Okay. So 2 times 4 is 8. If these two variables here become 8 times smaller, this variable has to become 8 times greater so that the whole thing is a constant. So this star would then spin 8 times faster, 8 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 know if you have any questions and let's get going.

# Conservation of Angular Momentum - Online Tutor, Practice Problems & Exam Prep

### Conservation of Angular Momentum

#### Video transcript

### Ice skater closes her arms

#### Video transcript

Hey, guys. So let's check out this classic example of conservation of angular momentum. Okay? Remember, conservation of angular momentum involves objects that are spinning that will change either their m, and that will cause a change in omega, or they will change the radius 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 an ice skater that has a moment of inertia of 6. So I equals 6 when she spins with her arms open. So her I is 6 when she spins with arms will open. And 4, if she closes her arm. It says here if she spins with 6 with a 120 RPM with her arms open. So RPM open is 120. What RPM will she have as a result of closing her arms? What will be RPM closed?

So you can think of open as initial because that's where we start, and you can think of close as final. K. So we're going to 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 going to be this is going to be just I omega for one person. I_{initial}, omega_{initial} equals I_{final}, omega_{final}.

The i's are given. So this is going to be initial is 6 and then I have omega. I'm gonna make a little space here and then this is 4 and then I'm gonna make 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 equation is 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 rewrite this equation here in terms of RPM. So let's do that real quick.

Remember, omega is 2 pi f or 2 pi over T or 2 pi. I'm gonna plug in f here and it's gonna be RPM over 60. So what I want to do real quick is I want to show you that there are three variations of this question. Okay? So let's do that real quick. So this is like the official legit version number 1. Here's version number 2. Instead of omega, I'm going to write 2 pi f. And look what happens. I_{initial}, 2 pi frequency_{initial} equals I_{final}, 2 pi frequency_{final}. Notice that the 2 pis cancel, and you end up with I_{initial}f_{initial} = I_{final}f_{final}.

If you do this with a period, this is version 2. I_{initial}2 pi period_{initial} equals I_{final}, 2 pi period_{final}. These guys will cancel, and you end up with I_{initial}period_{initial} = I_{final}period_{final}. And you can do the same thing for RPM. Okay. And this is the last one. That's the one we're going to use here. You can say I_{initial}. Now instead of 2 pi f, we're going to use 2 pi RPM over 60. So 2 pi RPM over 60 equals I_{final}, 2 pi RPM over 60. Here, I can cancel the 2 pis and the RPM over 60, and you're left with I_{initial} RPM_{initial} = I_{final} RPM_{final}.

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, by basically briefly rewriting the equation. So one point that I want to make here is that a way to know how to make these exchanges very quickly is to look at omega. Omega is on the top. It's on the denominator, up numerator up here. F is on the numerator up here. 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. K?

So in this question, we're actually we don't have to convert the RPMs into omega and then back into RPM. We can just actually plug in the RPM. So I'm gonna plug in RPM initial, RPM final. K? It would have been quick to just replace stuff, but I wanted to show you that we can do this. So RPM initial is 120, and then this is for RPM final. So RPM final will be 6 times 120 divided by 4. Okay. And the answer here is 180 RPM.

Now, the last point I want to make is notice that our I went from 6 to 4. It changed by a factor of 1.5. It went down by a factor of 1.5, and then the RPM went from 120 to 180. It went up by a factor of 1.5. And that's because conservation of angular momentum here, regular momentum, is a linear relationship. There's no squares or whatever. So if one goes down 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 to do a little bit of analysis, and I wanted to introduce you to these three alternative versions of the conservation equation so you can solve some of these questions faster. Alright? 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.

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?

_{2}

_{2}

_{2}

_{2}

### Star collapses

#### Video transcript

Hey guys. So let's check out this example here. A very common example of conservation of angular momentum questions is that of a star dying. And that's because a star spins around itself. And when the star dies or collapses, it will change not only its mass, it will become lighter but also reduce in size, radius, and volume. Okay? So, this is a good setup for a conservation of angular momentum because angular momentum will be conserved. So let's check it out.

When a star exhausts all of its stellar energy, it dies. That's why it's sad. Poor thing. 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 sun 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 called the period. So, the period of the sun spinning around itself is 24.5 days. If our sun were to collapse and shrink 90% mass and 90% in radius, in other words, our new mass, the new mass of the sun, I'll call this m′, is going to be it's shrinking 90% in mass, meaning my new mass is 10% of the original mass. Okay? And the new radius is 10 percent of the original radius. I want to know how long would its new period of rotation take in days. In other words, if it has taken 24.5 days for the sun to spin around itself, how long would it take for the sun to spin around itself once these changes happen? In other words, what is my t_final? Right. Think of this as t_sun_initial. I want to know what is my t_final and what I want to do here is, instead of writing l_initial = 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 going to write it actually like this. I'm going to say l_initial = a constant. I'm going to expand this. Okay. I'm going to expand this. The gonna be I_initial Omega_initial is a constant I so a sun the sun can be treated as, as a solid sphere even though it's actually like a huge ball of gas. So, treating the solid sphere is kind of kind of bad, but it won't matter as I show you in a second. So I have, let's just do that for now, half m r², and then Ω. I want not Ω but I want period and remember Ω is 2π/T. So I'm going to rewrite this as ½mr2 and 2π/T.

So, this is a proportional reasoning question, 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 model the sun as a solid sphere, it doesn't matter because that fraction goes away anyway. So just write m r² and you're good. It's sort of what I did earlier, what I had like box m r². Right? Omega. So something like that because the fraction doesn't matter.

Here's what's happening. This guy here is decreasing by 90 percent. 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 r. Now, the r is squared so if you square 0.1 you get 0.01. 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 a 1000 times smaller. Therefore, the right side of the equation has to become a 1000 times greater. Okay? So the right side of the equation has to become a 1000 times greater. So this side here grows by a 1,000. Now, the problem here is it's a little bit complicated because t is in the bottom. So if the whole thing grows by a 1,000 that means that t, which is in the denominator, actually goes down by a factor of a 1,000. Okay? If your fraction goes up, your denominator went down and that's how your fraction goes up. Right? Imagine, for example, you have a 100 divided by 10, that's 10. A 100 divided by 2, that's 50. Okay. Your entire thing went up because your denominator went down. So if this goes down by a lot, the right side right here goes up by a lot, and then the denominator goes down by a lot. So basically, your new period is a 1000 times smaller than your old period. So I can write t_final is one over a 1000 t_initial. So it's basically 24.5 days divided by a 1,000.

And then what you want to do is you want to convert this into hours. Okay? So we're going to do here is that, one day has 24 hours. This cancels this cancels and you multiply some stuff, and I actually have this in minutes. Okay? I actually have this in minutes. So let me turn that into minutes. So it's going to be 24 hours times 60 minutes. And when you do this, you get that the answer is about 35 minutes. I did minutes because otherwise you end up with about 0.55 hours and minutes makes more sense. It's easier to make sense of it. Okay? So imagine this, the sun takes 24 days to go around itself and after it collapses and it shrinks significantly, it's going to spin a 1000 times faster. So it's going to 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.

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.

_{f}= 1.5 rad/s

_{f}= 4 rad/s

_{f}= 6 rad/s

_{f}= 24 rad/s

### Landing and moving on a disc

#### Video transcript

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 on that disk. The person or object will either move closer to the center of the disc or away from the center of the disc. In doing this, you are going to change how fast the disc spins. Okay. So it's conservation of angular momentum. Let's check it out. You're moving on a rotating disc. So we have a disc with a mass equal to 200 kg. Its radius is 4 meters. Now this is a disc which means that the moment of inertia we're going to use for the disc is the moment of inertia of a solid cylinder, which is 12mr2. It spins about a perpendicular axis through its center. If you imagine a disc, a perpendicular axis means that the axis of rotation, an imaginary line, is perpendicular and makes a 90 degrees angle with the face of the disc so that the disc spins around this bottom axis. And there's going to be a person walking around on it. Right? So through its center at 2 radians per second. The disc is initially rotating with an angular velocity ωinitial=2 radians per second. And then you have a person. So this is big M. We're going to say that the person has mass little m equal to 80 kg, that falls onto the disc with no horizontal speed. Here is the disc, and the person will just like parachute into the disc, land on top of the disc. You're probably imagining that this will cause the disc to rotate at a lower rate because he added mass to it; it's heavier now. That's actually what's going to happen. So I'm going to draw a top view of the disc. Let's make that a little rounder. Alright. So here is the disc. The radius of the disk R is 4. The person is going to land somewhere over here at a distance of 3. So remember, radius is big R, distance from the center is little r, little r is 3. So the person lands there. Initially, there's 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 disc's new angular speed? So omega initial for the disc is 2. Omega final for the disc is what we're looking for, so question mark. Okay. So conservation of angular momentum. So I'm going to write that angular momentum initial equals angular momentum final. I'm going to expand this. This is Iinitialω and then this is Ifinalω. However, initially, there's only a disc. So I'm going to say that this is IbigM and ωbigM. But at the end, there are two things. There is big M and there's also little m, so we're going to do this, Ifinalωfinal of little m little m. Okay? So we added mass to the system. The system now has 2 I's instead of just 1. 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 omega of the disc is the same as the final omega of the person because if you land on a disc that's spinning, right, if the disc is spinning, you land on it, you're going to rotate with the disc. Right? So you're rotating with the disc so you have the same omega. So these two guys here are actually the same which means you can do something like this. Instead of saying omega final big M and omega final little m, you can just call this omega final. And instead of having 2 variables, you have one variable, which is simpler. So you can do omega final in both of these. You can factor it out and then have IfinalbigM+Ifinallittlem. Okay?

Then you have IinitialbigMωinitialbigM, which we have. So we're looking for this. We have this. So all we got to do is calculate all the I's. Okay. So let's do that. The moment of inertia of the disc, in the beginning, the disc is by itself. So you have 12mr2, and I have all these numbers, so this is easy to calculate. 12 m is 200 and r is 4 squared. And when you do all of this, you get 1600. So this is going to be 1600 right here. Okay. So this number is 1600. The initial speed of the disc is 2 equals omega final, and then these two numbers here. Now at the end, after you land on it, well, the disc still has the same moment of inertia. Right? The mass of the disc didn't change. The radius of the disc didn't change. What changed 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 the final moments of inertia of this person. We're treating the person as a point mass. So we're going to use the moment of inertia of a point mass, which is mr squared where r is the distance from the center. The person is 80. It is a distance of 3, not 4 but 3 squared. So if you multiply, all of this, you get that this is 720. This is 720, and that's the number that goes right here, 720. Cool. So if you solve here, you have this is 3200 on the left. The stuff on the right side adds up to 2320. And if you divide, you get an omega of 1.38 radians per second. Now this should make sense. You started off at 2. You added mass. The disk became hea

## Do you want more practice?

More sets### Your Physics tutor

- The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is 1.20 * 10^-...
- 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...
- (II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5 s to a fina...
- A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet...
- A satellite follows the elliptical orbit shown in FIGURE P12.77. The only force on the satellite is the gravit...
- A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250...
- During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each...
- (II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and mom...
- (II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and mom...
- (II) A uniform disk turns at 4.1 rev/s around a frictionless central axis. A nonrotating rod, of the same mas...
- (II) A woman of mass m stands at the edge of a solid cylindrical platform of mass M and radius R. At t = 0, t...
- (II) A woman of mass m stands at the edge of a solid cylindrical platform of mass M and radius R. At t = 0, th...
- Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revoluti...
- Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revoluti...
- A 70.0-kg person stands on a tiny rotating platform with arms outstretched.(c) If one rotation takes 1.2 s whe...
- A 70.0-kg person stands on a tiny rotating platform with arms outstretched.One rotation takes 1.2 s when the p...
- (II) Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s w...
- A 70.0-kg person stands on a tiny rotating platform with arms outstretched.(e) From your answer to part (d), w...
- (III) On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it l...
- (III) On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it l...
- (III) On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it l...
- (II) Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s w...
- (II) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total mome...
- CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless c...
- (III) Suppose a 65-kg person stands at the edge of a 9.8-m-diameter merry-go-round turntable that is mounted o...