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Ch 16: Angular MomentumWorksheetSee all chapters

# Conservation of Angular Momentum

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Sections
Angular Momentum & Newton's Second Law
Opening/Closing Arms on Rotating Stool
Spinning on String of Variable Length
Angular Momentum of Objects in Linear Motion
Intro to Angular Collisions
Conservation of Angular Momentum
Angular Collisions with Linear Motion
Intro to Angular Momentum
Jumping Into/Out of Moving Disc
Angular Momentum of a Point Mass

Concept #1: Conservation of Angular Momentum

Transcript

Hey guys! In this video we're going to start talking about the conservation of angular momentum. Let's check it out. Remember that when we talked about linear momentum that the most important part about linear momentum was the fact that it is conserved. It's conserved in certain situations. Let me talk about that in just a bit. The same thing is gonna happen for angular momentum. Most angular momentum problems are actually going to be about the conservation of angular momentum. They're going to be about conservation. They're going to be conservation problems. What I want to do here is do sort of a compare and contrast between linear momentum and its angular equivalent angular momentum. Linear momentum little p is mass times velocity, angular momentum big L is I _ moment of inertia and angular speed. Linear momentum is conserved if there are no external forces and angular momentum is conserved if there are no external torques. This should make sense. Angular momentum is the rotational equivalent of linear momentum. Torques is the rotational equivalent of forces. Even better description is it's actually not that there are no forces, it's just that there are no external forces or that if there are external forces, they at least cancel each other out. An even better definition is if the sum of external forces, they could exist they just have to add up to zero. 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. One difference between these is that most problems for linear momentum involve two objects, pretty much all of them involve two objects colliding against each other. The conservation equation will look like this: p initial = p final right so itÕs saying that momentum doesn't change, this is of the system. I can expand this equation and I have two objects, so p initial becomes p initial 1 + p initial 2 = p final 1 + p final 2. It's going to be this very familiar equation, m1v1 initial + m2v2 initial = m1v1 fiinal + m2v2 final. Angular momentum is a little bit different that there's a lot of angular momentum problems that involve just a single object. Probably the most classic conservation of angular momentum question is when you have an ice skater. Let's say you have a girl ice skating and she is spinning with her arms open and she closes their arms and sheÕs 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 spin. The conservation equation will be similar. I'm going to have that L initial = L final because L doesn't change. That's the whole deal. L is I _, so I'm going to say that I initial _ initial is not going to change, itÕs constants. But what I want to do here is I want to expand this equation a little bit to show you something. Moment of inertia I for a point mass is something like mr^2. For a shape, it's something like let's say for a solid cylinder, it would be _ MR^2. For another object for like a solid sphere, it'd be 2/5 MR^2. The point that I want to make here is that it's something MR^2. What changes is that here you have _, you have 2/5, here thereÕs a 1 that hides in there. That's implicit we don't have to write. I'm going to say that this takes the shape, this I, takes the shape of box which is some fraction MR^2 and then I have _. I'm just expanding I _ to show this. I'm going to say that this is a constant. Meaning this number doesn't change.

Really the kind of problems you're going to have, there's two basic types of problems. In one type, the mass will change. I'm going to put a little _ here on top of M. The mass will change which will cause a change in _. On the other type of problem, the R will change and cause the change in _. If the mass of the system changes, the system will slow down. 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 velocity of rotation will go down as well. The opposite case of what I just mentioned with the girl spinning is if sheÕs spinning like this and then she opens her arms, she slows down. That's because her total r, you can see that these things are going away from the axis of rotation, so the r grows therefore the _ becomes smaller. The two types of changes we're going to have for one object is that either M or R will change and those will cause a change in _. When we have two objects, we have problems where you're essentially adding or removing mass. The classic example here is there's a disc that's spinning, you add a little block to it, what happens? The disc is now gonna spin a little bit slower and we can calculate that. When we had linear momentum, the two 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. Push away, two things are going away from each other; collision, two things are coming into each other. We also had these types of problems where you add or removing a mass in linear motion, which if you think about it, adding a mass is a collision. One mass joins the other and removing a mass is really a push away problem is if you jump out of a skate or something. That's it for that.

I have an introductory example here talking about a bunch of different situations to see so we can discuss what happens in these situations. We want to figure out whether the angular speed w will increase or decrease. An ice skater, we just mentioned this, an ice skater spins in 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 gonna use the equation here. What I'm going to do is IÕm gonna say L is a constant, L which is I_ is a constant. I'm going to expand I_ into something (MR^2) _, and this is a constant. What's happening here is that by closing her arms, her R is decreasing therefore here _ is going to increase. The answer is that _ increases. B, a large horizontal disc spins around itself. What happens to discÕs angular speed if you land on it? There's a disc spinning around itself like this. You land on it right here. This is you. You got added to the disc. What happens to the discÕs speed? L = I_ is constant. I'm going to expand I_ to be something (MR^2) _. These are constants. What's happening here is thereÕs mass being added to the system, therefore the system will slow down. IÕm gonna write here that _ will decrease. C, an object is tied to a point via a string that spins horizontally around it. 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. An object is going like this because it's connected to a string. What we want to know is what happens if you shorten the string. Again, L = I_ is a constant. I_, IÕm going to expand to be something MR^2. It doesn't matter what that something is for these problems. We're just doing a quick analysis of what would happen. If you shorten the length of the string, you're shortening the radius of rotation of this object. Therefore the _ will increase. You can imagine that if you spin something in a really long cable, the second you pull the cable in, it's going to instead of spin like this, it's going to get faster like this. The last one, a star, like the Sun, spins around itself. I want to know what happens if it collapses and loses half of its mass and half of its radius. You may know this. Stars live for obviously billions of years. Eventually they went out of star fuel and they collapse. What that means is that they're going to significantly shrink in size, in volume and in mass. That's going to happen to our Sun like in ten billion years. You're safe, don't worry. What happens if it collapses and loses half of its mass and half of its radius? L = I_ = c. It's an object that spins but its angular momentum is conserved even though this thing is blowing up. This is going to be something MR^2 _ and that's a constant. Here we actually have precise numbers, half and half. If this goes down by a factor of two, and then this goes down by a factor of two, notice that R is squared. I'm going to actually square the factor of 2. The net result of this going down by a factor of 2 is that it actually goes down, the whole thing goes down by a factor of 4. I have this going down by a factor of 2, this going down by a factor of 4. I multiply those two and I have this thing growing by a factor of 8. 2*4 is 8. If these two variables here become 8 times smaller, this variable has to become 8 times greater. The whole thing is a constant. This star would then spin 8 times faster than it was before it collapsed. 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. Let's keep going.

Example #1: Ice skater closes her arms

Transcript

Hey guys! Let's check out this classic example of conservation of angular momentum. 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 w or they will change the radius of some sort and that will cause a change in w. Those are the two types that you have to look out for. Here we have an ice skater that has a moment of inertia of 6, so I = 6, when she spins with her arms open. Her I is 6 when she spins with arms open and 4 if she closes her arms. I close = 4. It says here if she spins with 120 RPM with her arms open, so RPM open is 120, what RPM we should have as a result of closing her arms? What will be rpm close? You can think of open as initial because that's where we start and you can think of close as final. We're going to use the conservation of angular momentum equation, which is Li = Lf. In this case we have one person, so this is going to be just Iw for one person. I initial w initial = I final w final. The IÕs are given so this is going to be initial is 6 and then I have w, and then this is 4. Notice that this is w and this is omega but I gave you one rpm and I asked you for another RPM. This whole question is in terms of rpm but the equation in terms of w like usual. All of our equations are in terms of w. You always have to convert RPM into w. But what I want to show you is that you can actually rewrite this equation here in terms of rpm. Let's do that real quick. Remember, w is 2¹f or 2¹ / T or 2¹ IÕm going to plug in f here and it's going to be RPM/60. What I want to do real quick is I want to show you that there are three variations of this question. Let's do that real quick. This is like the official legit version number one. Here's version number two. Instead of w, IÕm going to write 2¹f. Look what happens. I initial 2¹frequency initial = I final 2¹frequency final. Notice that the 2¹ cancel and you end up with I f = I f. This is another version of this equation. You can just basically replace w by frequency and they are both on the top. If you do this with period, this is what you get. This is version two. I 2¹period initial = I final 2¹ period final. These guys will cancel and you end up with I initial period initial = I final period final. You can do the same thing for rpm. This is the last one. That's the one we're going to use here. We can say I initial now instead of instead of 2¹f, we're going to use 2¹ RPM/60, over 2¹ RPM/60 = I final 2¹ RPM/60. Look what I can do here. I can cancel the 2¹ and the 60 and you're left with I initial RPM initial = I final rpm final. 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. One point that I want to make here is that a way to know how to make these exchanges very quickly is look at w. w is on the top, it's on the numerator up here, f is in the numerator up here. They're both up top that's why they both show up top here. T is on the denominator that's why T shows up at the bottom when you replaced it. RPM is at the top that's why rpm shows up at the top here. In this question, we don't have to convert the RPMs into w and then back into rpm. We can just actually plug in the RPM. I'm going to plug in rpm initial rpm final. It would have been quick to just replace stuff but I wanted to show you that we can do this. rpm initial is open which is 120, and then this is 4 rpm final. rpm final will be 6*120 / 4. The answer here is 180 rpm, so this is the final answer. 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. 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 is in linear relationship. There's no squares or whatever. If one goes down by 1.5, the other one has to go up by 1.5. 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. That's it for this one let me know if you need any help, if you have any questions and let's keep going.

Practice: Suppose a diver spins at 8 rad/s while falling with a moment of inertia about an axis through himself of 3 kg•m2. What moment of inertia would the diver need to have to spin at 4 rad/s?

BONUS: How could he accomplish this?

Example #2: Star collapses

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