Practice: 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?

Subjects

Sections | |||
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Intro to Angular Momentum | 15 mins | 0 completed | Learn |

Jumping Into/Out of Moving Disc | 24 mins | 0 completed | Learn |

Opening/Closing Arms on Rotating Stool | 18 mins | 0 completed | Learn |

Spinning on String of Variable Length | 20 mins | 0 completed | Learn |

Conservation of Angular Momentum | 46 mins | 0 completed | Learn |

Angular Collisions with Linear Motion | 8 mins | 0 completed | Learn |

Angular Momentum & Newton's Second Law | 11 mins | 0 completed | Learn |

Intro to Angular Collisions | 15 mins | 0 completed | Learn |

Angular Momentum of a Point Mass | 22 mins | 0 completed | Learn |

Angular Momentum of Objects in Linear Motion | 7 mins | 0 completed | Learn |

Additional Practice |
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!! Conservation of Angular Momentum with Energy |

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 m^{2} . 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

**Transcript**

Hey guys. So let's check out this example here. So a very common example of conservation of angular momentum questions is that they start 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, in radius and volume, okay? So this is a good setup for a conservation of angular momentum because angular momentum movement concerns. So let's check out. When a start exhaust 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, just tell me what it 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 complete a full revolution of any sort is period. So period of the Sun is around itself 24.5 days. If our sun were to collapse and shrink 90 percent in mass and 90 percent in radius, in other words, our new mass, the new mass of the Sun, I'm going to call this M prime, is going to be, it's shrinking 90% of mass, meaning my new mass is 10 percent 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's taken 24 and a half 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 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 going to do is I'm going to write actually like this, IÕm gonna say, L initial equals a constant, I'm going to expand this, okay? I'm going to expand this. Going to be I initial Omega initial is a constant I. So the Sun can be treated as a solid sphere, even though it's actually like a huge ball of gas. So treat it as a solid sphere is kind of bad but it won't matter as IÕll show you in a second. So I have let's just do that for now. half MR squared and then Omega, I want not Omega but I want period and remember Omega is 2 PI over T. So I'm going to rewrite this as 2 PI over T and it's a constant. Because it 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 R squared and you're good, it's sort of what I did earlier when I had like box MR squared, right? Omega. So something like that because the fraction doesn't matter. So here's what's happening. And this guy here is decreasing by it 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 of 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 thousand times smaller, therefore the right side of the equation has to become a thousand times greater, okay? So the right side of the equation has to become a thousand times greater. So this side here grows by 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 1,000, that means that T which is a denominator actually goes down by a factor of 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 100 divided by 10, that's 10, 100 divided by 2 that's 50, okay? Your entire thing went up because your denominator went down. So this goes down by a lot the right side, right here goes up by lots and then the denominator goes down by a lot. So basically your new period is a thousand times smaller than your old period. So I can write T final is 1 over 1,000 T initial. So it's basically 24.5 days divided by 1,000 and then what you want to do is you want to convert this into hours, okay? Or some measure of that sort. So we're going to do here 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 change 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 like point 55 hours and minutes makes more sense, it's easier to make sense out 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 thousand 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.

Practice: 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.

Example #3: Landing and moving on a disc

**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 disk that spins and then there's either a person or an object that disk and the personal object will either move closer to the disk or closer to the center of the disk or away from the center of the disk 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 disk so we have a disk of mass 200, mass equals 200 radius is 4 meters. 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 mr squared, it spins about a perpendicular axis through its center, if you have a disk, here is 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 spins like this about an axis and there's going to be like a person here walking around, right? So through its center at 2 radians per second. So the disk is initially rotating with an Omega initial of 2 radians per second and then you have a person, so this is Big M, we're going to say the person has mass little m equals 80 that falls into the disk with no horizontal speed. So, here is the disk and the person will just like parachuting to the disk, 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 it so it's heavy enough, 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 the disk, 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 sensor 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 person which means our Omega will change, what we want to know is, what is the discs new angular speed? So Omega initial for the disk is 2, Omega final for the disk 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 final, I'm going to expand this as I initial, Omega initial and then this is I final, Omega final however in the beginning there is only a disk. So I'm going to say that this is I of Big M and Omega of big M but at the end, at the end there are two things, there is big m and there's also little m, so we're going to do this I f Omega f of little m, little m, 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 disk which is this. Now, let me point out to you that the final Omega 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 disk spinning, you land on it, you're going to rotate with the disk, 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 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 two 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 I final m plus I final little m, okay? And then you have I initial Big M and then Omega initial Big M which we have. So we're looking for this, we have this. So all we got to do is calculate all the Is, okay? So let's do that the, I'm going to do this off to the side over here and the moment of inertia of the disk in the beginning, the disk is by itself. So you have 1/2 mr squared and I have all these numbers so this is easy to calculate 1/2 m is 200 and r is 4 squared and when you do all of this, when you do all of this to get 1600 so this is going to be 1600 right here, okay? So this number is 1600, the initial speed of the disk is 2 equals Omega final and then these two numbers here. Now at the end after you land on it, well, the disk, the disk 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 the final, the final moment of inertia of this person, we're treating the person as a point mass. So we're gonna use the moment of inertia of a point mass which is mr squared where r is distance from the center, the person is 80, it is a distance not 4 but 3, 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 3,200 on the left, the stuff on the right side adds up to 2320 and if you divide you get an Omega 1.38 radians per second. Now, this should make sense you started off at 2 you added mass the disk became heavier therefore the disk slows down a little bit and now it has an Omega final of 1.38, okay? So it went from 2 to 1.38, this is part a, I'm going to make some space here and then solve Part B, Part B is calculate a person's new tangential speed okay, if you are a point mass going around in a circular path not only do you have an Omega but you have an equivalent tangential speed that is tied to your Omega and that's given by r Omega. So that's all we got to 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 r 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 4 but you're spinning around a circular path of radius of 3. So you put a 3 here. Alright, 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 disk, 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 we're moving within the disk after you land. So, it says, the person starts walking towards a disk Center, calculate the disks speed once the person reaches it. So I want to draw here a quick diagram, you got the person here at r equals 3, okay? I'm going to do this and the person is initially at r equals 3 right there. So that's my r think of this as your r initial, okay? And then the person is going to walk in this direction here until gets to the center. So the person is going to have an r final of 0, okay? So do you think the disk is going to 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 going to spin faster, the effective total radius of this system is going to be small, they're more masses 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 or the mass then we're going to write 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 disk, we have disk plus guy, when the guy is at r equals 3 and at the end we also have disk plus guy but here the guy is at a position of r equals 0, the guy sits on the axis of rotation, okay? So let's expand this equation here I initial, Omega initial I final, Omega final and for both of these, we have guy, and we have guy and disk. So what you can do is you can write I as I disk plus I guy is the moment of inertia of the system, you can do it together like this, Omega initial equals I disk plus I guy Omega final, so this is obviously I disk initial, I guy initial, I disk final, I guy final, the moment of inertia of the disk doesn't really change because the disk 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 mr squared. Remember, the moment of inertia of the guy since he truly has a point mass is M little r squared, this is little r initial, this is the number that's changing, this is a 3 in the beginning and a 0 at the end Omega initial. So the guy lands at this thing and he automatically picks up a speed of 1.38. So that is the Omega initial of the guy in the disk together 1.38 it's not 2, 2 was the speed of the disk before you landed on it, okay? And then you hear on the other side you have same thing half mr squared, that doesn't change but, this is going to be mr final squared, in this final here r final is 0, okay? And Omega final is what we're looking for, we have all the numbers so we should be able to plug everything in and solve for Omega final. So the rest here is just algebra, most important step was getting here, okay? So let's do this, half the mass of the disk is 200, actually already calculate this whole thing here before, this is just 1600, right? Yeah, this is 1600 plus this thing here which, de we have this before? and the mass of the guy is 80 distance is three squared, I think we had that before yeah, that's 720, right? So we have 1600 plus 720 and the whole thing is 1.38, we had these two guys from before, from Part A, the end of Part A, and then here we have the same 1600 but now, check it out, this whole thing will be 0 because the r is 0, he is sitting of the axis of rotation, a point mass sitting on your axis of rotation doesn't contribute towards your moment of inertia, right? It says, if it wasn't there it's basically the same thing is that this guy just got like picked off from the disk. Alright, so I'm going to put a plus 0 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 3,200 divided by 1600 and I get that Omega final is 2 radians per second. Now, where have we seen 2 before, 2 was the original speed that the disk 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'd back to the original part of the problem which is disk without the person was spinning with 2, you add a person, you're slower, the person walks to the center which means that the person's contribution to moment of inertia disappears. So you're back to where you were before the person was on the disk, cool? So 2 is not a coincidence, it should make sense, hope the whole question makes sense, let me know you have any questions and let's keep going.

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A horizontal platform in the shape of a uniform disk (I = 1/2 MR 2 for an axis at its center) is rotating without friction about a vertical axis at its center. The platform has mass 8.00 kg and radius 0.400 m. The platform is initially rotating at an angular velocity of 0.300 rev/s. Then a small bag of sand with mass 6.00 kg is dropped from a small height onto the platform at its rim. The bag of sand can be treated as a point mass. What is the final angular velocity of the platform after the bag of sand has been dropped onto it?

A man holding a heavy object in each hand stands on a small platform that is free to rotate about a vertical axis. Initially he is standing with his arms outstretched and he and the platform are rotating with an angular velocity of 0.600 rad/s. With his arms outstretched, the moment of inertia of the system (man + platform + weights) is 4.00 kg•m2. Then he pulls the weights in close to his chest and the angular velocity of the rotating system becomes 1.80 rad/s. What is the moment of inertia of the system after he has pulled his arms in?

A figure skater on ice spins on one foot so there is minimal friction. She pulls in her arms and her rotational speed increases. Choose the best statement below:
1.When she pulls in her arms, her rotational kinetic energy is conserved and therefore stays the same.
2. When she pulls in her arms, the work she performs on them turns into increased rotational kinetic energy.
3. When she pulls in her arms, her angular momentum decreases so as to conserve energy.
4.When she pulls in her arms, her rotational kinetic energy must decrease because of the decrease in her moment of inertia.
5. When she pulls in her arms, her moment of inertia is conserved.
6. When she pulls in her arms, her rotational potential energy increases as her arms approach the center.

A merry-go-round spins freely when Janice moves quickly to the center along a radius of the merry-go-round. It is true to say that
(A) the moment of inertia of the system decreases and the angular speed increases.
(B) the moment of inertia of the system decreases and the angular speed decreases.
(C) the moment of inertia of the system decreases and the angular speed remains the same.
(D) the moment of inertia of the system increases and the angular speed increases.
(E) the moment of inertia of the system increases and the angular speed decreases.

A playground merry-go-round has a radius of 3 m and a rotational inertia of 600 kg · m2 . It is initially spinning at 0.8 rad/s when a 20 kg child starts crawling from the center toward the rim. When the child reaches the rim, what is the new angular speed of the merry-go-round? Ignore the friction force on the axle of the merry-go-round.
A. 1.1 rad/s
B. 0.73 rad/s
C. 0.8 rad/s
D. 0.62 rad/s
E. 0.89 rad/s

The sketch shows the top view of a merry-go-round which is rotating clockwise. A child jumps on the merry-go-round in three different ways: the child lands the same distance from the center. Compare the magnitudes of the final angular momenta of the merry-go-round.
I) from the left,
II) from the top, and
III) from the right. For all three cases,
1. LI = LII < LIII
2. LI > LII > LIII
3. LI = LII > LIII
4. LIII > LI > LII
5. LI = LIII < LII
6. LII > LI > LIII
7. LI = LIII > LII
8. LIII > LII > LI
9. LII > LIII > LI
10. LI > LIII > LII

The angular momentum of a system remains constant when: A. the total kinetic energy is constantB. when no net external force acts on the systemC. when the linear momentum and the energy are constantD. when no torque acts on the systemE. all the time since it is a conserved quantity

A merry-go-round is a playground ride that consists of a large disk mounted so that it can freely rotate in a horizontal plane. The merry-go-round is initially at rest, has a radius of R = 1.4 meters, and a mass M = 216 kg. A small boy of mass m = 41 kg runs tangentially to the merry-go-round at a speed of v = 1.9 m/s, and jumps on.(A) Calculate the moment of inertia of the merry-go-round.(B) Immediately before the boy jumps on the merry go round, calculate his angular speed about the central axis of the merry-go-round.(C) Immediately after the boy jumps on the ride, calculate the angular speed of the merry-go-round and the boy

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