Angular Momentum of a Point Mass - Video Tutorials & Practice Problems

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Angular Momentum of a Point Mass

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Hey, guys. So in this video, I want to talk about the angular momentum l of a point mass. Now, remember that in rotational problems, we can have either a point mass, which is a tiny object of no shape of no radius, no volume. Or you can have a rigid body and extended body, which is a an object of a known shape that has a rated. This has a volume, and depending on whether you have a point mass or rigid body, uh, questions, you're gonna be a little bit different. So let's look at the angular momentum of a point mass. Let's check it out. So first, I wanna I wanna talk about a point mass in a circular path. A point mass, Tiny little em in a circular path like this about a central axis right there. It's gonna have a rotational speed omega, and it has a linear equivalence. V 10 V 10 is a tangential velocity, um, that you have when you're spinning. Now, this does not mean this does not mean that you have two types of motions. You Onley have one type of motion, which is rotational motion. Okay, because just because you have a w n a V. It doesn't mean that you have linear and angular motion. You Onley have rotational motion. That's why I call this a linear equivalence. Okay, um so that's the first distinction I wanna make. Also want to point out that linear momentum in rotational momentum are not going to give you the same number are not going to give you the same number. So because I have a new omega, I confined a rotational momentum Rotational momentum. L equals Iomega. So if I have Omega, I have this well, is if I'm going in a circle. I also have a equivalent velocity V 10. So because I have a V that's equivalent to the W I confined and linear momentum a swell. So if I have a w, I can calculate l If I have a V, I can calculate P, but these two will not give you the same number. And that's because linear momentum, remember is absolute. But angular momentum is relative. It depends on the axis. Soapy is always going to be the same if your view is the same. But if you change the radius of your rotation, you're l will change. So these numbers will not be equivalent to each other. They will not be the same number. Okay, if you calculate linear and angular Momenta Momentous. Just plural for momentum. Linear momentum is people angular. Momentum is l for a point mass. You're going to get different equations in different numbers. We're going to a quick example here. So I can I can show you what I mean by that. And you see how this stuff works. So I have a small 2 kg object. So Massey goes to small, tells me that this is going to be a point mass. It's not gonna have a shape. Um, it spins horizontally around the vertical axis. So spinny horizontally means you spin like this around the vertical axis is just imagine that you have, um, imaginary line that you spin around that line, and that line is vertical. Okay, so I'm gonna draw, um, the axes here, and you're sort of spinning like this. Okay, so here's the object. Mm. Right there. Um, you do this at a rate of three radiance per second. Three radios per second is going to be your omega. You know that because of the units radiance per second. So omega equals three. And it says, maintaining a constant distance off four to the axis. So it means that your distance to the axis is going to be four. Distance to the axis is little are So I'm gonna write that little are is four. All right, So we want to know what is the objects? Linear momentum. And what is the objects? Angular momentum about it's central axis. So for a we're looking for P p equals M V and then for B, we're looking for l l equals I omega. Okay. Now, for the purposes of solving this question before I find the numbers, I actually wanna change up these equations. I can change l to make it look kind of like p. And I want to do this to show you that even though l can look like pee, um, it's not going to be exactly the same. Okay, you see what I mean once I get there. So what I'm gonna do here is I'm gonna replace I with the moment of inertia equation. Remember, Moment of inertia of a point mass is given by M r Squared where r is distance, um, to the center, to the axis of rotation. So when we write this as M R Square now Omega is related to V right? So if you're spinning like this and you have a new mega and you have a V, they are related by here's North. Our vector is the length the distance right there related by V equals R omega. So if I rewrite Omega, Omega is going to be V over r. Okay, So instead of omega, I'm gonna have the over our again. I'm doing this to show you something. Look, what happens here are cancers with our and then I have m r v. I'm gonna change that around. I'm gonna write instead of m V are on the point that I wanna make here Is that p is M V. P is M V and l is m V are therefore not the same equation and not the same number. Okay, so now we can plug stuff in masses to, um, the velocity the velocity would have to calculate that so v equals r w. So our is ours. Distance four w is three. So V is 12. So Emma's too And then this is 12. So this is gonna be 24. 24 it's kilograms times meters per second. Now for l. I have mass, which is true. I have 12 is the velocity and ours distance, which is four. So this is going to be 96 kilograms times meters squared over seconds. Okay, so notice how these are different equations, and they are different answers. Okay, The only case in which there will be the same is if they are happens to be one, which is just a gigantic coincidence on been. In that case, they will happen to be the same. But there's nothing special about articles one other than you get the same numerical value. Okay, so that's it for this one. The last point I wanna make here is that this equation that we got here that l so l is Iomega. But you can also have l as m v r for a point mass. So this equation l m v R is also used for angular momentum. L often object and linear motion about an axis of rotation. So if you ever get asked the question that says, find the find the angular momentum oven object about an axis of rotation. This is how you would do it. Okay, We're gonna do more of this soon, but I want to quickly show you a new example. Let's say this is one of the classic ones. Let's say you have this. A bar being supported by a folk from here, and then a little object is falling here. Right, So the object is about to hit. So when the object is about to hit has a velocity, let's say of 10. Okay. So even though this object is moving in linear motion, you can ask what is the angular momentum of this object relative to the axis of rotation? The axis of rotation is here because this is a point about which this could spin. Um, you would you would writes that l is m v are you would be given the three m. V is 10 and our would be this distance here. Okay, so that this equation here can be used if you want to find the angular momentum of an object that has linear motion. Okay, so that's it for this one. Hope this makes sense. Let me know if you have any questions and let's get going

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Problem

Problem

The Earth has mass 5.97 × 10^{24} kg, radius 6.37 × 10^{6} m. The Earth-Sun distance is 1.5 × 10^{11} m. Calculate its angular momentum as it spins around itself. Treat the Earth as a solid sphere of uniform mass distribution.

BONUS 1:Treating the Earth as a point mass, calculate its angular momentum as it spins around the Sun.

BONUS 2:Does the Earth have linear momentum as it spins around (i) itself; (ii) the Sun?

A

L_{E,E} = 4.23×10^{33} kg•m^{2}/s

B

L_{E,E} = 7.05×10^{33} kg•m^{2}/s

C

L_{E,E} = 1.18×10^{34} kg•m^{2}/s

D

L_{E,E} = 2.68×10^{34} kg•m^{2}/s

3

Problem

Problem

A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod. The system rotates with 180 RPM about an axis perpendicular to the rod and through one of its ends, as shown. Calculate the system's angular momentum about its axis.

A

782 kg•m^{2}/s

B

905 kg•m^{2}/s

C

1030 kg•m^{2}/s

D

1404 kg•m^{2}/s

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