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, or no volume, or you can have a rigid body, an extended body, which is an object of a known shape, that has a radius, has a volume. And depending on whether you have a point mass or a rigid body, questions are going to be a little different. So let's look at the angular momentum of a point mass. Let's check it out. So first, I want to talk about a point mass in a circular path, a point mass, tiny little \( m \) in a circular path like this, about a central axis right there, it's going to have a rotational speed, \( \omega \), and it has a linear equivalent, \( V_{\text{tan}} \). \( V_{\text{tan}} \) is a tangential velocity, that you have when you're spinning. Now, this does not mean that you have 2 types of motions. You only have one type of motion, which is rotational motion. Okay. Because just because you have \( \omega \) and \( v \), it doesn’t mean that you have linear and angular motion. You only have rotational motion. That's why I call this a linear equivalent. Okay. So that's the first distinction I want to make.
I also want to point out that linear momentum and rotational momentum are not going to give you the same number, are not going to give you the same number. So because I have an \( \Omega \), I can find a rotational momentum. Rotational momentum, \( l \) equals \( I \omega \). So if I have \( \omega \), I have this. Well, if I'm going in a circle, I also have an equivalent velocity, \( V_{\text{tan}} \). So because I have a \( v \), that's equivalent to the \( \omega \), I can find a linear momentum, as well. So if I have \( \omega \), I can calculate \( l \). If I have \( v \), I can calculate \( p \). But these 2 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. So \( p \) is always going to be the same if your \( v \) is the same. But if you change the radius, of your rotation, your \( l \) will change. So these numbers will not be equivalent to each other. They'll not be the same number. Okay. If you calculate linear and angular momenta, momenta just plural for momentum, linear momentum is \( p \), angular momentum is \( l \). For a point mass, you're going to get different equations and different numbers. We're going to do a quick example here so I can show you what I mean by that. And you'll see how this stuff works.
So I have a small 2-kilogram object, so mass equals 2. Small tells me that this is going to be a point mass. It's not going to have a shape. It spins horizontally around the vertical axis. So spinning horizontally means you spin like this. Around the vertical axis is just imagine that you have an imaginary line that you spin around that line and that line is vertical. Okay. So I'm going to draw, the axis here and you're sort of spinning like this. Okay. So here's the object, \( m \), right there. You do this at a rate of 3 radians per second. 3 radians per second is going to be your \( \omega \). You know that because of the units, radians per second. So \( \omega \) equals 3. And it says, maintain a constant distance of 4 to the axis. So it means that your distance to the axis is going to be 4. Distance to the axis is little \( r \). So I'm going to write that little \( r \) is 4. Alright. So we want to know what is the object's linear momentum and what is the angular momentum about its central axis. So for \( a \), we're looking for \( p \), \( p \) equals \( mv \). And then for \( b \), we're looking for \( l \), \( l \) equals \( I \Omega \). Okay. Now for the purposes of solving this question, kind of like kind of like \( p \). And I want to do this to show you that even though \( l \) can look like \( p \), it's not going to be exactly the same. K. You'll see what I mean once I get there. So what I'm going to do here is I'm going to replace \( I \) with the moment of inertia. Equation, remember moment of inertia of a point mass is given by \( mr^2 \) where \( r \) is distance to the center, to the axis of rotation. So I'm going to rewrite this as \( mr^2 \). Now \( \omega \) is related to \( v \). Right? So if you're spinning like this and you have an \( \omega \) and you have \( v \), they are related by there's an \( r \). Our vector is the length, the distance. Right? They're related by \( v = r \omega \). So if I rewrite \( \omega \), \( \omega \) is going to be \( v \) over \( r \). Okay. So instead of \( \omega \), I'm going to have \( v \) over \( r \). Again, I'm doing this to show you something. Look what happens here. \( r \) cancels with \( r \) and then I have \( mrv \). I'm going to change that around and I'm going to write instead of \( mv r \). And the point that I want to make here is that \( p = mv \), \( p = mv \) and \( l = mv r \). Therefore, not the same equation and not the same number. Okay. So now we can plug stuff in, the mass is 2, the velocity, the velocity would have to calculate that. So \( v = r\omega \). So \( r \) is 4, \( \omega \) is 3. So \( v \) is 12. So \( m \) is 2 and then this is 12. So this is going to be 24 \( \text{kg}\cdot\text{m/s} \). Now for \( l \), I have mass, which is 2, I have 12 as the velocity, and \( r \) is the distance which is 4. So this is going to be 96 \( \text{kg}\cdot\text{m}^2/\text{s} \). Okay. So notice how these are different equations and they are different answers. Okay. The only case in which they will be the same is if the \( r \) happens to be 1 which is just a gigantic coincidence. And then in that case, they will happen to be the same. But there's nothing special about an \( r = 1 \) other than you get the same numerical value.
