Hey guys. So, in this video, I'm going to talk about the center of mass, which is a pretty straightforward concept in physics. Now, the idea of the center of mass is that when you have a bunch of objects in a system, you can simplify that system and represent it as a single object. For example, if you have 1,000 or even millions of planets and stars in the galaxy spread all over the place, you could treat the entire system of millions of things as one object. Okay? Let's check it out.

So, as I said, sometimes it's useful to simplify a system of objects, a collection of objects, by replacing all of them with a single equivalent object. For example, instead of having a bunch of things moving this way, I can simplify this and just say that there's a single object that goes that way. Okay? Now, this single object will have mass \( m = \sum \) of all the individual masses. And the system will be located, right, at the system's center of mass. Here’s a really simple example of that. Let's say I want to combine the system made up of 2 masses, 2 kilograms and 2 kilograms, into a single object. So, first of all, the total combined mass will be 4 kilograms.

Now, where would it go? Well, I hope you're thinking that if I wanted to simplify this into one thing, the center of this whole combination of things is actually right down the middle. Okay. So, I would have something like this where this gap here is 5 meters. And this would be a 4-kilogram object. Now, the reason it's down the middle is that this is a balanced system. The left and the right have exactly the same masses. But if, for example, I had something like a 2 here and a 10 here, the center of mass would be much closer to the 10. It’d be somewhere over here. Okay? So that’s the idea.

Now, the middle thing only works if they are the same. You’re not going to get that. So you’re going to need to use an equation. You’re going to need to use the center of mass equation to figure out where the center of this combination of masses is. And you’re going to use the x-position. So where along the horizontal the center of mass is, is going to use this equation: \( \text{Sum of } m \times x \div \text{Sum of } m \). And I’ll explain what that means. I’ll give you an example where you have, let's say, 3 objects. Sum of \( m \times x \) looks like this: \( m_1 \times x_1 + m_2 \times x_2 + m_3 \times x_3 \). That’s what it means to do the summation of mx. You’re going to have \( m_1 \times x_1, m_2 \times x_2, m_3 \times x_3 \). And the sum of m's is just \( m_1 + m_2 + m_3 \).

Now, this is in the case like this, where both objects are on the x-axis, you draw a line between them and the center of mass will be somewhere along that line. But if you have objects in a 2-dimensional plane, something like this, and notice that I’m intentionally drawing the balls of different sizes. If you have a system made up of 4 objects, the center of mass will be somewhere in the middle of the system. In this case, the left side balances with the right. So it's going to be somewhere down the middle. But the bottom is much heavier than the top. So you're not going to have the center of mass be along a line here. It's going to be further down somewhere here. Okay? A little bit closer. This is where the center of mass is. So, I hope you see how this is 2 dimensions because I have x and y. So, in that case, you're going to use not only the y equation, but you’re also going to use an x and y, and the equation is the same, sum of \( m \) instead of \( x \), sum of \( my \div sum \) of \( m \). Okay. So, this looks like this: \( m_1 \times y_1 \, m_2 \times y_2 \, m_3 \times y_3 \, m_1 + m_2 + m_3 \). Now, \( x \) and \( y \) is the \( x \) and \( y \) position of these objects.

Let’s do 2 quick examples to see what this looks like in practice. First, I have 2 masses placed along the x-axis. So here’s the x-axis. One mass is mass a at 0 meters. So let’s say 0 is over here. So, I got at 0 meters, I got a, which is 10 kilograms. And at 4 meters, I have b, which is 20 kilograms. And I want to know what is the center of mass. This is x-axis only, so I’m going to say that the center of mass is the x center of mass, and the equation is this. So, it’s \( m_1 \times x_1 + m_2 \times x_2 \). So, what I’m going to do is write the masses here: \( 10 \times 20 \times 10 + 20 \). Okay. Now, all I got to do is plug in the numbers. What is the x-position of the 10? The 10 is at 0 and the 20 is at 4. Okay. So, I put a 4 here and that’s it. Very straightforward. This cancels. I have 80 over 30, and the x-position of this thing is 2.67 meters.

Now, the middle between \(0\) and \(4\) is \(2\). You should have been expecting that the actual center of mass is somewhere to the right of the middle because the system is heavier on the right side. And that’s what it is. \(x\) center of mass, that’s what we get, \(2.67\). So, you're able to, if you think about it a little bit, sometimes you’d be able to sort of guess where the answer will be. Okay?

I want to make a quick point, and then we’re going to jump into the next example. There are 2 terms that are similar. One is center of gravity, and the other one center of mass. Now in this video, we’re only talking about center of mass. We’re not talking about center of gravity, and we’re not really going to do problems with center of gravity. But I do want to clarify that there are 2 different terms and they mean different things. Now, without getting into what center of gravity is, I will just tell you that they are actually the same thing if the gravitational field is constant.

So, this is a conceptual point that I want you to remember: If the gravitational field is constant, then center of mass and center of gravity are the same thing. We can use them interchangeably. So, what does it mean for the gravitational field to be constant? Well, let me draw something real quick. You don't have to draw this. This is just conceptual. But let's say here’s the earth, and, you're here, and your sister's here, whatever. You're very close to each other and I want and then you form a system. The gravity where you are is, let’s say, 9.8 and it points straight down. Your sister is right next to you, so the gravity that she feels is also 9.8 and it's also straight down. If you guys are close enough together, the gravities will be almost identical. In fact, they'll be so close that we can consider them to be the same. So, because the two objects feel the same gravity, the gravitational field is the same for both. The center of mass is the same as the center of gravity. That’s the idea.

Now if 2 people on earth are really far apart, they will feel different gravities because even if it’s 9.8, one's being pulled this way, the other one's being pulled this way. This is a conceptual point for you to know. A lot of professors don’t even get into this. So, if you didn't really hear the distinction between the 2, don’t even worry about it. It's not that big of a deal. Okay? The last point I want to make here is that unless otherwise stated, we're just going to assume that this here is the case.

We’re going to assume that the gravitational field is constant. So what does that mean? Well, it means that these two things mean the same, or the same thing. So, if the gravitational field is constant, these two words mean the same. We’re going to assume that it’s constant. So we’re going to assume that these terms are the same thing and we can use them interchangeably. Okay? Again, not going to do problems with the center of gravity, but I do want to touch up on the conceptual point there. Alright.

So, let me quickly do example 2, and we’ll be done with this concept. That's all we're doing. So, 3 masses are placed on the xy-plane. So I’m going to draw a little xy plane like this: \(y, x\). Mass \(A\) is placed at \(0, 0\) is right here. Remember, guys, coordinate systems are \(x, y\). So this means \(x = 0\) and \(y = 0\). This is object \(a\), which has a mass of 10. \(B\) is at \(3, 0\). So this is \(x\) and \(y\). \(X = 0\) is on this line here. Okay. And \(3\), I’m sorry. \(x = 3\) going up will be somewhere here. So this is \(0\) on the y in the x-axis and \(3\). Okay? So, \(0\) on the x-axis means you haven’t moved left or right, you’re in the middle. And then \(3\) on the y-axis means you go up \(3\), \(B\), 8 kilograms. And then \(c\) is at \(4, 0\). So the x-value is \(4\). I go \(4\) this way, but I stay on the on the x-axis. \(4, 0\) looks like this. And this is \(c\), which has a mass of \(6\) kilograms. Here’s the diagram.

I want to know what are the x and y coordinates of the center of mass of the system. You might be thinking the center of mass of the system is somewhere here. Okay? You can actually think about this in terms of x and y. Let’s try to look at this. On the y-axis, I have this guy on the y-axis and these 2 guys on the y-axis. Okay. Don’t make all these scribbles but notice how this one is 8 and these 2 here combine to be 16. So the y-axis is heavier toward the bottom. So I’m going to predict that this thing will be, somewhere like around \(1\). Okay? Not \(3\), not the middle, but closer to \(0\).

Now on the x-axis, you have these 2 here and this here. So the center of mass will be somewhere in the middle. The left, I have \(10 + 8\), that’s \(18\). And to the right, I have \(6\). Okay. To the right, I have \(6\). So, this thing is much heavier toward the left. So instead of being in the middle between \(4\), which would be \(2\), I’m going to guess it's going to be to the left of \(2\). I’m going to guess that’s going to be around \(1\). Again, I’m just doing rough estimates so that we can later see if it makes sense with what we expected. Okay? We're just going to plug it in and we’re done. \(x\) center of mass is going to be, the masses are \(10\). So I’m going to do this, \(8\) and \(6\) divided by \(10 + 8 + 6\).

And then the \(y\) center of mass is the same thing. \(10, 8, and 6\) divided by \(10 + 6\). The only difference is that here I’m going to add x values and then the other one I’m going to add y values. Okay. So, what's the x-value of the \(10\)? Just look right here. It’s \(0\). of the \(8\) is \(0\). And of the \(6\) is \(4\). What about the \(y\)-value? The \(y\)-value of the \(a\) is \(0\), of \(b\) is \(3\), and of \(c\) is \(0\) as well. So when you do this, when you do this, you get \(6 \times 4, 24\) divided by \(24\). Okay. So, it’s a coincidence that the numbers happen to divide so neatly, \(1\) meter. And then here, this cancels and this cancels. I get \(8 \times b, 24\) as well. Again, a coincidence that this happens to be the same as that divided by \(24\), \(1\) meter. So, I actually, sort of my rough estimate would happen to be dead on. But usually, you just know that it's roughly a number. It's \(11\). So you can say that the center of mass of the system is at position \(1, 1\) meter. Okay. That’s the final answer. Alright, that's it for the center of mass. Let me know if you have any questions.