Center of mass equation | Impacts and linear momentum | Physics | Khan Academy

So let's say you wanted to know where the center of mass was between this 2-kilogram mass and the 6-kilogram mass. Now, they're separated by 10 centimeters, so it's somewhere in between them, and we know it's going to be closer to the larger mass because the center of mass is always closer to the larger mass.

But exactly where is it going to be? We'd need a formula to figure this out. The formula for the center of mass looks like this: it says the location of the center of mass (that's what this is). This x_cm is just the location of the center of mass; it's the position of the center of mass. It's going to equal...

You take all the masses that you're trying to find the center of mass between; you take all those masses times their positions, and you add up all of these m times x's until you've accounted for every single m times x there is in your system. Then, you just divide by all of the masses added together. What you get out of this is the location of the center of mass.

So let's use this. Let's use this for this example problem right here and see what we get. We'll have that the center of mass, the position of the center of mass, is going to be equal to—alright, so I'll take m1, which you could take either one as m1, but I already colored this one red, so we'll just say the 2-kilogram mass is m1. We're going to have to multiply by x1, the position of mass one.

At this point, you might be confused. You might be like, "The position? I don't know what the position is; there's no coordinate system up here." Well, you get to pick. So you get to decide where you're measuring these positions from. Wherever you decide to measure them from will also be the point where the center of mass is measured from. In other words, you get to choose where x equals zero.

Let's just say, for the sake of argument, the left-hand side over here is x equals zero. Let's say right here is x equals zero on our number line, and then it goes this way—it's positive this way. So if this is x equals zero, halfway would be x equals five, and then over here would be x equals ten. We're free to choose that.

In fact, it's kind of cool because if this is x equals zero, the position of mass one is 0 meters, so it's going to be this term is just going to go away, which is okay. We're going to have to add to that m2, which is 6 kilograms, times the position of m2. Again, we can choose whatever point we want, but we have to be consistent. We already chose this as x equals zero for mass one, so that still has to be x equals zero for mass two. That means this has to be ten centimeters now.

And those are our only two masses, so we stop there and we just divide by all the masses added together, which is going to be 2 kilograms for m1 plus 6 kilograms for m2. What we get out of this is 2 times 0, which is 0, plus 6 times 10 is 60 kilograms centimeters, divided by two plus six, which is going to be eight kilograms. This gives us 7.5 centimeters.

So it's going to be 7.5 centimeters from the point we called x equals zero, which is right here. That's the location of the center of mass. In other words, if you connected these two spheres by a rod—a light rod—and you put a pivot right here, they would balance at that point right there.

And just to show you, you might be like, "Wait, we can choose any point as x equals zero; won't we get a different number?" You will. So let's say you did this. Instead of picking that as x equals zero, let's say we pick this side as x equals zero.

Let's say we say x equals zero is this 6-kilogram mass's position. What are we going to get then? We'll get the location of the center of mass for this calculation. It's going to be... well, we'll have 2 kilograms, but now the location of the 2-kilogram mass is not zero. It's going to be—if this is 0, and we're considering this way as positive—it's going to be negative 10 centimeters because it's 10 centimeters to the left.

So this is going to be negative 10 centimeters plus 6 kilograms times now the location of the 6-kilogram mass is zero using this convention. We divide by both of the masses added up, so that's still 2 kilograms plus 6 kilograms.

And what are we going to get? We're going to get 2 times negative 10 plus 6 times zero. Well, that's just zero, so there's going to be negative 20 kilograms centimeters divided by 8 kilograms, which gives us negative 2.5 centimeters.

So you might be worried. You might be like, "What? We got a different answer. The location can't change based on where we're measuring from." And it didn't change; it's still in the exact same position because now this negative 2.5 centimeters is measured relative to this x equals zero. So what's negative 2.5 centimeters from here? It's 2.5 centimeters to the left, which, lo and behold, is exactly at the same point since this was 7.5 and this is negative 2.5.

The whole thing is 10 centimeters, and it gives you the exact same location for the center of mass. It has to; it can't change based on whether you're calling this point zero or this point zero. But you have to be careful and consistent with your choice. Any choice will work, but you have to be consistent with it, and you have to know at the end where this answer is measured from.

Otherwise, you won't be able to interpret what this number means at the end. So recapping: you can use the center of mass formula to find the exact location of the center of mass between a system of objects. You add all the masses times their positions and divide by the total mass. The position can be measured relative to any point you call x equals zero, and the number you get out of that calculation will be the distance from x equals zero to the center of mass of that system.