1

concept

## Intro to Center of Mass

6m

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Hey guys, so some problems going to give you a system of objects, multiple objects, and they're gonna ask you to calculate the center of mass. So in this video I'm going to introduce you what the center of mass is and the equation that we used to calculate it. It's very straightforward. So let's go ahead and take a look here. So the idea here is that sometimes it's useful to simplify a group or a system of objects by replacing them with a single equivalent objects. So the idea here is that a lot of times in problems, instead of having to deal with lots of small little masses and having your equations to be really long and sort of tedious, you can actually replace it or simplify your problem by just considering a single object. There's a couple things, interesting things that happen. So let's take a look at our example here. So we want to do is we want to take these two objects, Right? So we have 10 and 10. These two masses M. One, I'll call them M. One and M. Two. And want to replace this system here with a single equivalent objects. So how do we do that? Well, it turns out that when you combine objects, the mass of the combined object is just the sum of all the masses of the individual objects that make it up. It's just going to the sum of all the little Ems. So really what we want to do here is that we want to replace this system of objects with a single M. And that's just gonna be 10 plus 10. And so that's just gonna equal 20 kg. So, we want to replace this system here with a single 20 kg object. The problem here is, I don't know where exactly that 20 kg object is going to be located along this number line. And that's what I want to figure out. So these two objects here are X equals zero and X equals four. I want to figure out where along this number line, I'm going to replace these two objects with a single 20 kg object and that's where the center of mass comes in. So the mass is the sum of all the masses of the objects, but the center of mass is going to be the average position of all of the objects in the system. It's basically tell you where the location of that single equivalent object is going to be. So let's go ahead and take a look at some very quick examples here. So, a very quick conceptual examples. So if you have a single object, the center of mass is pretty easy to find a single object. The center of mass is going to be basically right in the center of that object. If you have two objects, it's gonna be a little bit more complicated. A little bit more interesting. So that imagine we have these two blocks of mass. M like this, the center of mass is going to be somewhere between them. But if they are equal mass, the center of mass is actually gonna be directly in between them. The center of mass for two equivalent or equal mass objects is going to be directly in between them. That's where the center of all of the mass is concentrated in the system is when you have two objects of unequal mass, it's actually a little bit even more complicated. So now, let's say you have a block of em and then another block of two M. Now, what happens is that the center of mass is not actually going to be directly in between them. The center of mass is actually gonna be skewed towards the heavier one because that's where more of the mass is central is located in the system. All right. So how do we actually calculate the center of mass? That actually brings me to the equation and I'll just give it to you. The center of mass here is gonna be the sum of all the masses times the positions of all the objects in the system, divided by the sums of all the masses. So basically the way it works is if you have M. One, you're just gonna multiply M one times its position and one X one and then M two times X two and then M three times X three and then so on and so forth. However many masses you have in the system divided by the sum of all the masses, so M one, M two M three and so on and so forth. All right, so this is the center of mass equation. Let's go ahead and take a look at our examples and finish them off. So, you know, we want to calculate the uh the center of mass here because we want to figure out where we can put the single equivalent 20 kg block. So to do that, we're gonna use the center of mass equation. So this is gonna be the sum of all masses times the positions divided by the simple masses. There's only two here, we have M one X one plus M two X two divided by M one plus M two. So we have the masses, right? They're both just 10 and 10. So we're going to set this up the way we would normally. So we we kind of set up a momentum equation. We have the masses. So we're just gonna plug in those masses and then we have to figure out what goes inside the parentheses here. So 10 plus 10 on the bottom. So what goes inside these parentheses? Well, this is just gonna be the position of this 1st 10 kg box and that's actually X equals zero, so there's zero that goes here and it goes away and then the X equals four is for the 2nd 10 kg mass. That's what goes inside this parentheses here and then we're done, there's only two masses. So it's gonna be 40 divided by 20. So your center of mass is going to be two m. And this should make some sense if you have these two boxes, X equals zero and four. The center of mass because they are equal mass is actually just going to be directly in between them. So your center of mass is actually going to be right over here and that's actually exactly what we said for the situation right here. If they're equal mass, the center of mass is directly in between them. Let's take a look at our other example. It's the same setup. Now, we have a 10 and a 30 kg box here. So just before we start, where do you think the center of mass is going to be? Do you think it's going to be at X equals two? Or do you think it's going to be to the left or right? Well, hopefully you guys realize that this object is heavier and so the center of mass is going to be somewhere over here. Let's go ahead and check this out. So the total mass of this object is just gonna be 40 kg. How do we calculate the center of mass? Again? We just use the equation again, there's only two objects. So I'm just gonna use em one X one plus M two X two divided by M one plus M two. So here I have 10 times something plus 30 times something divided by 10 plus 30. So what goes inside here will remember that this 10 kg block is gonna be at X equals zero. So we're just gonna plug in zero there and this 30 kg block is at X equals four. So I plug in a ford here. What you end up getting here is, you know, beginning 120 divided by 40 and you're gonna get an X center of mass position to be three m. So, if we look at our number line turns out we were right, this is X equals two, but this is not the center of mass because the heavier one sort of skews the center of mass towards towards the rights. So this is actually your center of mass right here at X equals three. So the center of mass is not necessarily in the middle of objects. As we've said before. If you have different masses, it's usually going to be closer towards the heavier objects. All right, guys, So that's it for this one. Let me know if you have any questions.

2

example

## Center of Mass on Y Axis

3m

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Alright, everyone hopefully got a chance to solve this one. We have these three weights and we're given the coordinates of where they are on the Y access and we want to find the center of mass of these three weights. Remember the center of mass is just if you could collapse all of these into a single object, where would that be located? So what I'm gonna do is I'm going to draw a Y axis, that's my Y axis and this is the X. Like this. And basically I'm gonna draw out where each of these things are. So the first one is a one kg block er waits at two m positive. So I'm gonna say this is the plus Y axis. So I'm just gonna draw this out. It doesn't have to be perfect. So this is going to be um my one kg And my wife value is two. The next one is a 1.5 kg at the origin. So in other words at the origin here, we have 1.5 kg and finally we have a 7.5 kg at negative 1.5 m. In other words, if this is negative one, this is negative two, it's gonna be somewhere in the middle here and I'm gonna draw a really, really, really big dot. Alright, so this is gonna be 7.5 kg. Alright, so if you could collapse all of these into a single object, where would the center of mass B before we even start calculating anything. Remember the center of mass tends to sort of skewed towards the higher mass. So in other words, this one here which has a higher mass than anything else, is probably going to have the center of mass to be very close to it. Right, rather than the other two. So how do we calculate this? Remember we just have an equation with this for X. The center of mass. Where you just do em one X one, M two, X two. As many objects as you have. The only thing that's different is that we're just doing in the Y axis now. So we have Y. X. For our Y axis, center of mass. We're going to do M. One, Y. One, M. two, Y 2 & M. three, Y 3 And then divide by the total mass. So m. one M. Two, M. three. All right. So basically this M. One here, we're gonna have to just assign which one is M. One which is M. Two. Which one's M. Three? It really doesn't matter the order in which you do it. So, this one could be M. One or it could be M. Three? It doesn't matter. You still get the right answer. I'm just gonna call this M. One. What does matter is that you keep the positions consistent? So, if this is M. One, this has to be Y one. If this is your M. Two, this is going to be your Y. Two which is zero And then your M. three And then your Y three is going to be negative 1.5. All right. So that has to be consistent. So just plugging this stuff in, we usually have one times two plus and then we have 1.5 and if it's the origin then the Y. Apple Y. Value is just zero and that whole term goes away and then we have 7.5 times negative 1.5. The negative does matter in this case because that's the position. Right? So now we divide by the total mass which is one plus 1.5 plus 7.5. All right. So when you work this out, what you're gonna get is you're gonna get let's see two plus. And this is gonna be negative 11.25 divided by a grand total of 10. And when you work this out, what you're gonna get is a negative 0.93 m. That is your final answer. So, that's where the center of mass is located. And if you were to draw this out, this would be somewhere around here, somewhere like right under right above the negative one mark. This is where your center of mass is. And that makes some sense. Again, the center of mass should be sort of in the middle of all of them, but it's gonna be skewed towards the higher mass on the bottom. Alright folks, so that's it for this one

3

example

## Coordinates of Center of Mass of 4 objects

5m

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Hey everyone. So let's work this problem out together, we're gonna calculate the X and y coordinates of the center of mass of this system of objects here in centimeters. So here's the idea up until now, all of the problems that we've looked at have had objects on either the X or the y axis. But in this problem it's sort of arranged in a square. So we have both X and y. It turns out that we can still use the same exact center of mass equations for the X and the Y axis. All we really are doing here is we're just trying to figure out if you take this sort of system of objects, where's the one center where you could basically just replace all of the mass as if we were located at that point, is it gonna be over here is gonna be over here or over here or over here or something like that? Well, a good guess is it's going to be that it's going to be closer towards the heavier objects. So probably it's going to be somewhere over here. And basically, what you need to do is figure out what is the X coordinates. And then what is the Y coordinate, some of the words, the X center of mass and the y center of mass. So really, that's all we're doing here. And this is basically that's that's gonna be the location of that center of mass and to do that. We just need to go ahead and use our X and Y equations remember with a two dimensional sort of problem, you can always break it down into one dimensional problems. So basically we just need to figure out the extent of mass and that's going to be M A X A plus MB XB plus EMC XC plus M D X D. Divided by the total mass. Right? This is this is the big M basically remember this M one plus M two. All of this really is just big M. So just to make the math a little bit simpler, we can actually just calculate that right then. So we don't have to plug in all these numbers. If you work this out you're gonna get is 0.25 plus 0.4 plus 0. plus 0.6. Just all the masses added up together. And you're gonna get 1.65 kg. So in other words, the X coordinate here is really just going to be 0.25. And then what's the X coordinate for A Well actually the X coordinate is just gonna be 000 because it's at the origin. And in fact I'm going to go ahead and just write the coordinates for each one of the points. So this is going to be 008. Um This is going to be 8208. This is gonna be eight comma zero. Remember this is the X coordinate and that's the y coordinate for each one of the respective points. So then to get back to this, uh the X coordinate for A. Is zero. So, what about the mass for B. It's 0.4. What about the X coordinate for B. It's still zero, right? Because it's still basically just along the Y axis. It has no X coordinates. So that goes away as well. What about this one? This is going to be 0.6 times this is going to be eight and this is gonna be 0.4 times eight again. All right, so the only two that really are going to be non zero are gonna be these two. These are the only two that actually have an X coordinate. That makes sense. Now, we're just going to divide by 1 61.65. What you should get here is an X coordinate of 4. cm. All right. So that is your X coordinate. And not going to make some sense here. If this side length here is 8cm right? Because it's over here, then the center of mass is going to be not halfway. Now, this would be like four. So, this would be four. It's actually gonna be a little bit beyond that because remember this big mass over here at that corner is sort of pulling or skewing all of the center of mass towards it. So it's not gonna be before it's actually gonna be a little bit more than that. Okay, so what about the Y center of mass? We actually just do the exact same process, but we're just going to do it with Wise instead. So M. A. Y. A plus M B, Y B plus M C Y C plus M. D. Y. D. All divided by 1.65. So in other words this is going to be 0.25. And then what's the Y coordinate for A the Y coordinate is zero because it's still at the origin. What about the Y coordinate for B. It's gonna be eight. Right, That's that corner plus the Y coordinate for C. So that's gonna be eight as well. And then finally four D. It's going to be just zero. Right? That's the white corner for zero. So then if you work this out, what happens is these two terms will cancel out, divided by 1.65. And when you work this out you're gonna get 4.85 cm again. Now this isn't sort of this isn't just a coincidence, it's actually just because if you think about this, these two blocks over here, these four the 0.4 kg blocks are sort of symmetrical about this axis of the square. So really what happens is the center of mass is really just determined by these two blocks over here, it's these two. So what happens here is that this is the smaller one and this is the larger one. So the center of mass sort of gets shifted towards it, and it's going to end up over here. So, because of the symmetry over here, that's why you ended up with the same coordinates for the X and Y axis. That's it for this one. Guys let me know if you have any questions.