Intro to Center of Mass: Videos & Practice Problems

The center of mass is the average position of a system of objects, allowing simplification by replacing multiple masses with a single equivalent mass. The equation for calculating the center of mass is (∑mi × ⁢xi) / ∑mi. This concept is crucial for understanding how mass distribution affects the overall behavior of a system, particularly in physics and engineering contexts.

Intro to Center of Mass

Guys, some professors are going to give you a system of objects, multiple objects, and they're going to ask you to calculate the center of mass. In this video, I'm going to introduce what the center of mass is and the equation that we use to calculate it. It's very straightforward. So let's go ahead and take a look here. 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 object. 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 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 of interesting things that happen. So let's take a look at our example here. We want to take these two objects. Right? So we have 10 and 10, these two masses, m₁ and m₂, and we'll replace this system here with a single equivalent object. 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 be the sum of all the little m's. 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 going to be 10 plus 10, and so that's just going to equal 20 kilograms. We want to replace this system here with a single 20-kilogram object. The problem here is I don't know where exactly that 20-kilogram object is going to be located along this number line, and that's what I want to figure out. These two objects here are at x equals 0 and x equals 4. I want to figure out where along this number line I'm going to replace these two objects with a single 20-kilogram 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 going to 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 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 just going to be basically right in the center of that object. If you have two objects, it's going to be a little bit more complicated, a little bit more interesting. So last, imagine we had these two blocks of mass m like this. The center of mass is going to be somewhere in between them. But if they are equal mass, the center of mass is actually going to 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 that's 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 m_, and then another block of 2 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 going to be skewed towards the heavier one because that's where more of the mass center is located in the system.

Alright. 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 going to 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 that it works is if you have m₁, you're just going to multiply m₁ times its position, m₁ x₁, and then m₂ times x₂, and then m₃ times x₃, and then so on and so forth, however many masses you have in the system, divided by the sum of all the masses. So m₁, m₂, m₃, and so on and so forth. Alright? So this is the center of mass equation. Let's go ahead and take a look at our examples and finish them off. So we know we want to calculate these, with the center of mass here because we want to figure out where we can put the single equivalent 20-kilogram block. So to do that, we're going to use the center of mass equation. So this is going to be the sum of all masses times the positions divided by the sum of all masses. There's only two here. We have m₁ x₁ plus m₂ x₂ divided by m₁ plus m₂. So we have the masses. Right? They're both just 10 and 10. So we're going to set this up the way we'd normally so the way we kinda settle up a momentum equation. We have the masses. So we're just going to 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 this parenthesis? Well, this is just going to be the position of this first 10-kilogram box, and that's actually x equals 0. So there's a 0 that goes here and it goes away. And then the x equals 4 is for the second, 10-kilogram mass. That's what goes inside this parenthesis here, and then we're done. There's only two masses. So this is going to be 40 divided by 20. So your center of mass is going to be 2 meters, and this should make some sense. If you have these two boxes, x equals 0 and 4, 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 except now we have a 10 and a 30-kilogram block 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 2, 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 these total mass of this object is just going to be 40 kilograms. How do we calculate the center of mass? Again, we just use the equation. Again, there's only two objects, so so I'm just going to use m₁ x₁ plus m₂ x₂ divided by m₁ plus m₂. So here I have 10 times something plus 30 times something divided by 10 +30. So what goes inside here? Well, remember that this 10-kilogram block is going to be at x equals 0. So I'm just going to plug in 0 there. And this 30-kilogram block is at x equals 4, so I plug in a 4 here. What you end up getting here is you end up getting a 120 divided by 40 and you're going to get an x_{center of mass} position to be 3 meters. So if we look at our number line, it turns out we were right. This is x equals 2, but this is not the center of mass because the heavier one sort of skews the center of mass towards the right. So this is actually your center of mass right here at x equals 3. 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. Alright, guys. So that's it for this one. Let me know if you have any questions.

Center of Mass on Y Axis

Alright, everyone. Hopefully, you 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-axis. 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 going to do is I'm going to draw a y-axis. That's my y-axis, and this is the x-axis like this. And, basically, I'm going to draw out where each of these things are. So, the first one is a 1-kilogram block or weight at 2 meters positive. So I'm going to say this is the plus y-axis. So I'm just going to draw this out. It doesn't have to be perfect. So this is my 1 kilogram, and my y-value is 2. The next one is a 1.5-kilogram weight at the origin. So, in other words, at the origin here, we have 1.5 kilograms. And finally, we have a 7.5-kilogram weight at negative 1.5 meters. So, in other words, if this is negative 1, this is negative 2, it's going to be somewhere in the middle here, and I'm going to draw a really, really big dot. Alright. So this is going to be 7.5 kilograms. Alright. So, if you could collapse all of these into a single object, where would the center of mass be? Before we even start calculating anything, remember that the center of mass tends to skew 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 be very close to it, right, rather than the other 2. So how do we calculate this? Remember, we just have an equation for x as the center of mass, where you just do m1 x1, m2 x2, as many objects as you have. The only thing that's different is that we're just doing it on the y-axis now. So, we have ycm for our y-axis center of mass. We're going to do m1 y1, m2 y2, and m3 y3. And then, just divide by the total mass. So m1, m2, m3. Alright? So, basically, this m1 here, we're going to have to just assign which one is m1, which one is m2, which one's m3. It really doesn't matter the order in which you do it. So this one could be m1 or it could be m3. It doesn't matter. You'll still get the right answer. I'm just going to call this m1. What does matter is that you keep the positions consistent. So if this is m1, this has to be y1. If this is your m2, this is going to be your y2, which is 0, and then your m3 and then your y3 is going to be negative 1.5. Alright. That has to be consistent. So just plugging this stuff in, we usually have 1 times 2 plus, and then we have 1.5. And if it's at the origin, then the y-value is just 0, and that whole term goes away. And then we have 7.5 times −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 1 + 1.5 + 7.5. Alright. So when you work this out, what you're going to get is you're going to get, let's see, 2 plus and this is going to be −11.25 divided by a grand total of 10. And when you work this out, what you're going to get is a negative 0.93 meters. 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 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 going to be skewed towards the higher mass on the bottom.

Alright, folks. So that's it for this one.

Coordinates of Center of Mass of 4 objects

Hey, everyone. So let's work this problem out together. We're going to 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 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 squares. 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 it were located at that point? Is it going to be over here? Is it going to 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 object. So probably it's going to be somewhere over here. And basically, what you need to do is figure out what is the x coordinate and then what are the y coordinates. So, in other 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 going to 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 2 dimensional sort of problem, you can always break it down into 1 dimensional problems. So basically, we just need to figure out the x center of mass and that's going to be \( m_a x_a + m_b x_b + m_c x_c + m_d x_d \), divided by the total mass. Right? This is the big M. Basically, remember this \( m_1 + m_2 \), 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, what you're going to get is 0.25 plus 0.4 plus 0.4 plus 0.6, just all the masses added up together, and you're going to get 1.65 kilograms. 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 going to be 0,0 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 \(0, 8\). This is going to be \(8, 8\). This is going to be \(8, 0\). 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, the x coordinate for a is 0. So what about the mass for b? It's 0.4. What about the x coordinate for b? It's still 0, 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 8, and this is going to be 0.4 times 8 again. Alright. So the only 2 that really are going to be non-zero are going to be these 2. These are the only 2 that actually have an x coordinate, and that makes sense. Now we're just going to divide by 1.65. What you should get here is an x coordinate of 4.85 centimeters. Alright. So that is your x coordinate, And that kind of makes some sense here. If this side length here is 8 centimeters, right, because it's over here, then the center of mass is going to be not halfway. Now this would be like 4. So this would be 4. It's actually going to 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 going to be a 4. It's actually going to be a little bit more than that. Okay. What about the y center of mass? We actually just do the exact same process, but we're just going to do it with y's instead. So \( m_a y_a + m_b y_b + m_c y_c + m_d y_d \), 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 0 because it's still at the origin. What about the y coordinate for B? It's going to be 8, right? That's that corner, plus the y coordinate for C, so that's going to be 8 as well. And then finally for d, it's going to be just 0. Right? That's the y coordinate for 0. 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 going to get 4.85 centimeters again. Now this isn't sort of a this isn't just a coincidence. It's actually just because, if you think about this, these two blocks over here, these 4 these 0.4 kilogram 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 2. 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 this 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.

Here’s what students ask on this topic:

What is the center of mass and why is it important in physics?

The center of mass is the average position of all the mass in a system of objects. It allows us to simplify complex systems by replacing multiple masses with a single equivalent mass located at the center of mass. This simplification is crucial in physics and engineering because it makes analyzing the motion and behavior of systems more manageable. The center of mass helps in understanding how mass distribution affects the overall dynamics, such as in rotational motion, stability, and collisions.

How do you calculate the center of mass for a system of objects?

The center of mass for a system of objects is calculated using the equation:

$\frac{\sum m_i\times x_i}{\sum m_i}$

where $m_i$ is the mass of the i-th object and $x_i$ is its position. You sum the products of each mass and its position, then divide by the total mass of the system. This gives the average position of the mass in the system.

How does the center of mass change with unequal masses?

When dealing with unequal masses, the center of mass is not located at the midpoint between the objects. Instead, it is skewed towards the heavier object. This is because the heavier object contributes more to the total mass, pulling the center of mass closer to its position. For example, if you have a 10 kg mass at x = 0 and a 30 kg mass at x = 4, the center of mass will be closer to the 30 kg mass, specifically at x = 3.

What is the significance of the center of mass in rotational motion?

The center of mass is crucial in rotational motion because it is the point around which an object or system naturally rotates. Understanding the center of mass helps in predicting the behavior of rotating systems, such as determining the stability and balance of objects. In engineering, knowing the center of mass is essential for designing stable structures and machinery that can rotate or move efficiently without tipping over.

Can the center of mass be outside the physical boundaries of an object?

Yes, the center of mass can be outside the physical boundaries of an object. This occurs in cases where the mass distribution is uneven or the object has an irregular shape. For example, a ring or a hollow sphere has its center of mass at the geometric center, which is a point in space not occupied by the material of the object. This concept is important in understanding the behavior of such objects in motion.