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Lesson Video: Center of Gravity of Laminas Mathematics

In this video, we will learn how to find the position of the center of gravity of a lamina composed of standard shapes in 2D.

17:01

Video Transcript

In this video, we’re talking about the center of mass of laminas. As we’ll see, a lamina is a two-dimensional surface, and by finding the center of mass of such surfaces, we’re able to determine important physical quantities. As we get started, let’s look at a few examples of laminas. As we said, these are two-dimensional surfaces. A lamina could be a circle or a triangle or a hexagon or really any arbitrary shape where its important quality is that it exists in two dimensions in a plane, but not in three. In other words, a lamina is not a solid object.

Say that we have a lamina, this rectangle here, at rest on a surface. In this video, we’re interested in understanding the center of mass of such laminas. We can think of a two-dimensional shape’s center of mass as the average position of mass in that object. For a given shape, the center of mass is the point at which we could balance that mass in a horizontal plane on the end of some small surface, say, the end of a pencil. Physically, that’s what center of mass means. And this is related to a term we sometimes hear called center of gravity. When an object exists in a uniform gravitational field, then these two points, the center of mass and the center of gravity, are the same. In this video, we’ll assume that they are, and we’ll use the phrase center of mass to refer to this point.

But back to our rectangle, how do we actually find the center of mass of this lamina? One key fact we can know, and which we’ll assume to be true all throughout this lesson, is that the mass of a given lamina is evenly distributed throughout the whole surface. This means if we find the geometric center of a given shape, that matches the center of mass of that lamina. And we see we can find the geometric center of this rectangle by considering it in two dimensions. First, there’s the vertical dimension, which has a midpoint right about here, and then the horizontal dimension has a midpoint here. If we create dashed lines coming from these midpoints and perpendicular to the edges, then their point of intersection is our shape’s overall center of mass. This is where, on average, the mass of this rectangle is concentrated.

Typically, the center of mass of a given lamina is reported in reference to a certain set of coordinates. In other words, it’s not uncommon to calculate an object’s center of mass in reference to a certain 𝑥- and 𝑦-plane. This way, if our rectangle had a height ℎ and a width 𝑤, then the coordinates of its center of mass, we can call these 𝐶𝑂𝑀 𝑥 and 𝐶𝑂𝑀 𝑦, would be 𝑤 over two and ℎ over two. This approach of finding the geometric center of a shape to locate its center of mass is a good first step to apply when we can. Often, this approach works well when our lamina is in the shape of a rectangle or a circle.

However, for less-regular shapes, it’s not always quite so easy to find their geometric center. For example, say that we’re given this triangle here and we want to calculate its center of mass. Recalling that for any lamina, this is the average position of mass in that object, we might guess that this point would be about here. But in terms of actually calculating these coordinates, how might we do that? Well, it turns out that for any triangle, if we know the coordinates of its three vertices — here we’ve called these vertices 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two; and 𝑥 three, 𝑦 three — then we can use this information to solve for the center of mass coordinates.

The center of mass 𝑥-coordinate is equal to the sum of 𝑥 one, 𝑥 two, and 𝑥 three all divided by three and similarly for the center of mass 𝑦-coordinate. This approach is very helpful because it applies to all triangles. Knowing this, let’s consider the center of mass for another fairly common shape. This shape is the semicircle. Recalling that center of mass is equivalent to the geometric center of a shape, we might think that if we know the diameter of the semicircle, then the center of mass must lie somewhere along the midpoint of that distance. And this is true. Our intuition is correct. But then we need to figure out where along this line the center of mass lies.

It turns out that if we know the radius of our semicircle, then that can help us solve for the center of mass coordinate, in this case, in the vertical or 𝑦 direction. It’s equal to four times the semicircle’s radius divided by three 𝜋. So, since the radius of this semicircle is four units, we can say that the coordinates of its center of mass are four and four times four over three 𝜋 or four and sixteen-thirds over 𝜋.

Now that we’ve considered the centers of mass of individual laminas we might encounter, let’s clear a bit of space and consider how we might calculate the center of mass not of one lamina, but of two together. In this case, we see our two laminas are both rectangles. For each one individually then, we know how to find the geometric center and therefore its center of mass. But to report the center of mass of this combination of shapes, we’ll somehow need to combine these two centers of mass. Let’s imagine that we know not only the center of mass coordinates for each of these two rectangles, but we also know the area of each shape. We’ve called them 𝐴 one and 𝐴 two. If we have all this information, then we can enter a process of determining the center of mass of multiple laminas.

The general formulas for the 𝑥- and 𝑦-coordinates of that center of mass are given here. The idea is that we have some collection of laminas, and each of those shapes has an area we’ve called 𝐴 one, 𝐴 two, and so on up to 𝐴 𝑛. For those individual areas, each one has a center of mass 𝑥- and 𝑦-coordinate. We’ve called the center of mass 𝑥-coordinate for area one 𝑥 one, that for area two 𝑥 two, and so on up to 𝑥 𝑛. And the same thing holds true for the center of mass 𝑦-coordinate. So, however many shapes we have, we multiply each area by its respective center of mass coordinate, 𝑥 or 𝑦, and then add up all of those products for as many shapes as we have. We then divide this value by the sum of all the areas.

That process overall will give us the center of mass 𝑥- and 𝑦-coordinates for a collection of laminas. So, in the case of our two rectangles here, we would say that the center of mass 𝑥-coordinate for these two rectangles combined is given by 𝐴 one times 𝑥 one plus 𝐴 two times 𝑥 two divided by the sum of the areas and similarly for the combined center of mass 𝑦-coordinate. In this way, we could stack as many laminas as we want together. And so long as we knew the center of mass and area for each lamina individually, we could find the combined center of mass using this technique. Knowing all this about laminas and their center of mass, let’s look now at an example exercise.

In the figure shown, find the position of the center of mass of the uniform triangular lamina 𝐴𝐵𝐶, considering 𝐴 to be the origin point.

In this figure, we see the triangle with vertices 𝐴, 𝐵, and 𝐶 positioned on this 𝑥𝑦-coordinate plane. We want to find the center of mass of this triangular lamina, and we knew that that corresponds with the centroid or geometric center of the shape. Just estimating by eye, we might put the geometric center of this triangle here. But to know the 𝑥- and 𝑦-coordinates of this point accurately, we’ll call them 𝐶𝑂𝑀 𝑥 and 𝐶𝑂𝑀 𝑦, we’ll need to recall a more precise approach. Whenever we’re working with triangles seeking to find their center of mass, the key information for doing this is the coordinates of the triangle’s three vertices.

If we know these, then regardless of the shape of the triangle, we can calculate the 𝑥- and 𝑦-coordinates of its center of mass using these relationships. Basically, they involve solving for the average 𝑥-coordinate and the average 𝑦-coordinate among the vertices. If we apply these relationships to our scenario with triangle 𝐴𝐵𝐶, then we can note that the coordinates at the vertex 𝐵 are zero, five 𝑎; those at vertex 𝐶 are four 𝑎, zero. And because vertex 𝐴 is positioned at the origin, those coordinates are zero, zero.

To solve then first for the center of mass 𝑥-coordinate, we’ll add together zero, zero, and four 𝑎 and divide all that by three, which gives us four 𝑎 over three. And then to solve for the center of mass 𝑦-coordinate, we’ll add together five 𝑎, zero, and zero, the 𝑦-coordinates of the three vertices of our triangle, and divide all that by three, giving us five-thirds 𝑎. These, then, are the coordinates of the center of mass of this uniform triangular lamina.

Let’s now look at an example where we calculate center of mass for a combination of laminas.

A uniform square lamina 𝐴𝐵𝐶𝐷 has a side length 𝑙. Another uniform lamina 𝐵𝐶𝐸 of the same density, shaped as an isosceles triangle, is attached to the square such that 𝐸 lies outside the square and 𝐵𝐸 is equal to 𝐶𝐸. Given that the square’s side length is five-thirds times the length of the triangle’s height, find the center of mass of the system.

Taking a look at our diagram, we see our square 𝐴𝐵𝐶𝐷 and the isosceles triangle attached to one side of the square. In this exercise, we want to find the center of mass of the square and triangle combined. To help us do that, we’re told that the square has a side length 𝑙 and that the length of the triangle’s height, this distance here, is related to the square side length by this relationship: five-thirds times ℎ equals 𝑙. Having written this down, let’s clear some space on screen and begin working on our solution of what is the center of mass 𝑥- and 𝑦-coordinate for these combined shapes.

For our overall strategy, we can begin by solving for the center of mass of the shapes individually, the square and the triangle by themselves. Once we’ve done that, we can use the fact that a two-lamina center of mass, that is, the center of mass of two individual laminas combined together, is given by these relationships for the 𝑥- and 𝑦-coordinates of that center of mass. Here, 𝐴 one and 𝐴 two are the respective areas of our two laminas involved. 𝑥 one and 𝑥 two are the center of mass 𝑥-coordinates of these respective shapes. And 𝑦 one and 𝑦 two are their corresponding 𝑦 center of mass coordinates. In order to apply these relationships then, we’ll need to solve for the centers of mass of our square and triangle as well as their areas.

Let’s start by considering our square. And if we consider the fact that this square’s geometric center is located at its center of mass, we know that that is at a point here with both 𝑥- and 𝑦-coordinates of 𝑙 over two. So let’s write those down as 𝐶𝑂𝑀 𝑥 square and 𝐶𝑂𝑀 𝑦 square coordinates. And then let’s consider the area of our square. Since it has side length 𝑙, that area is simply 𝑙 squared. Now let’s move on to our triangle, calculating the center of mass 𝑥- and 𝑦-coordinates for it. We can recall that for such a shape, the center of mass 𝑥- and 𝑦-coordinates are equal to the average 𝑥- and 𝑦-coordinates, respectively, of the triangle’s vertices. Considering the coordinates at vertex 𝐵, we can see that these are 𝑙, 𝑙. At vertex 𝐶, they are 𝑙 and zero.

But then what about at vertex 𝐸? We were told in our problem statement that this is an isosceles triangle. This means that the 𝑦- coordinate of point 𝐸 is midway between that of points 𝐶 and 𝐵. This tells us that it’s 𝑙 over two. To solve for the 𝑥-coordinate though, we’ll need to add the height ℎ to the length 𝑙 that we already have as the side length of our square. Considering this relationship between ℎ and 𝑙, if we multiply both sides of the equation by three divided by five, then we find that ℎ, the height of this isosceles triangle, is three-fifths 𝑙, which means that the 𝑥-coordinate of vertex 𝐸 will be 𝑙 plus three-fifths 𝑙. Combining these gives us eight 𝑙 divided by five.

Now that we know the coordinates of each of the three vertices of our triangle, we can use the fact that the center of mass 𝑥-coordinate is equal to the average of these three 𝑥-values and the center of mass 𝑦-coordinate is equal to the average of these three 𝑦-values. And note that this is true for any triangle whether isosceles or not. Beginning with our 𝑥-coordinate, this is equal to 𝑙 plus 𝑙 plus eight 𝑙 over five all divided by three. This is equal to 18𝑙 over five divided by three or 18 over 15𝑙, simplifying to six-fifths 𝑙. So, we’ll record this value as our center of mass 𝑥-coordinate for the triangle and then clear this away so we can start to work on the center of mass 𝑦-coordinate.

This is equal to the average 𝑦-coordinate of the three vertices. That’s 𝑙 plus zero plus 𝑙 over two divided by three, which is three-halves 𝑙 over three or simply 𝑙 over two. And we’ll then enter this value into our coordinate pair. The last thing we need to do before we can apply these two center of mass equations is calculate the area of our triangle. In general, triangle area is one-half base times height. Here we’ve said that our height is three-fifths times 𝑙. And with that orientation, we can see that our base here is 𝑙. The area then is one-half 𝑙 times three-fifths 𝑙 or three-tenths 𝑙 squared. We’re now completely prepared to apply these two relationships to solve for our overall center of mass.

The center of mass of our two laminas combined has an 𝑥-coordinate given by the area of the square multiplied by the center of mass 𝑥-coordinate of the square plus the area of the triangle times the center of mass 𝑥-coordinate of that shape all divided by the areas of these two shapes added together. When we plug in all these values, we can note that a factor of 𝑙 squared appears in both numerator and denominator in all terms. This means we can factor it out and then cancel it out, leaving us with this expression for our center of mass 𝑥-coordinate.

Note that in the numerator we can now factor out one power of 𝑙. And then if we add together one-half and three-tenths times six-fifths, that’s equal to 43 divided by 50. And then in our denominator, one plus three-tenths is equal to 13 divided by 10. If we then multiply top and bottom by 10 divided by 13, our denominator simplifies to one, and we have 𝑙 times forty-three fiftieths times ten thirteenths. This simplifies to forty-three sixty-fifths 𝑙. That’s the center of mass 𝑥-coordinate of our combined two-lamina shape. So we enter this value into that coordinate pair, and now we clear space to calculate our overall center of mass 𝑦-coordinate.

Similar to our 𝑥-coordinate, this equals the area of the square times the center of mass 𝑦-coordinate of that shape plus the area of the triangle times the center of mass 𝑦-coordinate of the triangle, and all this is divided by the sum of the areas of those two shapes. With those values substituted in, once again, we see that 𝑙 squared can be factored out of numerator and denominator. Just like before then, that factor cancels out, and we see that, once more, one power of 𝑙 can be factored out of the numerator.

When we go to add one-half to three-tenths times one-half, we get 13 over 20. And just like before, one plus three-tenths simplifies to thirteen-tenths. Multiplying numerator and denominator once more by 10 over 13, we see that our denominator simplifies to one while our numerator becomes 𝑙 times one-half. This then is the center of mass 𝑦-coordinate of our two laminas together, which means that our final answer is that the center of mass coordinates of these two laminas is forty-three sixty-fifths 𝑙 and 𝑙 over two.

Let’s finish up our lesson now by summarizing a few key points. In this video, we learned that a lamina is a two-dimensional shape. And for a uniform lamina, its center of mass is its geometric center, also called its centroid. When it comes to triangles, we saw that this geometric center, and therefore center of mass, can be found by solving for the average 𝑥- and 𝑦-values of its vertices. We learned further that for a semicircle positioned so its flat side is on the 𝑥-axis, if it has a radius of 𝑟, then the 𝑦-coordinate of its center of mass is four 𝑟 over three 𝜋.

And lastly, we saw that when we want to solve for the center of mass of a combination of laminas, we can do so by solving for the area of each lamina individually as well as each one’s center of mass coordinates and using that information in these two formulas for any number of laminas to calculate the center of mass 𝑥- and 𝑦-coordinates of the collection.

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