Lesson Video: The Negative Mass Method Mathematics

In this video, we will learn how to find the center of gravity of a lamina that contains holes using the negative mass method.

17:34

Video Transcript

Everyday objects all have positive mass or no mass at all. Nevertheless, we can conceive of an object with negative mass mathematically by simply using a negative number to represent its mass instead of a positive number. In this video, we will learn that objects with negative mass are actually a valuable mathematical tool for finding the center of mass of real objects, even though those real objects have positive mass.

Each coordinate of the center of mass of a collection of objects is given by this formula, here written for the coordinate π‘₯. To use this formula, we add up the mass of each object in the collection multiplied by the π‘₯-coordinate of its location and then divide by the total mass. This gives us the π‘₯-coordinate of the center of mass of the entire collection. For the point masses in our collection of objects, the mass is just the mass of the point mass and the coordinate is just the coordinate of its location. If any of our objects are extended, like, say, two-dimensional laminae, we use the total mass of the object as the mass. And for the coordinate location, we use the center of mass of that individual object.

In addition to using this formula to find the center of mass of a collection of disconnected objects, we can also use it to find the center of mass of a more complicated object by breaking that complicated object up into smaller, simpler objects. For example, this is one large lamina with an irregular shape. However, if we imagine separating the lamina along this dashed line, we see that it is also the combination of two regularly shaped laminae, the triangle shaded with blue and the rectangle shaded with magenta. We can then use known results for the center of mass of triangles and rectangles along with our center of mass formula to easily find the center of mass of the resulting shape.

We can even take this one step further. Instead of a lamina with an irregular shape, let’s consider a lamina with an irregular mass distribution. Here we have a large rectangle, and the mass distribution is uniform except for the square area crosshatched with magenta. In fact, let’s give this a quantitative value. Let’s say that the magenta crosshatched square has three times as much mass as it would have had if the entire lamina was uniform. To break this up into multiple laminae that are easier to deal with, we’ll start with a large rectangular lamina with a uniform mass distribution. We now need to add at least one more lamina to triple the density of this particular square portion of the uniform lamina, which corresponds to the crosshatched area we had before.

Since we’re trying to increase the mass of this square region, let’s affix another square lamina with the same size and orientation to the middle of that region. To figure out the proper mass for this second lamina, let’s call the mass of the original square region 𝑀. Since our end result needs to be three times as much mass as we would otherwise have in this region, let’s make the mass of our second square lamina two 𝑀. Now when we affix this uniform square with mass two 𝑀 to the uniform square with mass 𝑀, the total mass will be 𝑀 plus two 𝑀 or three 𝑀. Since three 𝑀 is three times 𝑀, we have succeeded in recreating our original nonuniform lamina from two uniform laminae, a rectangle and a square. We can then easily use the masses and centers of mass of these two laminae to plug in to our center of mass formula to find the overall center of mass of the nonuniform lamina.

We’re one step away from seeing why negative mass can greatly simplify certain types of calculations. Before we do this, though, let’s be very clear about why the method we just used succeeded. When we began splitting up our nonuniform lamina, we started with a uniform lamina that had a mass of 𝑀 in the particular region where the nonuniform lamina had mass three 𝑀. In other words, there was a discrepancy between the mass of that region in the uniform lamina and the final mass that we were aiming for. We were successful because we were able to find a mass, in this case, two 𝑀, that would make up that discrepancy between the uniform lamina and the nonuniform lamina. It is this idea, making up a discrepancy between the mass of a uniform lamina and a nonuniform lamina, that will lead us directly to objects with negative mass.

Let’s slightly adjust our nonuniform lamina. Instead of having a denser region, let’s have a hole. A hole has no mass. So instead of this region having three times as much mass as it would otherwise have, it has zero times as much mass. Even though we’re now thinking about less mass instead of more mass, let’s try to do what we did before, take a uniform rectangular lamina and use a small square lamina to correct the discrepancy in that particular region. The mass of the square region in the uniform lamina is still 𝑀, but now the total mass is zero. So we need to find a mass for the square lamina that will satisfy the condition that the total mass of the new region is zero. Well, any number plus its negative is zero.

So if we ignore for a moment the fact that masses are usually positive, if we allow the mass of the square lamina to be negative 𝑀, then the total mass of the resulting region is 𝑀 plus negative 𝑀. And 𝑀 plus negative 𝑀 is just zero. So we can represent this nonuniform lamina with a hole in it using two uniform laminae as long as we allow one of those laminae to have a negative mass. To use this in the center of mass formula, we simply plug in a negative value for the appropriate term in the numerator and the denominator. In fact, aside from the introduction of a few negative numbers, these calculations are identical to the calculations we would do in any other situation where we find center of mass. Let’s now see some examples of where objects with negative mass are useful. We’ll start with an example that’s similar to the one we just looked at, a rectangular lamina with a rectangular hole cut out of it.

Find the coordinates of the center of mass of the following figure, which is drawn on a grid of unit squares.

The shape that we’re interested in is the green shaded area on this graph here. Since the shape is nonuniform because it has a big hole in the middle, we’ll need to break it up into several uniform shapes and then use the center of mass formula. The center of mass formula allows us to calculate the coordinates of the center of mass for a collection of objects using the mass and location of each individual object. For extended objects like two-dimensional laminae, we define their location as their center of mass. Okay, now we just need to find the shapes that will make our calculation easiest. We can actually do this calculation using only two laminae if we use the negative mass method.

In the negative mass method, we treat laminae with holes as solid laminae with positive mass combined with a lamina the shape of the hole, but with negative mass. When we fix the lamina with negative mass to the lamina with positive mass, the positive and negative masses cancel, giving us the effective result of zero mass or a hole. We then find the center of mass of these two laminae, the positive mass and the negative mass, using the exact same center of mass formula. To apply the negative mass method to our particular shape, we’ll start with a positive mass lamina with the same perimeter as the outer perimeter of the shape that we’re interested in. To account for the hole, we’ll then include a lamina with negative mass with the same perimeter as the inner perimeter of our shape.

As we can see, the shape that we’re interested in is exactly the portion of the positive mass lamina that does not coincide with the negative mass lamina. Before we worry about particular values for these two masses, let’s find the centers of mass of these two laminae. Recall that for any uniform rectangular lamina, the center of mass is exactly at the point where the two diagonals meet. So to find the centers of mass of these two laminae, we’ll just draw the diagonals. Here, we’ve drawn the two diagonals of the positive mass lamina, and they meet right here. When we include the diagonals of the negative mass lamina, we see that they meet at exactly the same point. So both of these laminae have the same center of mass.

Whenever we combine two objects and those two objects have the same center of mass, the resulting combination also has that center of mass. We can see this mathematically by noting that if all of the π‘₯ sub 𝑖’s have the same value, call it π‘₯, then the numerator is just π‘₯ times the denominator. And π‘₯ times something divided by that something is just π‘₯. Okay, so the center of mass is this center point here. This is four and a half units along the π‘₯-axis and exactly four units up on the 𝑦-axis. Four and one-half is nine-halves. So the center of mass of this figure is located at nine-halves comma four. The ease with which we found this answer shows the power of the negative mass method. Not only were we able to find the centers of mass entirely graphically, we didn’t even need to plug in values into the center of mass formula to get the answer.

Let’s work through another example where we’ll perform a similar procedure but this time actually use the formula to calculate the center of mass.

The diagram shows a uniform lamina 𝐴𝐡𝐢 from which a triangle 𝐺𝐡𝐢 has been cut out. 𝐴𝐡𝐢 was an equilateral triangle with a side length of 93 centimeters and center of mass 𝐺. Find the coordinates of the new center of mass of the resulting lamina. Round your answer to two decimal places if necessary.

The statement refers to this arrow-shaped lamina formed from the equilateral triangle 𝐴𝐡𝐢 with the triangle 𝐺𝐡𝐢 cut out of it. The specific information we’re given is that 𝐴𝐡𝐢 has a uniform mass distribution, its side length is 93 centimeters, and 𝐺 is its center of mass. Since we’re looking for the center of mass of this resulting shape and we’re explicitly told that it was an equilateral triangle with 𝐺𝐡𝐢 cut out of it, this suggests that we should use the negative mass method. To use the negative mass method, we first start with this diagram here that shows the equilateral triangle 𝐴𝐡𝐢 shaded in orange and the triangle 𝐺𝐡𝐢 shaded over it in magenta.

Separating this out into its two pieces, we have the original lamina 𝐴𝐡𝐢 and the lamina 𝐡𝐺𝐢 that represents the part that was cut out. The uniform lamina 𝐴𝐡𝐢 is a regular object, so its mass is positive. However, since the combination of 𝐺𝐡𝐢 and 𝐴𝐡𝐢 results in a lamina with no mass at the bottom, where 𝐴𝐡𝐢 clearly does have mass, 𝐺𝐡𝐢 must have negative mass because negative plus positive is zero. We find the resulting center of mass the way we would find the center of mass of any two laminae. It’s just one of these laminae now has a negative number instead of a positive number for its mass. This is why we call it the negative mass method.

We now need to find the mass and the location of the center of mass of these two laminae. Since these laminae are uniform, we can express their total mass as their area times a constant, which is the mass per unit area of the material. For the lamina with positive mass, we need the area of the equilateral triangle 𝐴𝐡𝐢. From the diagram, we know the length of the base is 93 centimeters and the height is the length of the blue dashed line, which passes through the points 𝐴 and 𝐺. Because this dashed line is an altitude, it is perpendicular to the line 𝐡𝐢. Furthermore, because 𝐴𝐡𝐢 is equilateral, this angle here at point 𝐡 is 60 degrees. So the triangle formed by 𝐴𝐡 and the point where the altitude meets the side 𝐡𝐢 is a 30-60-90 right triangle.

Either directly using trigonometry or using facts about 30-60-90 right triangles, since the hypotenuse of this triangle is 93 centimeters, the height must be 93 times the square root of three divided by two. We’ll actually calculate a definite value for this later, but for now, let’s just call it π‘Ž. So the triangle 𝐴𝐡𝐢 has a base of 93 and a height of π‘Ž. So its area, one-half base times height, is one-half 93π‘Ž. If we call the constant uniform density π‘š, then one-half times 93 times π‘Ž is the area of 𝐴𝐡𝐢, and one-half 93π‘Ž times π‘š is the total mass of the lamina 𝐴𝐡𝐢.

What about 𝐺𝐡𝐢? Well, recall that 𝐺 is the center of mass of the equilateral triangle. For any uniform triangle, the center of mass is one-third of the way along any of the medians. In any collateral triangle, the median which connects any vertex to the midpoint of the opposite side is also the altitude of the triangle. So the distance from the midpoint of 𝐡𝐢 to the point 𝐺 is one-third of the total height of the triangle 𝐴𝐡𝐢. But we know what that value is; we called it π‘Ž. So the height of 𝐡𝐺𝐢 is one-third of π‘Ž. 𝐡𝐺𝐢 and 𝐴𝐡𝐢 have the same base. So the area of 𝐡𝐺𝐢 is one-half times 93 times one-third of π‘Ž. Now remember, 𝐡𝐺𝐢 needs to have negative mass but have the same kind of mass distribution as 𝐴𝐡𝐢. So the density of 𝐡𝐺𝐢 will be negative π‘š.

Let’s rearrange our expression for the mass of 𝐡𝐺𝐢 by factoring out the negative one and the factor of one-third. Written this way, we see that the mass of 𝐡𝐺𝐢 is negative one-third times the mass of 𝐴𝐡𝐢. The factor of negative one-third is present because 𝐴𝐡𝐢 has three times the area of 𝐡𝐺𝐢, and the mass of 𝐡𝐺𝐢 is negative. Just to make our life easier and clean up some space, let’s call one-half times 93 times π‘Ž times lowercase π‘š capital 𝑀 so that the mass of 𝐴𝐡𝐢 is capital 𝑀 and the mass of 𝐡𝐺𝐢 is negative one-third times capital 𝑀.

Now that we know the masses, let’s find the centers of mass of 𝐴𝐡𝐢 and 𝐡𝐺𝐢. We know that the center of mass of 𝐴𝐡𝐢 is located at the point 𝐺. And we’ve already determined the 𝑦-coordinate of point 𝐺; it’s one-third of π‘Ž. 𝐺 lies along the altitude of 𝐴𝐡𝐢. And we know that this altitude also being a median intersects the π‘₯-axis halfway between 𝐡 and 𝐢. Well, since the length of 𝐡𝐢 is 93, halfway between them is 46.5. So the coordinates of point 𝐺 are 46.5 comma one-third π‘Ž. For triangle 𝐡𝐺𝐢, note that 𝐡𝐺𝐢 is an isosceles triangle. We can see this because the dashed line is the perpendicular bisector of 𝐡𝐢. So any point on this bisector including the point 𝐺 is equidistant from point 𝐡 and point 𝐢. So the sides 𝐡𝐺 and 𝐺𝐢 are congruent, and therefore 𝐡𝐺𝐢 is isosceles.

In an isosceles triangle, the altitude drawn from the vertex in between the two congruent sides is also a median, just like in the equilateral triangle. But that means that the altitude of 𝐡𝐺𝐢, which is the dotted line that we’ve already drawn, is also a median. Since the center of mass is exactly one-third of the way along the median, the center of mass of 𝐡𝐺𝐢 is right around here. Although we’ve only estimated the location of this point in the figure, we can still find its location exactly. It’s on the same vertical line as 𝐺, so its π‘₯-coordinate is also 46.5. The 𝑦-coordinate is one-third of the height of 𝐡𝐺𝐢, and the height of 𝐡𝐺𝐢 is one-third of π‘Ž. So the 𝑦-coordinate is one-third of one-third or one-ninth of π‘Ž.

Now we have the mass and center of mass of 𝐡𝐺𝐢 and 𝐴𝐡𝐢. To use this information to find what we’re looking for, we can use the center of mass formula to find each coordinate of the resulting center of mass of the combination of 𝐴𝐡𝐢 and 𝐡𝐺𝐢. For each coordinate, we plug in the mass and corresponding coordinate of the center of mass for each of our laminae into the numerator and the mass of each lamina into the denominator. Let’s start with the 𝑦-coordinate. The mass of 𝐴𝐡𝐢 is 𝑀, and the 𝑦-coordinate of the center of mass is one-third of π‘Ž. So the first term of the sum of our numerator is 𝑀 times one-third π‘Ž. The second term in the numerator is negative one-third 𝑀 times one-ninth π‘Ž, which is the mass of 𝐡𝐺𝐢 times the 𝑦-coordinate of its center of mass.

The denominator is the sum of the two masses, 𝑀 plus negative one-third 𝑀. To start simplifying this fraction, let’s note that both terms in the numerator have an 𝑀, a one-third, and an π‘Ž. So we’ll factor one-third times 𝑀 times π‘Ž out of both of these terms. In the denominator, 𝑀 plus negative one-third 𝑀 is just two-thirds 𝑀. So we have one-third times 𝑀 times π‘Ž times one minus one-ninth all divided by two-thirds 𝑀. 𝑀 divided by 𝑀 is one, and one-third in the numerator divided by two-thirds in the denominator leaves just two in the denominator. In the numerator, one minus one-ninth is eight-ninths. So we have π‘Ž times eight-ninths divided by two.

Note that the actual value of the mass of 𝐴𝐡𝐢 does not factor into this calculation because the only thing that’s important is the relative mass of 𝐴𝐡𝐢 and 𝐡𝐺𝐢 not their absolute masses. Eight-ninths divided by two is four-ninths. So π‘Ž times eight-ninths divided by two is four-ninths times π‘Ž. Plugging back in our definition for π‘Ž, we get that the 𝑦-coordinate of the center of mass of our new lamina is four-ninths times 93 times the square root of three divided by two. If we plug this into a calculator, we get 35.7957 and several more decimal places. Since we’re asked to round to two decimal places, the third decimal place is a five. So we add one to the second decimal place, which is nine. One plus nine is 10, so 35.7957 et cetera rounds to 35.80.

To find the π‘₯-coordinate of the center of mass, instead of going through this whole calculation again, we note that the π‘₯-coordinate of the center of mass for 𝐴𝐡𝐢 and 𝐡𝐺𝐢 is the same. This means that when we combine them, we won’t change the π‘₯-coordinate of the center of mass. So the π‘₯-coordinate of the center of mass of our resulting lamina will also be 46.5. So rounding as specified, we find that the center of mass of our new lamina is at 46.5 comma 35.80.

Alright, now that we’ve seen a few examples, let’s review what we’ve learned about the negative mass method. In this video, we learned how regular laminae with holes or other shapes cut out of them can be treated as two or more uniform laminae with regular shapes as long as we allow the lamina representing the hole to have a negative mass. This ensures that when we combine it with the totally uniform lamina with positive mass, we get a hole that is zero mass, exactly where we’re meant to. Even though there are no negative mass objects in the real world, mathematically, negative mass values are just as valid as positive mass values. This means we can use them in things like our center of mass formula. So finding the center of mass of this nonuniform lamina is reduced to the much easier problem of combining two uniform laminae as long as we allow one of those laminae to have negative mass.

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