Video Transcript
Six masses of 70, 30, 70, 50, 70, and 10 kilograms are placed at the vertices 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and 𝐹 of a regular hexagon of side length 30 centimeters. Find the distance between the center of the hexagon and the center of gravity of the system.
To start with, to find the distance between the center of the hexagon and the center of gravity of the system, we’ll need a way to find the center of gravity. For this particular situation, we are told the mass of our objects rather than their weights. And we’re not told anything about the local gravitational field. So we can assume that the center of gravity is the same as the center of mass.
The center of mass of a system is the weighted average of the position of all of the objects that make up that system, where the weight of a particular position is the mass of the object at that location. As a formula, we can calculate the 𝑥-coordinate of the center of mass of a system by multiplying the mass of each object by the 𝑥-coordinate of its location, adding up all of these products, and dividing by the total mass of the system. To find any other coordinate, we simply replace 𝑥 in the formula with the coordinate that we’re looking for.
So our plan is to use this formula to find the coordinates of the center of mass and then find the distance from that point to the center of the hexagon. We aren’t given the coordinates of each of the masses of our system. But we are told that they’re at the vertices of a regular hexagon with side length 30 centimeters. This is enough information to work out what the coordinates are. All we have to do is draw a diagram.
Here is our regular hexagon, with its side length labeled, along with the letter corresponding to each vertex, as well as the corresponding mass. We’ve also included a set of axes. Let’s call the horizontal axis 𝑥 and the vertical axis 𝑦. The orientation of the hexagon doesn’t affect our final answer. So we’ve arbitrarily placed vertices 𝐴 and 𝐷 on the 𝑥-axis and also aligned the center of the hexagon to the origin just to simplify our calculations.
At this point, we can plow forward by simply finding the coordinates of each vertex and then plugging in to the center-of-mass formula and calculating. However, instead of using that approach, we will reduce the problem to a simpler one through a series of steps by making use of some of the symmetries that are already present. This will reduce the calculations we need to do in the end, as well as the number of vertices whose coordinates we need to identify.
In order to do these simplifications, we’ll need to understand a little bit more about the center of mass. In particular, we can use the notion of a center of mass to replace a collection of objects with different masses in different locations with a single object located at the center of mass and whose mass is the total mass of the system. This replacement also holds in our center-of-mass formula itself. That is, the mass and corresponding location don’t need to necessarily refer to a single object but could instead refer to the total mass of a system and the center of mass of that system.
Concretely, say we have a system of three masses: 𝑚 one, 𝑚 two, and 𝑚 three. To find the center of mass of this system, we simply use our formula. But now consider the subsystem made up of only mass two and mass three. Say the location of the center of mass of these two masses is right here. Then, if we place a mass at this point, whose mass is 𝑚 two plus 𝑚 three, the center of mass of the system consisting of the first mass and this new mass will be identical to the center of mass of the system consisting of all three original masses. What this means is that we can break up our system of six masses into several smaller subsystems, find the center of mass of each subsystem, and then combine those centers of mass to find the center of mass of the overall system.
We’ll start by noting that the vertices 𝐴, 𝐶, and 𝐸 all have the same mass of 70 kilograms. We also note that since these are every other vertex of a regular hexagon, they also form the vertices of an equilateral triangle. Let’s look at this triangle more closely. Here’s our equilateral triangle 𝐴𝐶𝐸 with 70 kilograms at each vertex. It turns out it’s very easy to find the center of mass of a regular polygon with the same mass at each vertex.
To see this, we first observe that the center of mass of this shape is somewhere inside of the triangle. Now imagine we rotate the triangle so vertex 𝐴 goes to vertex 𝐶, 𝐶 goes to 𝐸, and 𝐸 goes to 𝐴. When this happens, the center of mass also rotates around one-third of a turn because its position relative to the vertices must remain unchanged. But observe that the masses on the triangle are in identical positions to where they were before. We’ve just relabeled the vertices. But this means that the center of mass of the triangle must be the same as in the previous orientation because the two orientations are identical. And that means that this point and this point are actually the same point. The only point inside of a regular polygon, like this equilateral triangle, that doesn’t move no matter how we rotate the polygon itself is the exact center of the polygon.
So without doing any calculation, we know that the center of mass of this equilateral triangle is right here at the exact center of the triangle. And this is because the triangle itself is symmetric about this point. This is, in general, a very powerful argument. And in this particular case, it tells us that an equilateral triangle with 70 kilograms at each vertex is equivalent to 210 kilograms, that’s three times 70, located at the exact center.
Returning to our diagram, the center of the equilateral triangle is also the center of the hexagon, which is located at the origin of our coordinate system. So let’s replace the 70 kilograms at 𝐴, 𝐶, and 𝐸 with 210 kilograms at the origin. This actually helps us in two ways. First, it drops us from six points to four points, which greatly reduces the number of calculations we do. And, moreover, one of those points is at the origin. Points at the origin are very helpful because they appear as zeros in the numerator of the center-of-mass formula. These zeros then even further simplify our calculations.
It’s worth mentioning that although it may seem like to reach this simplification we had to go through a lot of effort with symmetry, arguments, and equilateral triangles, once we’re familiar with these sorts of arguments, we can actually go from our initial diagram to our simplified diagram in a single step. That is, once we observe that points 𝐴, 𝐶, and 𝐸 all have the same mass and are equally spaced about the origin, we can immediately replace them with their combined mass located at the origin itself. This substitution is a fairly significant simplification in our system.
Let’s do one more substitution to simplify our system just a little bit more. For this substitution, we’ll need to know how to break apart a mass at a single point into two or more different masses. Say we have a two-mass system, one with mass 𝑚 one plus 𝑚 two and one with mass 𝑚 three. If we wanted to though, we could split one of the masses, let’s say the first one, into two masses as long as they’re still located at the same point and the overall mass is the same.
Let’s now see how this sort of splitting can help us substitute a simpler system in place of the system formed by the 30-kilogram vertex at 𝐵 and the 10-kilogram vertex at 𝐹. Here are our two points, and we’ve also included a horizontal axis just to illustrate the ease and usefulness of this substitution. Note that the two points have the same 𝑥-coordinate and their 𝑦-coordinates are equal in magnitude but opposite in sign. In fact, the only thing breaking the perfect symmetry about the horizontal axis is the fact that the masses are different.
Remember though, as we saw before, symmetry is a very powerful tool. So let’s try to make this situation as symmetric as possible. Since the smaller mass is 10 kilograms, we can create the symmetry we’re looking for by splitting the larger mass into two pieces, one with mass 10 kilograms and the other with the remaining 20 kilograms. We now have perfect symmetry about the horizontal axis for these two 10-kilogram masses.
Using the same sorts of symmetry arguments we used for the equilateral triangle, we know that we can replace these two 10-kilogram masses with a single 20-kilogram mass at the exact center, which is right here on the horizontal axis. Including this substitution in our diagram, we’re left with 20 kilograms at the vertex 𝐵, 20 kilograms on the horizontal axis directly below 𝐵, 210 kilograms at the origin, and 50 kilograms at the vertex 𝐷.
This is a comparatively small simplification since we have just as many points as we had before. It does, however, carry two key benefits. First of all, now, all but one of the points is on the horizontal axis, which means that the only 𝑦-coordinate that we have to deal with is the 𝑦-coordinate of the point 𝐵, since the 𝑦-coordinate of the other three points is zero. This will greatly simplify our calculation for the 𝑦-coordinate of the center of mass.
Secondly, we now only need to establish three values for coordinates, specifically the 𝑥- and 𝑦-coordinates of vertex 𝐵 and the 𝑥-coordinate of vertex 𝐷. Of the remaining five coordinates, the 𝑥-coordinate of the 20-kilogram mass on the horizontal axis is the same as the 𝑥-coordinate of the vertex at 𝐵. And the other four coordinates are all zero.
It’s worth mentioning that, just like for the equilateral triangle, this simplification took some working out. But once we’re familiar with this sort of process, we can proceed directly from the original diagram to the substituted diagram in one or two steps.
Alright, now let’s work out the coordinates of our points. Let’s start by finding the coordinates of point 𝐷. Back in our original diagram, since the whole diagram shows a regular hexagon, the triangle formed by 𝐶𝐷 and the origin is an equilateral triangle. But one of the sides of this equilateral triangle is also a side of the hexagon, which we know to have a length of 30 centimeters, which means that all of the sides of the triangle have a length of 30 centimeters. And since 𝐷 is on the horizontal axis, 𝐷 is 30 centimeters away from the origin. So the coordinates of point 𝐷 are 30 comma zero.
The coordinates of the origin are, of course, zero comma zero. Point 𝐵, just like point 𝐷, is the vertex of an equilateral triangle, this one formed by 𝐴, 𝐵, and the origin. Using the rules for the side lengths of triangles with angles of 30, 60, and 90 degrees, we can determine that the 𝑥-coordinate of the point 𝐵 is negative one-half times the length of the side of the triangle, where the coordinate is negative because it is to the left of the origin. Also, the 𝑦-coordinate is the square root of three times half the length of the side of the triangle. And, again, since the triangle has side length 30 centimeters, the coordinates of 𝐵 are negative 15 comma 15 times the square root of three.
Our last point shares an 𝑥-coordinate with point 𝐵, and the 𝑦-coordinate is zero. So its coordinates are negative 15 comma zero. Now all we need to do to find the center of mass is plug in and calculate. First, we note that the total mass of our system is 210 plus 20 plus 20 plus 50, or 300 kilograms. Let’s start with the 𝑥-coordinate of the center of mass. The sum in the numerator is 50 times 30 plus 210 times zero plus 20 times negative 15 plus 20 times negative 15. The denominator, as we’ve already calculated, is just 300.
To simplify this calculation, we’ll note several things. First, 210 times zero is just zero. Second, 20 times 15 is 300. So both of the last two terms are equal to negative 300. Finally, 50 times 30 is the same as five times 300. Now we see that every term in the numerator is divisible by 300. Specifically, five times 300 divided by 300 is five, zero divided by 300 is still zero, and negative 300 divided by 300 is negative one. So this whole expression is equal to five minus one minus one, which is equal to three. So the 𝑥-coordinate of the center of mass is three.
Let’s now clear some space so that we can do a much simpler calculation for the 𝑦-coordinate of the center of mass. For the 𝑦-coordinate, we calculate the numerator the same way by replacing 𝑥 with 𝑦. As we saw before, the only point with a nonzero 𝑦-coordinate is the 20-kilogram mass with a 𝑦-coordinate of 15 times the square root of three. So the only nonzero term in our numerator is 20 times 15 root three. And as before, our denominator is still 300.
This is where the power of our second substitution really comes in. We only need to consider one term in the numerator because all of the other terms are zero. As we saw before, 20 times 15 is 300. So the numerator is 300 times the square root of three. And finally, 300 divided by 300 is one. So the 𝑦-coordinate of the center of mass is just the square root of three.
The last thing to do is calculate the distance from three comma root three, the center of mass of the hexagon, to the origin, which is the center of the hexagon. The distance from any point to the origin is just the square root of the sum of the squares of its 𝑥- and 𝑦-components. So we have the square root of three squared plus the square root of three squared. And since three squared is nine and the square root of three squared is three, this is the square root of nine plus three, or the square root of 12. And since 12 is two squared times three, the square root of 12 is two times the square root of three.
And so, at last, we find that the distance between the center of gravity of our system and the geometric center of the hexagon is two times the square root of three centimeters.
