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 𝑥. To use this formula, we add up the
mass of each object in the collection multiplied by the 𝑥-coordinate of its
location and then divide by the total mass. This gives us the 𝑥-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 𝑥 sub 𝑖’s have the same value, call it 𝑥, then the
numerator is just 𝑥 times the denominator. And 𝑥 times something divided by
that something is just 𝑥. Okay, so the center of mass is this
center point here. This is four and a half units along
the 𝑥-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
𝐴𝐵𝐶 from which a triangle 𝐺𝐵𝐶 has been cut out. 𝐴𝐵𝐶 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 𝐴𝐵𝐶 with the triangle
𝐺𝐵𝐶 cut out of it. The specific information we’re
given is that 𝐴𝐵𝐶 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 𝐺𝐵𝐶 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 𝐴𝐵𝐶 shaded
in orange and the triangle 𝐺𝐵𝐶 shaded over it in magenta.
Separating this out into its two
pieces, we have the original lamina 𝐴𝐵𝐶 and the lamina 𝐵𝐺𝐶 that represents the
part that was cut out. The uniform lamina 𝐴𝐵𝐶 is a
regular object, so its mass is positive. However, since the combination of
𝐺𝐵𝐶 and 𝐴𝐵𝐶 results in a lamina with no mass at the bottom, where 𝐴𝐵𝐶
clearly does have mass, 𝐺𝐵𝐶 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 𝐴𝐵𝐶. 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 𝐵𝐶. Furthermore, because 𝐴𝐵𝐶 is
equilateral, this angle here at point 𝐵 is 60 degrees. So the triangle formed by 𝐴𝐵 and
the point where the altitude meets the side 𝐵𝐶 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 𝑎. So the triangle 𝐴𝐵𝐶 has a base
of 93 and a height of 𝑎. So its area, one-half base times
height, is one-half 93𝑎. If we call the constant uniform
density 𝑚, then one-half times 93 times 𝑎 is the area of 𝐴𝐵𝐶, and one-half 93𝑎
times 𝑚 is the total mass of the lamina 𝐴𝐵𝐶.
What about 𝐺𝐵𝐶? 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 𝐵𝐶 to the point 𝐺 is one-third of the total height of the triangle 𝐴𝐵𝐶. But we know what that value is; we
called it 𝑎. So the height of 𝐵𝐺𝐶 is
one-third of 𝑎. 𝐵𝐺𝐶 and 𝐴𝐵𝐶 have the same
base. So the area of 𝐵𝐺𝐶 is one-half
times 93 times one-third of 𝑎. Now remember, 𝐵𝐺𝐶 needs to have
negative mass but have the same kind of mass distribution as 𝐴𝐵𝐶. So the density of 𝐵𝐺𝐶 will be
negative 𝑚.
Let’s rearrange our expression for
the mass of 𝐵𝐺𝐶 by factoring out the negative one and the factor of
one-third. Written this way, we see that the
mass of 𝐵𝐺𝐶 is negative one-third times the mass of 𝐴𝐵𝐶. The factor of negative one-third is
present because 𝐴𝐵𝐶 has three times the area of 𝐵𝐺𝐶, and the mass of 𝐵𝐺𝐶 is
negative. Just to make our life easier and
clean up some space, let’s call one-half times 93 times 𝑎 times lowercase 𝑚
capital 𝑀 so that the mass of 𝐴𝐵𝐶 is capital 𝑀 and the mass of 𝐵𝐺𝐶 is
negative one-third times capital 𝑀.
Now that we know the masses, let’s
find the centers of mass of 𝐴𝐵𝐶 and 𝐵𝐺𝐶. We know that the center of mass of
𝐴𝐵𝐶 is located at the point 𝐺. And we’ve already determined the
𝑦-coordinate of point 𝐺; it’s one-third of 𝑎. 𝐺 lies along the altitude of
𝐴𝐵𝐶. And we know that this altitude also
being a median intersects the 𝑥-axis halfway between 𝐵 and 𝐶. Well, since the length of 𝐵𝐶 is
93, halfway between them is 46.5. So the coordinates of point 𝐺 are
46.5 comma one-third 𝑎. For triangle 𝐵𝐺𝐶, note that
𝐵𝐺𝐶 is an isosceles triangle. We can see this because the dashed
line is the perpendicular bisector of 𝐵𝐶. So any point on this bisector
including the point 𝐺 is equidistant from point 𝐵 and point 𝐶. So the sides 𝐵𝐺 and 𝐺𝐶 are
congruent, and therefore 𝐵𝐺𝐶 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
𝐵𝐺𝐶, 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 𝐵𝐺𝐶 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 𝑥-coordinate is also 46.5. The 𝑦-coordinate is one-third of
the height of 𝐵𝐺𝐶, and the height of 𝐵𝐺𝐶 is one-third of 𝑎. So the 𝑦-coordinate is one-third
of one-third or one-ninth of 𝑎.
Now we have the mass and center of
mass of 𝐵𝐺𝐶 and 𝐴𝐵𝐶. 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 𝐴𝐵𝐶 and
𝐵𝐺𝐶. 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 𝐴𝐵𝐶 is 𝑀, and the
𝑦-coordinate of the center of mass is one-third of 𝑎. So the first term of the sum of our
numerator is 𝑀 times one-third 𝑎. The second term in the numerator is
negative one-third 𝑀 times one-ninth 𝑎, which is the mass of 𝐵𝐺𝐶 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 𝑎. So we’ll factor one-third times 𝑀
times 𝑎 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
𝑎 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 𝑎 times eight-ninths
divided by two.
Note that the actual value of the
mass of 𝐴𝐵𝐶 does not factor into this calculation because the only thing that’s
important is the relative mass of 𝐴𝐵𝐶 and 𝐵𝐺𝐶 not their absolute masses. Eight-ninths divided by two is
four-ninths. So 𝑎 times eight-ninths divided by
two is four-ninths times 𝑎. Plugging back in our definition for
𝑎, 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 𝑥-coordinate of the
center of mass, instead of going through this whole calculation again, we note that
the 𝑥-coordinate of the center of mass for 𝐴𝐵𝐶 and 𝐵𝐺𝐶 is the same. This means that when we combine
them, we won’t change the 𝑥-coordinate of the center of mass. So the 𝑥-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.
