- Today we're going to take
a look at another application of integrals, and that is
answering the question of how do we find the center of mass. The idea here is kind
of like a teeter-totter. If you wanted to balance
perfectly-- or maybe spinning plates would
be a better example. If you want it to
balance perfectly, where is the center of the mass
that will balance perfectly on a point? We're going to start by
looking at the simplified one dimensional option,
first, looking at the discrete situa
tion,
discrete basically meaning we've just got
points on the x-axis. And the way we find
the center of balance with a bunch of
points is first, we need to identify the
mass, the entire mass. What we will do is
we will sum or add up all the individual
values of the mass, and that calculates
our total mass. To find the center
of that mass, we're going to find M sub
y, which is going to represent the moment
away from the y-axis. M sub y , the distance
from a y-axis, that moment from the y-axis
is
the sum of the product of the individual masses
times the individual x values. And that will give
us the moment. And then to actually
calculate the point where the center of balance is,
we will take the moment function and divide
by the mass function. So we have three
equations here to help us build the center of
mass, with the key one being at the end when we
divide the two pieces. So in two dimensions,
we'll first find-- I'm sorry, one dimension-- we'll first find
the total mass, then we'll fi
nd the total moment
away from the y-axis. And then we'll find that x
bar, that center of mass, by dividing those values. So for example, if I put my
first mass of 12 kilograms at x equals negative 4, my
second mass of 12 kilograms at x equals 4-- and this is in meters-- and my third mass has 30
kilograms, and I put it at-- we should be numbering
these x's 1, 2, and 3-- we put it at x equals
2 meters, and we put our fourth mass
of 6 kilograms, and we put it at x 4
equals negative 4 meters. We wan
t to know where to put
the center of the balance so that this balances
out perfectly. First, calculate the total mass. We find the total
mass by finding the sum of the individual
masses, which is 12 plus 12 plus 30 plus 6. And so that's going
to come out to 60 kilograms for our total mass. Next, we'll find
the moment function for the y, which is the sum
of the mass times the x value. Notice we multiply
by the opposite coordinate from what
the moment asks for. So the moment of y, we're
going to m
ultiply by the x. So we'll do 12 times the
first x of negative 4, plus the second mass 12 times
the second x of four, plus 30-- the next mass--
times its location 2, plus 6 times negative
4, multiplying the mass times the location. And then we add all that up,
we get 24 kilogram meters. And now to find the
center of balance, x bar. We just have to divide here. My, 24, by your m of 60. And when we reduce
that, we'll find out that it is 2/5 of a meter
away from the origin, will cause this situatio
n
to balance perfectly. So that's
one-dimensional discrete, with our mass, moment,
and center function. But what if it's not discrete? What if it is continuous? With continuous, we'll see
some type of rho function that represents the mass
over that continuity. Well similarly, we'll
first calculate the mass, which is going to
be the sum of all of the masses
along our interval. Well, that would just
be the integral from a to b of the rho dx. We'll also fine the moment
function in the y direction,
and just as before, we took
the x's times the mass. We're going to also take the
x's times the mass function, and integrate it
from a to b to get all of our individual
moments added together. And then, to find
our center x bar, we simply will take that
moment and divide by the mass like before. Again, our goal is that x
bar, but all three of these help us get there. So let's try a
continuous example where we have to calculate
where is the center of balance? Let' say rho is going
to be equal to x
cubed on the x interval
from 0 to 3, and we want to know where this
is going to balance. Well first, we'll calculate
the total mass, which is the integral from 0
to 3 of our function, x cubed dx, which is x
to the fourth divided by 4 integrated from 0 to 3. And if we plug 3 in, we
get 81 over 4 minus 0. So our total mass
is 81/4 kilograms. Our moments away
from the y-axis, that's going to be the
integral from 0 to 3 of x times our function x cubed
dx, or the integral from 0 to 3 of x to the
fou
rth dx, which is x to the fifth divided by
5, integrated from 0 to 3. And 3 to the fifth is
243 divided by 5 minus 0. And so we have our moment. And the way we calculate
the center of balance is, we divide that moment 243
over 5 by the 81 over 4, and we can multiply by
the reciprocal 4 over 81. And when we reduce, we get 12/5. 12/5 is where this x cubed
will balance perfectly between 0 and 3. This has all been in
one dimension, though, So let's make this a little more
interesting in two dimensio
ns. We'll again first look
at the discrete case to set up the continuous case. With the discrete
case, now we're not just talking about
points on the x-axis, we're talking about
points in general. Again, we'll start by
calculating the mass, which is the sum
of all of the m's. And again, we'll
calculate the moments away from the y-axis, which
is the sum of the masses times the x-coordinates. And then, we'll also
calculate the moments away from the x-axis the
other direction, which is the sum of t
he moments
times the y-coordinates. Once we have those, we
can calculate x bar-- which is the y moment
divided by the mass-- and y bar-- which is the x
moment divided by the mass. And that will give us our center
point that we're looking for, of x bar comma y bar. So in two dimensions,
we just kind of extend that same formula considering
both the x and y-coordinates to get our formulas. So let's try, if I took a
single mass of 5 kilograms, and I put it at the point
negative 2, negative 3, and I
took a second
mass of three kilograms, and I put it at the point
positive 2, positive 3, and I took a third
mass of two kilograms, and I put it at the point
negative 3, negative 2. Where would the
center of balance be for these three masses? Well first, we know
the total mass. 5 plus 3 plus 2 equals 10
kilograms for the total mass. Next, we'll find the
moment in the y direction, which is the x-coordinate, the
mass times the x-coordinate. So we'll start with the
first mass was 5 times the x-coord
inate of negative 2,
plus 3 times this x-coordinate of 2, plus 2 times its
x-coordinate of negative 3. And that gives us negative
10 plus 6 minus 6, that's negative 10. We also need to find a
moment in the x direction where we multiply
by the y-coordinate. So the first mass of 5 times
its y-coordinate of negative 3 plus 3 times 3, plus 2 times
negative 2, that's negative 15 plus 9 minus 4, is
also negative 10. Interesting that's
it's the same. So then define our x bar. We use the M sub y-- I'm g
oing to label that. We use that one to get
the x bar, because they're both the same number. I don't want to lose
which when we use. Negative 10 divided by the
mass of 10 gives us negative 1. y bar, we use the
second one, which came from the moment of x,
negative 10 divided by 10 equals negative 1. So this time, our center
that we're looking for is at the point negative
1 comma negative 1. The continuous version of this,
then, as you might expect, just alters the
formula slightly. And with contin
uous
functions, if we're finding the center of balance
or the center of mass, we call it a lamina. And here we're going
to talk about laminas that go to the x-axis. They don't go to the x-axis,
it's a little different, and we'll talk about
that in a minute. So first, we have to
calculate our mass. As you might expect,
we have to add up all the individual masses. That's just an integral from
a to b of the function that represents the masses, the x. Then we'll find the
m sub y, which again, as you
might expect, similar
to the last continuous case, is x times f of x dx. But when we find the
moments from the x-axis, this one's quite different. The integral from a to
b up, and we actually take our function squared
divided by the 2 dx. It's a little different
than you might expect. It comes from having to
solve for the other variable, but we're not going to get
into the derivation right now. Once we find those
two moments, we can then find x bar again
by taking My divided by m, and we can fi
nd y bar by
taking Mx divided by m. And so again, our center
is x bar comma y bar. This is a continuous
version of our formulas. So let's see if we can
solve a continuous case. Let's say we've got a function
that is bounded above, or a region that's bounded above
by f of x equals x squared, and below by the
x-axis on 0 to 2. We want to find
its center of mass. Just to get a
visual idea of what we're looking at, this function
starts at 0, goes to 1, 1, goes to 2, 4,
bounded from 0 to 2. So we hav
e this
kind of a triangle. It's got a curved
left side to it. We want to see, where is
that shape going to balance? We're going to find
the exact point. To do that, first we need
the total mass, which is the integral from
0 to 2 of our function x squared dx, which
we know is x cubed over 3 integrated from 0 to
2, and 2 cubed is 8/3 minus 0. So we just have 8/3. Then, we need the
moment of y, which is the integral from 0 to 2
of x times our function, which is x squared dx, where the
interval from
0 to 2 of x cubed dx, which gives us x
to the fourth over 4 integrated from 0 to 2. 2 to the fourth power is
16 divided by 4, is 4. To find the M sub x,
This is the weird one. We're going to integrate from
0 to 2 of our function squared divided by 2, which gives us
the integral from 0 to 2 of x to the 4th divided by 2 dx,
which is x to the 5th divided by 5 times 2, or 10,
integrated from zero to 2. 2 to the 5th is 32. So we have 32 over 10,
which reduces to 16/5. Now we're ready to
find x bar a
nd y bar. X bar is the moment from y,
which is 4, divided by the 8/3, or 4 times the reciprocal
of 3/8 gives us 3/2. For the y's, we'll
take the 16/5, and we'll divide
by the mass of 8/3, which is equal to 16/5
times the reciprocal 3/8. Reducing gives us 6/5, and so
we found the center of balance at x equals 3/2, y equals 6/5. So 3/2 and 6/5, it's going
to be somewhere around there on our graph
will perfectly balance that triangle. Now I mentioned
that this is only going to work if we're
bounded
below by the x-axis. If we're not bounded
below on the x-axis, we need a slight
adjustment to our formulas. So let's look at
still continuous. So we still have
what's called a lamina. The lamina this time
is between functions. Similar to how we found the
area between two curves, we just need to subtract
the two functions. So our mass function is
still the integral from a to b, but this time we'll
subtract the function f of x minus g of x dx. For our moment function,
it's in a y direction. It's
still going to be
the integral from a to b, and we're still going
to do x times f of x. But this time, before
we multiply by the x, we have to subtract g of x. So x times the difference
in the functions dx. And the moments from
the x from a to b, it's still going to be the
function squared divided by 2, but we're going to
square the functions and then subtract them. Let's move the divide by 2 up
front as 1/2 just for the sake of space, and then it's
going to be f of x squared minus g of x square
d dx. And once you have those, you
can find x bar, as always, by dividing the moment
from y by the mass. Find y bar by dividing the
moment from x by the mass. These functions will help us
find the center of balance, or the center of mass between
two functions on an interval from a to b. So let's do one last example
where we do just that. Let's say we want
the space that's bounded above by f of x equals
6 minus x squared, and below, by g of x equals 3 minus 2x. And we get a picture of
what we're
looking at. That's-- we want a
little more center. The above function 6 minus
x squared looks like this. And the below function 3 minus
2x looks something like this. And so we've got
this mass that we're trying to find the
center of balance for. Looks like it's probably
going to be around here somewhere, that's
just kind of eyeballing it. A little to the right,
maybe it's a little too high in that picture. We want to find out where
is that center of balance? So to do that, first
we find the tota
l mass. The total mass is the
integral goal from a to b. Well, we have to find these
x-coordinates of where it's going to intersect,
where is this space going to start and end? You can derive that
by hand by setting them equal to each
other and solving. Let's do that kind of
up above in this space. 6 minus x squared
equals 3 minus 2x. Adding everything
to the right, we get x squared minus 2x minus
3, x minus 3 times x plus 1, so x equals 3 and negative 1. So we've got negative 1 to
the left, thr
ee to the right. So we're integrating
from negative 1 to 3 of the difference
in the functions. 6 minus x squared's on
top, minus the 3 minus 2x, remember that
changes both sides. Let's combine like terms before
we integrate the integral from negative 1 to 3, of
3 minus x squared plus 2x. The x is going to
equal 3x minus x cubed divided by
3 plus x squared integrated from negative 1 to 3. Which means, like I said, I made
a parentheses, 3 times 3 is 9, minus 3 cubed is 27, divided
by 3 is 9, plus
3 squared is 9, minus 3 times negative
1 is a positive 3. Being careful with
our signs, we've got to subtract a negative. But then, when I do
the negative 1 cubed, it also becomes a negative 1/3. Then plus 1 minus 1. And we put that all
together, 9 minus 9 plus 9 plus 3 minus 1/3, minus
1, we should get 32 over 3 for our total mass. All right, let's find
the moment for y, which is the integral
from negative 1 to 3 of x times the difference
in the functions. 6 minus x squared minus
the other func
tion, which is 3 plus 2x dx. Combine like terms
and distribute the x through 6 minus 3
is 3 times x is 3x, minus x squared times x is
x cubed, plus 2x times x is 2x squared. Integrating, we have
3x squared divided by 2 minus x to the
fourth divided by 4, plus 2x cubed divided by 3
integrated from negative 1 to 3. Plugging three in, 3 squared
is 9, times 3 is 27/2, minus 3 to the fourth is 81/4,
plus 3 cubed is 27 times 3 is 54/3, minus,
plugging negative 1 in, we get 3/2 minus
negative as positi
ve, negative squared to the fourth
power isl a positive 1/4. minus, but negative 1 cubed is
a negative again, positive 2/3, and we put that
in our calculator, we'll get 32/3 for our
moment in the y direction. Now for the moment
in the x direction, this one's going to be
probably the most involved, going from negative 1 to 3. We have the 1/2 out front times
the first function squared. The top function is 6
minus x squared, squared, minus the second function
squared, 3 minus 2x squared dx. Squarin
g everything gives us the
integral from negative 1 to 3 of 1/2 times 36 minus 12x
squared plus x to the fourth. And then we're subtracting,
so 3 squared is 9. Then we have 12x,
it's a negative 12x, but when we subtract it,
it becomes a positive 12x. And then negative 2x squared
becomes negative 4x squared dx. Which-- let's pull the
1/2 out because that's a constant from negative 1 to 3. Of combining like terms,
we've got an x to the fourth. We've got negative 12 minus
4 is negative 16x squared.
I keep making sure we
don't lose anything here. Next, we've got a 12x. And then we've also got a 36
minus 9 is a positive 27 dx. So when we take the internal, we
have 1/2 times x to the fifth, divided by 5, minus 16
x cubed, divided by 3, plus 12x squared,
divided by 2, gives us 6x squared, plus 27x integrated
from negative 1 to 3. So plugging 3 in, we have 1/2
times 3 to the fifth is 243/5, minus 3 to the third is 27. 27 divided by 3 is
9 times 16 is 144, plus 3 squared is 9 times 6
is 54, plus
3 times 27 is 81. And then subtracting, being
careful with signs, negative 1 to the fifth is negative 1/5. Subtracting negative 1/5
makes it positive 1/5. negative 1 cubed
gives us negative 16/3 with the negative
in front, which is a positive 16/3 and when we
subtract it, we get negative 16/3. Be very clear with our signs. Negative 1 squared
just gives us 6. So we subtract the six. And then we subtract negative
27, which is a positive 27. And if I put that
all in my calculator and tell my calcu
lator to
change that into a fraction, we get 416/15 for the
moment in the x direction. Lots of work, but we
finally got through it. It's mainly algebra. The calculus is pretty
straightforward. Now to figure out our x
bar, we take the moment from a y direction which is 32
over 3, and divide by the mass, which is also 32 over 3. So the center of
balance for x is 1. For the y bar, we
take that moment in the x direction
of 416/15 and we divide by 32/3 which is 416/15
times the reciprocal of 3/32, an
d I'm just going to do
that on the calculator. And that's going to equal 13/5. So finally, we have
the coordinates of our center of balance,
the center of mass, at 1 comma 13/5. So that's how we can do it if
they're between two functions. We just have to subtract the
functions just in our formulas. I do want to show you one little
trick that comes up sometimes. It doesn't always apply, but
when it does, it's really nice. We want to use symmetry
whenever possible. It can save you a lot of work. F
or example, if I have x plus 3
squared plus y minus 4 squared equals 4, and I were to
graph that, turns out that that's going to be a
circle with a radius of 2 centered at negative 3 comma 4. So I can draw a nice
little circle here. And if that's a
nice little circle, I know the center of
the circle, because it's symmetrical about
the x and y-axis. The center is going to be right
there at negative 3 comma 4, and we're done. We don't have to go
through all the equations to find those points,
alth
ough if we did, we'd get negative 3, 4. So that's really nice
and can save you time. More often, it'll give
you one coordinate. You'll notice it's
symmetrical around x equals 0. So x bar equals 0. And then you have to do
the work to find the y bar. Or you know the y bar,
because it's symmetrical and you have to
seek the x bar out. But that can save
you a lot of time if you catch some
symmetry to your shape before you go through all
the calculations, which is why I strongly
suggest you always gra
ph the function before you
go through the algebra. So take a look at these on
the homework assignment. Practice finding
some centers of mass, and we will see you in class to
discuss these centers of mass in more detail.
Comments