### Video Transcript

In this video, weβll learn how to
find the center of mass of uniform rods. In a uniform gravity field, the
center of mass, which is sometimes called the center of gravity, is the unique point
from which the force of the weight of the object acts. In other words, we can assume that
the objectβs mass is concentrated at the center of mass as if it were a
particle. And if we support a rigid body at
its center of mass, then it will be perfectly balanced at this single point. In a rigid body with constant
density, the center of mass is located at the geometric center of the body.

So how do we generalize this to
uniform rods? Well, a uniform rod is a linear
object with a constant linear density. And its center of mass is located
at its midpoint. This definition means that a
uniform rod would be perfectly balanced at its midpoint, as shown in the
diagram. And so this means that we can treat
the mass of a uniform rod as a particle located at its midpoint. Now, when dealing with a system of
masses involving one or more uniform rods, we can find the center of mass of this
system by treating those rods as particles and finding the center of mass of the
resulting system.

So letβs remind ourselves of the
formula for finding the center of mass for a system of particles in a coordinate
plane. The center of mass of a system of
particles is the average position of the particle weighted according to its
mass. In other words, given mass π sub
π with coordinates π₯ sub π, π¦ sub π, the π₯-coordinate of the center of mass is
given by the sum of π sub π times π₯ sub π divided by the sum of π sub π. And the π¦-coordinate is given by
the sum of π sub π times π¦ sub π divided by the sum of π sub π. So now that we have the formulae
and the various definitions, weβre going to look at an example of how we can find
the center of mass of a system that involves a rod with weights at each end.

Line segment π΄π΅ is a uniform
rod of length four centimeters and mass four kilograms. A mass of magnitude five
kilograms is fixed at π΄, and another mass of magnitude one kilogram is fixed at
π΅. Find the distance from the
center of gravity of the system to π΄.

The first piece of information
weβre given is that the rod, which is represented by a line segment π΄π΅, is
uniform. Now, of course, we know that a
uniform rod has a constant density. So the center of mass of this
rod is located at its midpoint. Letβs use this information and
the information about the masses that are fixed at π΄ and π΅ to represent this
in a diagram. The next thing that we can do
is draw a coordinate line to indicate the position of these masses with respect
to the coordinate plane. Letβs imagine then that point
π΄ is at the origin and the line segment π΄π΅ lies along the π₯-axis as
shown.

Since the length of the rod is
four centimeters, if we let centimeters be our length unit, then point π΅ must
have coordinates four, zero. Then the midpoint of the line
segment π΄π΅ is found by calculating the sum of the individual coordinates and
dividing by two. So thatβs zero plus four
divided by two and zero plus zero divided by two, which gives us the coordinates
two, zero. Now, in fact, weβre really only
interested in the value of the π₯-coordinates. But weβve completed the value
of the π¦-coordinates just for completeness.

Next, we recall that if a
system of particles has mass π sub π at coordinates π₯ sub π, π¦ sub π, then
the π₯- and π¦-coordinates of the center of mass are as shown. Since the three centers of mass
that weβre interested in lie along the π₯-axis, weβre just going to use the
calculation for the π₯-coordinates. Letβs begin by working out the
sum of the products of π sub π and π₯ sub π. The mass of particle π΄ is five
kilograms, and its π₯-coordinate is zero. The mass of the uniform rod is
four kilograms, and its center of mass is located at the point two, zero. And finally, the mass of the
weight at point π΅ is one kilogram, and itβs located four units along the
π₯-axis. And so the sum of π sub π
times π₯ sub π is 12 or 12 kilogram centimeters.

The denominator of the fraction
is the sum of π sub π. And thatβs simply the sum of
the masses of all the objects in the system. Itβs five plus four plus one,
which is equal to 10, or 10 kilograms. The center of mass is the
quotient of these, so itβs 12 divided by 10, or 1.2. And so the π₯-coordinate of the
center of mass is 1.2. And, in fact, we know that that
measurement is in centimeters. Since this point lies on the
π₯-axis and point π΄ lies at the origin, the distance from the center of mass or
center of gravity of the system to π΄ is 1.2 centimeters.

In this example, we were able to
identify the center of mass for this system by considering the mass of the uniform
rod to be concentrated at its midpoint. We can use this same strategy when
we have a system of multiple uniform rods.

When a number of uniform rods are
connected together, they form a rigid body, which we call a wireframe. Then we can find the center of mass
of a wireframe by treating the mass of each rod to be concentrated at its
midpoint. And then this leads us to a system
of particles whose center of mass coincides with that of the wireframe. Itβs also worth noting that if each
rod in the wireframe has the same density, then the value of this density doesnβt
influence the center of mass. And so we can disregard it from our
calculations altogether. Given this information then, letβs
demonstrate how to find the center of mass of a system of uniform rods.

The figure shows a uniform wire
π΄π΅πΆπ· of length 10 centimeters, where π΄π΅ equals π΅πΆ equals two πΆπ· equals
four centimeters. Find the coordinates of the
center of gravity of the wire with respect to axes π΅π΄ and π΅πΆ.

Weβll begin by recalling that
the center of mass of a uniform rod is located at its midpoint. Now we, in fact, have a system
of uniform rods. So what we can do is we can
treat the mass of each rod to be concentrated at its own midpoint. And we remember, of course,
that if each rod in the wireframe has the same density, then the value of that
density doesnβt actually affect the center of mass. Note that π¦ is uniform, so the
rods have a constant mass distribution, where weβre considering π΄π΅ to be one
rod, π΅πΆ to be a second, and πΆπ· to be the third. Now this means that their
masses are proportional to their length. And we know that each rod has
the same density. So if we know the length of
each rod, in other words the length of each line segment, we can quite easily
work out their mass. To work out their length, weβll
use the fact that π΄π΅ equals π΅πΆ equals two πΆπ· equals four centimeters.

Defining the linear density of
each rod to be one kilogram per centimeter, then that means the mass of line
segment π΄π΅ is four kilograms. Similarly, the mass of line
segment π΅πΆ, or rod π΅πΆ, is also four kilograms. The mass of line segment πΆπ·
is a half times four kilograms. And itβs a half because we know
that two πΆπ· equals four centimeters, so πΆπ· will be a half of four. And so the mass of πΆπ· is two
kilograms.

Our next job is to find the
center of mass of each individual wire segment. Now if we use coordinates as
the length unit in the coordinate system, we can find the coordinates of each
point. So since line segment π΄π΅ is
four centimeters, point π΄ can be considered as zero, four. Point πΆ has coordinates four,
zero. Point π· has coordinates four,
negative two. And, of course, point π΅ is at
the origin, so itβs got coordinates zero, zero. This now allows us to calculate
the midpoint of each of our line segments. The midpoint of line segment
π΄π΅ is the sum of each individual coordinate divided by two. So thatβs zero plus zero
divided by two and four plus zero divided by two, which is zero, two. In a similar way, the midpoint
of line segment π΅πΆ has coordinates two, zero and the midpoint of πΆπ· has
coordinates four, negative one.

Adding these coordinates to our
diagram, we can clear some space and define the center of mass of our
system. We know that the center of mass
will be given by the sum of π sub π, π₯ sub π over the sum of π sub π and
the sum of π sub π, π¦ sub π over the sum of π sub π. In other words, the coordinate
of the center of mass is the average of each particleβs coordinate weighted with
its mass. Letβs see what this looks like
for the π₯-coordinate. The numerator of this fraction
is the sum of the products of the mass and the π₯-coordinate of its center. So we work out four times zero,
another four times two, and then two multiplied by four. Then the denominator is simply
the sum of their masses, so four plus four plus two. That gives us 16 over 10, which
simplifies to eight-fifths.

And we can now repeat this
process for the π¦-coordinate. This time, we use the
π¦-coordinates of the center of gravity. So the numerator is four times
two plus four times zero plus two times negative one. And the denominator is still
the sum of the masses. Itβs the mass of the entire
wireframe. That gives us six over 10,
which simplifies to three-fifths. And so we see the coordinates
of the center of mass of this system are eight-fifths, three-fifths.

In our final example, weβll
demonstrate how to find the distance between the center of mass of a uniform wire
structure and a specific point.

A thin uniform wire is shaped
into trapezoid π΄π΅πΆπ·, where the measure of angle π΅ equals the measure of
angle πΆ, which equals 90 degrees; π΄π΅ equals 494 centimeters; π΅πΆ equals 105
centimeters; and πΆπ· equals 134 centimeters. Find the distance between the
center of gravity of the wire and point π΅, rounding your answer to two decimal
places.

Before we attempt to answer
this question, weβre going to begin by sketching out trapezoid π΄π΅πΆπ·. We know that the measure of
angle π΅ and the measure of angle πΆ are 90 degrees. This tells us that side lengths
π΄π΅ and πΆπ· must be the parallel sides of the trapezoid. Letβs add the trapezoid to the
coordinate plane. And weβll let point π΅ be at
the origin, since thatβs the point weβre going to be working with a little later
on.

Now, before we go any further,
weβre actually going to calculate the length of line segment π΄π·. And weβll see why thatβs
important in a minute. By adding a right triangle to
the diagram, we can in fact use the Pythagorean theorem to calculate the value
of π₯. This tells us that the square
of the hypotenuse, so π₯ squared, is equal to the sum of the squares of the two
shorter sides. Well, the base of the triangle
is in fact 105 centimeters. And its height is the
difference between 494 and 134. So π₯ squared is 140625. The square root of this gives
us a length of 375 centimeters.

Now this is really useful
because weβre told that the wire is uniform, so each rod has the same
density. And we can define their linear
density to be one kilogram per centimeter. And we can therefore say that
since the rods have a constant mass distribution, their masses are proportional
to their length. And so the mass of the piece of
wire between π΄ and π΅ is 494 kilograms, π΅πΆ is 105 kilograms, πΆπ· 134
kilograms, and π΄π· 375 kilograms.

The next thing that weβre going
to do is work out the center of mass of each rod. We know that the center of mass
is located at the midpoint of each rod. So weβll work out the
coordinates of each midpoint. Line segment π΄π΅ lies on the
π¦-axis, so the π₯-coordinate of its midpoint will be zero. Then its π¦-coordinate is the
average of the π¦-coordinate of π΄ and π΅. So thatβs 494 plus zero divided
by two, which is 247. Then, line segment π΅πΆ lies on
the π₯-axis. So the π¦-coordinates of its
midpoint is zero. Then, the π₯-coordinate of its
midpoint is half of 105, say 52.5.

To get to the midpoint of πΆπ·,
we move 105 units along the π₯-axis. And then we move a half of 134
units up. So its π₯-coordinate is 105 and
its π¦-coordinate is 67. Then, we have the midpoint of
π΄π·. The midpoint of its
π₯-coordinate is, in fact, 52.5. Once again, itβs a half of
105. Then the π¦-coordinate of its
midpoint is, in fact, the average of 494 and 134, which is 314. Now, we recall that the
coordinates of the center of mass are the average of each particleβs coordinates
weighted with its mass.

So weβll begin by calculating
the π₯-coordinate of the center of mass of the entire system. Then we find the numerator by
finding the sum of the products of each element in the first and second row of
our table. The denominator is just the
total mass of the system, so itβs the sum of all elements in the first row
itself. That gives us 39270 over
1108.

Letβs clear some space and
repeat this for the π¦-coordinate of the center of mass. To find the numerator of the
π¦-coordinate of the center of mass, we find the sum of the products of the
elements in the first row and the third row. And then, once again, we divide
that by the total mass. That gives us 248746. And so we have the coordinates
of the center of mass. Of course, we want to find the
distance of this point from point π΅. And we know that point π΅ is
the origin. And so weβre going to use the
distance formula, which is, of course, just a special version of the Pythagorean
theorem.

The distance is simply the
square root of the sum of the squares of the π₯- and π¦-coordinates, which is
227.28. The distance between the center
of gravity of the wire and point π΅, correct to two decimal places, is 227.28
centimeters.

Letβs now recap the key concepts
from this lesson. In this video, we learned that the
center of mass of a uniform rod is located at its midpoint. We also saw that a rigid body
comprised of multiple uniform rods is called a wireframe. The center of mass of such
wireframe is found by treating the center of mass of each rod in the wireframe as if
itβs concentrated at its midpoint. Once we have this information, we
can use the usual formulae to help us calculate the center of mass of the
system.