Video Transcript
The figure shows a uniform wire
𝐴𝐵𝐶𝐷 of length 10 centimeters, where 𝐴𝐵 equals 𝐵𝐶 equals two 𝐶𝐷 equals
four centimeters. Find the coordinates of the
center of gravity of the wire with respect to axes 𝐵𝐴 and 𝐵𝐶.
We’ll 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 doesn’t actually affect the center of mass. Note that 𝑦 is uniform, so the
rods have a constant mass distribution, where we’re considering 𝐴𝐵 to be one
rod, 𝐵𝐶 to be a second, and 𝐶𝐷 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, we’ll
use the fact that 𝐴𝐵 equals 𝐵𝐶 equals two 𝐶𝐷 equals four centimeters.
Defining the linear density of
each rod to be one kilogram per centimeter, then that means the mass of line
segment 𝐴𝐵 is four kilograms. Similarly, the mass of line
segment 𝐵𝐶, or rod 𝐵𝐶, is also four kilograms. The mass of line segment 𝐶𝐷
is a half times four kilograms. And it’s a half because we know
that two 𝐶𝐷 equals four centimeters, so 𝐶𝐷 will be a half of four. And so the mass of 𝐶𝐷 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 𝐴𝐵 is
four centimeters, point 𝐴 can be considered as zero, four. Point 𝐶 has coordinates four,
zero. Point 𝐷 has coordinates four,
negative two. And, of course, point 𝐵 is at
the origin, so it’s got coordinates zero, zero. This now allows us to calculate
the midpoint of each of our line segments. The midpoint of line segment
𝐴𝐵 is the sum of each individual coordinate divided by two. So that’s 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 𝐵𝐶 has coordinates two, zero and the midpoint of 𝐶𝐷 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 𝑚 sub 𝑖, 𝑥 sub 𝑖 over the sum of 𝑚 sub 𝑖 and
the sum of 𝑚 sub 𝑖, 𝑦 sub 𝑖 over the sum of 𝑚 sub 𝑖. In other words, the coordinate
of the center of mass is the average of each particle’s coordinate weighted with
its mass. Let’s see what this looks like
for the 𝑥-coordinate. The numerator of this fraction
is the sum of the products of the mass and the 𝑥-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 𝑦-coordinate. This time, we use the
𝑦-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. It’s 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.
