### Video Transcript

In this video, we’re going to learn
about center of mass. We’ll learn what center of mass is,
why it’s useful, and how to calculate it. To start out, imagine that you work
at a glass blowing facility that specializes in custom works and fine art. One day, as you’re moving one of
the glass sculptures from one spot in the shop to another, it begins to slip and
fall to the floor. Reacting quickly, you reach out to
grab the falling work, trying to hold on to it at the most stable place.

To know where on the work its point
of greatest stability is, it will be helpful to know something about center of
mass. The center of mass, or COM for
short, of an object is the average location of all of its mass. For example, the center of mass of
a square, triangle, and circle would be in roughly these locations. For two-dimensional shapes, one way
to think about center of mass is it’s the point at which if the shape was supported
there, it would balance.

Often center of mass is defined
relative to a reference coordinate system. Say, for example, that we had a
three-dimensional shape in an 𝑥𝑦𝑧-coordinate system. We could express this object’s
center of mass as a coordinate point relative to the origin of these axes.

Speaking of center of mass
coordinates, let’s talk a bit about how to actually calculate center of mass. Let’s consider a simple case where
we have two masses, 𝑚 one and 𝑚 two, both on a coordinate line. If we want to calculate the center
of mass of this system of two masses, the way we do that mathematically is in the
numerator to sum up the product of each individual mass, 𝑚 one and 𝑚 two, with its
position, what we’ve called 𝑥 one and 𝑥 two, respectively. These position values are measured
relative to our reference point, in this case our zero point on the line. That’s the numerator of this
fraction. And in the denominator, we have the
sum of our two masses, 𝑚 one and 𝑚 two.

This equation for center of mass is
specific to this scenario. But we can generalize this idea to
apply to any shape of any dimension. If we have a general distribution
of masses in some coordinate system, then along any one of our independent
coordinate directions, we can solve for the center of mass in that direction using a
similar formula. Mass element one times the distance
along that axis to that mass element plus mass element two times the distance along
that axis to that second mass element, and so forth and so on, for all the mass
elements along that line.

This numerator is divided by a
denominator equal to the sum of all the individual mass elements. We can also write this expression
as the sum over all of the mass elements 𝑖 of 𝑚 sub 𝑖 times 𝑑 sub 𝑖 divided by
the sum of the mass elements by themselves. Calculating the center of mass of a
system of masses can be very useful. Because once we’ve located the
center of mass of a system, we know that if we were to apply a force directly to
that point in the system, assuming it is a material point, then the system of masses
would move along without rotating at all. That is, it would translate.

Thinking back to our opening
example of trying to catch a falling glass sculpture, this is why we’d want to grab
the sculpture at its center of mass so that, after holding on to it there, the mass
wouldn’t rotate. One of the best ways to understand
center of mass is to calculate it in a few practical examples. Let’s do that now.

A 0.75-meter-long rod of iron with
a density of 8.0 grams per cubic centimeter is joined end to end with a
0.75-meter-long rod of copper with a density of 2.7 grams per cubic centimeter. If the rods have an equal
cross-sectional area to each other, how far from the unjoined end of the iron rod is
the center of mass of the object?

We can call this distance 𝑑 sub cm
and start off by drawing a diagram of the situation. In this example, we have a
0.75-meter-long rod of iron connected on its far right end with a 0.75-meter-long
rod of copper. And we’re given both the density of
iron and the density of the copper rod. Considering these rods as an entire
system in and of themselves, if the center of mass is located somewhere along this
1.50-meter length, we want to solve for the distance — we’ve called it 𝑑 sub cm —
of that center of mass from the unattached end of the iron rod.

We can recall the mathematical
relationship describing the center of mass along a particular axis. The center of mass of a
distribution of mass elements — we’ve called them 𝑚 sub 𝑖 — is equal to the sum of
each element multiplied by the distance it exists from a defined origin, all divided
by the sum of the masses by themselves. Looking at our particular example
of our system of two metal rods, we know that the center of mass of the iron rod
exists in its geometric center and the center of mass of the copper rod also exists
in its geometric center. So we can write that 𝑑 sub cm, the
center of mass of this system of two rods, is equal to the mass of the iron
multiplied by its distance from a reference point plus the mass of the copper times
its distance from a reference point, all divided by the sum of these two masses.

We’ve been told the densities of
our iron and copper rods, but not their masses. We can recall though that density,
𝜌, is equal to mass divided by volume. Or in other words, mass is equal to
density times volume. So in substitution for 𝑚 sub i and
𝑚 sub c in our equation for 𝑑 sub cm, we can write 𝜌 sub i times 𝑉 and 𝜌 sub c
times 𝑉, respectively. We know the volumes of the two rods
are the same because they had the same length as well as the same cross-sectional
area.

Looking at this expression for 𝑑
sub cm, we see that the volume, 𝑉, appears in every term. This means we can factor this term
out and then cancel it out from our expression. We see now that since we know 𝜌
sub i and 𝜌 sub c, all we need to solve for in order to determine 𝑑 sub cm is the
distance of the center of mass of the iron rod and the distance of the center of
mass of the copper rod from the origin. Since those two rods are uniform,
the distance on the center of mass of the iron rod from the leftmost end of that rod
is half of 0.75 meters, or 0.375 meters. Similarly, 𝑑 sub c is equal to
that same value plus the length of the iron rod, 0.75 meters.

We’re now ready to plug in and
solve for 𝑑 sub cm. When we plug in for all these
values — the density of iron and the position of its center of mass, the density of
copper and the position of its center of mass — and enter this expression on our
calculator, we find a result of 0.56 meters. That’s the distance from the
unattached end of the iron rod of this center of mass of this system of two
rods.

Now let’s look at an example
involving a center of mass that changes in time.

Two particles of masses 2.0
kilograms and 4.0 kilograms move in uniform circles with radii of 5.0 centimeters
and 𝑅 centimeters, respectively. The 𝑥-coordinate of the particle
moving in the 5.0-centimeter radius circle is given by 𝑥 as a function of 𝑡 equals
5.0 cos of two 𝑡. And the 𝑦-coordinate is given by
𝑦 as a function of 𝑡 equals 5.0 sin of two 𝑡. The 𝑥-coordinate of the center of
mass of the particles is given by 𝑥 sub cm as a function of 𝑡 equals 6.0 cos of
two 𝑡. And the 𝑦-coordinate of the center
of mass of the particles is given by 𝑦 sub cm as a function of 𝑡 equals 6.0 sin of
two 𝑡. Find 𝑅.

We want to solve for the radius of
rotation of our 4.0-kilogram particle. And to do that, we can start by
drawing a diagram of this scenario. In this example, we’re told about
two masses that are moving in circles. Mass one moves in a circle of
radius we’ve called 𝑟 of 5.0 centimeters. The second mass, mass two, moves in
a circle of radius capital 𝑅 that we want to solve for.

With respect to the smaller mass,
we’re told the 𝑥- and 𝑦-coordinates of that mass as a function of time. And we’re also told the 𝑥- and
𝑦-coordinates of the center of mass of this system of two masses as a function of
time. Even with the center of mass that
changes with time, the relationship that the center of mass is equal to the sum of
the product of each mass element multiplied by its distance from the center of mass
divided by the sum of the overall masses still applies to our situation.

Let’s look for a moment at the
𝑥-position of our particle and the 𝑥-position of the center of mass of the
system. We notice that these two
expressions have the same phase. And if we look at 𝑦 as a function
of 𝑡 and the center of mass 𝑦-coordinate, we see the same relationship that the
phase is the same. This means we can solve for the
radius of our second particle, capital 𝑅, by looking at either the 𝑥-coordinate or
separately the 𝑦-coordinate. Both will give us the same
information.

Just to choose one of the
coordinates, let’s write down the center of mass relationship in the
𝑥-direction. The 𝑥-coordinate of our system of
masses center of mass is equal to 𝑚 one times 𝑥 as a function of 𝑡 times 𝑚 two
times the function we’ve called 𝑥 sub two as a function of 𝑡, currently unknown,
all divided by the sum of 𝑚 one and 𝑚 two. Let’s plug in to this equation what
we know. We know 𝑥 sub cm, 𝑚 one, 𝑚 two,
and 𝑥 as a function of time.

When we write in all the
information we know, we see that the final form of our center of mass has an
expression cos of two 𝑡. We see that same form in the
expression of the exposition of our smaller particle. And knowing that the 5.0 that
precedes that phase information represents the radius of our smaller particle’s
rotation. This gives us a clue that the
expression for 𝑥 sub two as a function of 𝑡 will be the radius of the larger
particle, capital 𝑅, multiplied by the same phase, cos of two 𝑡.

We know that 𝑥 sub two of 𝑡 will
have this form because this is the exposition of the particle with a radius capital
𝑅. And we know the phase relationship
of this second particle’s position must be consistent with the phase relationship of
the overall center of mass of the system. So substituting this expression
into our center of mass equation, we see that the units of kilograms cancel out from
this expression. And if we multiply both sides of
our equation by 6.0 and then subtract 10.0 times the cos of two 𝑡 from both sides
and then seeing the cos of two 𝑡 appearing on both sides canceling that
trigonometric phase term out. We see that 26 is equal to 4.0𝑅 or
𝑅 is equal to 26 divided by 4.0, where 𝑅 implicitly is in units of
centimeters. To two significant figures as a
decimal, 𝑅 is equal to 6.5 centimeters. That’s the radius of the circle in
which the larger particle moves.

Let’s summarize what we’ve learned
about center of mass. We’ve seen that the center of mass
of an object is the average location of all its mass. One of the reasons knowing an
object’s center of mass is useful is because a force applied through an object’s
center of mass tends to translate it without rotating it at all. And finally, we saw that, as a
mathematical relationship, an object’s center of mass is equal to the sum of the
product of its mass elements, 𝑚 sub 𝑖, multiplied by each element’s distance from
a reference point, all divided by the sum of the individual mass elements. And we noted that each dimension of
an object’s center of mass, whether 𝑥 or 𝑦 or some other dimension, needs to be
computed separately.