Video: Center of Mass | Nagwa Video: Center of Mass | Nagwa

# Video: Center of Mass

In this video we learn what center of mass is, why it is useful, and how to compute center of mass of any mass distribution along any dimension.

12:58

### 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.

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