Video: Center of Mass of a Variable Density Object

A 0.75 m long rod of iron with a density of 8.0 g/cmΒ³ is joined end to end with a 0.75 m long rod of copper with a density of 2.7 g/cmΒ³. 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?

03:44

Video Transcript

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 β€” we’ve called it 𝑑 sub cm β€” 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 β€” we’ve called them π‘š sub 𝑖 β€” 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 π‘š sub i and π‘š 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 β€” the density of iron and the position of its center of mass, the density of copper and the position of its center of mass β€” 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.

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