Lesson Video: Center of Gravity of Uniform Rods | Nagwa Lesson Video: Center of Gravity of Uniform Rods | Nagwa

Lesson Video: Center of Gravity of Uniform Rods Mathematics

In this video, we will learn how to find the center of gravity of uniform rods.

15:52

Video Transcript

In this video, we’ll 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 object’s 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 let’s 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 𝑚 sub 𝑖 with coordinates 𝑥 sub 𝑖, 𝑦 sub 𝑖, the 𝑥-coordinate of the center of mass is given by the sum of 𝑚 sub 𝑖 times 𝑥 sub 𝑖 divided by the sum of 𝑚 sub 𝑖. And the 𝑦-coordinate is given by the sum of 𝑚 sub 𝑖 times 𝑦 sub 𝑖 divided by the sum of 𝑚 sub 𝑖. So now that we have the formulae and the various definitions, we’re 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 𝐴𝐵 is a uniform rod of length four centimeters and mass four kilograms. A mass of magnitude five kilograms is fixed at 𝐴, and another mass of magnitude one kilogram is fixed at 𝐵. Find the distance from the center of gravity of the system to 𝐴.

The first piece of information we’re given is that the rod, which is represented by a line segment 𝐴𝐵, 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. Let’s use this information and the information about the masses that are fixed at 𝐴 and 𝐵 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. Let’s imagine then that point 𝐴 is at the origin and the line segment 𝐴𝐵 lies along the 𝑥-axis as shown.

Since the length of the rod is four centimeters, if we let centimeters be our length unit, then point 𝐵 must have coordinates four, zero. Then the midpoint of the line segment 𝐴𝐵 is found by calculating the sum of the individual coordinates and dividing by two. So that’s zero plus four divided by two and zero plus zero divided by two, which gives us the coordinates two, zero. Now, in fact, we’re really only interested in the value of the 𝑥-coordinates. But we’ve completed the value of the 𝑦-coordinates just for completeness.

Next, we recall that if a system of particles has mass 𝑚 sub 𝑖 at coordinates 𝑥 sub 𝑖, 𝑦 sub 𝑖, then the 𝑥- and 𝑦-coordinates of the center of mass are as shown. Since the three centers of mass that we’re interested in lie along the 𝑥-axis, we’re just going to use the calculation for the 𝑥-coordinates. Let’s begin by working out the sum of the products of 𝑚 sub 𝑖 and 𝑥 sub 𝑖. The mass of particle 𝐴 is five kilograms, and its 𝑥-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 𝐵 is one kilogram, and it’s located four units along the 𝑥-axis. And so the sum of 𝑚 sub 𝑖 times 𝑥 sub 𝑖 is 12 or 12 kilogram centimeters.

The denominator of the fraction is the sum of 𝑚 sub 𝑖. And that’s simply the sum of the masses of all the objects in the system. It’s five plus four plus one, which is equal to 10, or 10 kilograms. The center of mass is the quotient of these, so it’s 12 divided by 10, or 1.2. And so the 𝑥-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 𝑥-axis and point 𝐴 lies at the origin, the distance from the center of mass or center of gravity of the system to 𝐴 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. It’s also worth noting that if each rod in the wireframe has the same density, then the value of this density doesn’t influence the center of mass. And so we can disregard it from our calculations altogether. Given this information then, let’s demonstrate how to find the center of mass of a system of uniform rods.

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.

In our final example, we’ll 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 𝐴𝐵𝐶𝐷, where the measure of angle 𝐵 equals the measure of angle 𝐶, which equals 90 degrees; 𝐴𝐵 equals 494 centimeters; 𝐵𝐶 equals 105 centimeters; and 𝐶𝐷 equals 134 centimeters. Find the distance between the center of gravity of the wire and point 𝐵, rounding your answer to two decimal places.

Before we attempt to answer this question, we’re going to begin by sketching out trapezoid 𝐴𝐵𝐶𝐷. We know that the measure of angle 𝐵 and the measure of angle 𝐶 are 90 degrees. This tells us that side lengths 𝐴𝐵 and 𝐶𝐷 must be the parallel sides of the trapezoid. Let’s add the trapezoid to the coordinate plane. And we’ll let point 𝐵 be at the origin, since that’s the point we’re going to be working with a little later on.

Now, before we go any further, we’re actually going to calculate the length of line segment 𝐴𝐷. And we’ll see why that’s 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 𝑥. This tells us that the square of the hypotenuse, so 𝑥 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 𝑥 squared is 140625. The square root of this gives us a length of 375 centimeters.

Now this is really useful because we’re 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 𝐴 and 𝐵 is 494 kilograms, 𝐵𝐶 is 105 kilograms, 𝐶𝐷 134 kilograms, and 𝐴𝐷 375 kilograms.

The next thing that we’re 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 we’ll work out the coordinates of each midpoint. Line segment 𝐴𝐵 lies on the 𝑦-axis, so the 𝑥-coordinate of its midpoint will be zero. Then its 𝑦-coordinate is the average of the 𝑦-coordinate of 𝐴 and 𝐵. So that’s 494 plus zero divided by two, which is 247. Then, line segment 𝐵𝐶 lies on the 𝑥-axis. So the 𝑦-coordinates of its midpoint is zero. Then, the 𝑥-coordinate of its midpoint is half of 105, say 52.5.

To get to the midpoint of 𝐶𝐷, we move 105 units along the 𝑥-axis. And then we move a half of 134 units up. So its 𝑥-coordinate is 105 and its 𝑦-coordinate is 67. Then, we have the midpoint of 𝐴𝐷. The midpoint of its 𝑥-coordinate is, in fact, 52.5. Once again, it’s a half of 105. Then the 𝑦-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 particle’s coordinates weighted with its mass.

So we’ll begin by calculating the 𝑥-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 it’s the sum of all elements in the first row itself. That gives us 39270 over 1108.

Let’s clear some space and repeat this for the 𝑦-coordinate of the center of mass. To find the numerator of the 𝑦-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 𝐵. And we know that point 𝐵 is the origin. And so we’re 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 𝑥- and 𝑦-coordinates, which is 227.28. The distance between the center of gravity of the wire and point 𝐵, correct to two decimal places, is 227.28 centimeters.

Let’s 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 it’s 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.

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