Video Transcript
A uniform rod 𝐴𝐶 of length of 36 centimeters was bent from point 𝐵, where 𝐴𝐵 equals 36 over five centimeters, and the measure of 𝐴𝐵𝐶 is 90 degrees. The rod was then freely suspended from 𝐴. Find the tangent of the angle that 𝐵𝐶 makes to the horizontal.
Let’s start with the diagram of this scenario. We have a rod 𝐴𝐵𝐶 that is bent at a point 𝐵. The measure of the angle 𝐴𝐵𝐶 is a right angle and the length of 𝐴𝐵 is 36 over five centimeters. Since the total length of the rod is 36 centimeters, the length of 𝐵𝐶 is 36 minus 36 over five centimeters. This simplifies to 144 over five centimeters. The rod has a center of mass at the point 𝑃, which we can roughly guess is about here. When the rod is freely suspended from 𝐴, its center of mass will hang directly below it. The straight line through 𝐴 and 𝑃 is therefore the vertical. The straight line perpendicular to this line is therefore the horizontal.
We are asked to find the tangent of the angle that 𝐵𝐶 makes to the horizontal, which is this angle here, 𝜃. Since the horizontal is perpendicular to the vertical, this angle in here is a right angle. These three angles above the line 𝐵𝐶 must add up to 180 degrees. Therefore, the angle here is 90 minus 𝜃. And since the angles in this triangle must also add up to 180 degrees, this angle here, 𝐵𝐴𝑃, is 𝜃.
Let’s define a coordinate system with the origin at 𝐴 and positive 𝑥 going to the right and positive 𝑦 going vertically downwards. So, for example, the 𝑦-coordinate of the point 𝐵 is 36 over five. Now, the tangent of the angle 𝜃 will be given by the 𝑥-coordinate of the center of mass, COM 𝑥, over the 𝑦-coordinate of the center of mass, COM 𝑦. We therefore need to find the horizontal distance of the center of mass from 𝐴 and the vertical distance of the center of mass from 𝐴. Recall that the center of mass of a uniform rod lies at its midpoint, so uniform rod of length 𝑙 will have its center of mass 𝑙 over two from both ends.
In our case, the uniform rod is bent, but we can model it as two separate uniform rods, 𝐴𝐵 and 𝐵𝐶. The centers of mass of each of these rods will be at their midpoints. In this coordinate system, 𝐴𝐵 is vertical, so the 𝑥-coordinate of its center of mass will be zero and its 𝑦-coordinate will be at its midpoint, which is halfway along its length of 36 over five, so 18 over five. Likewise, the center of mass of the rod 𝐵𝐶 clearly has a 𝑦-coordinate of 36 over five and an 𝑥-coordinate at its midpoint, so half of 144 over five, which is 72 over five.
We can now model these two uniform rods as particles from their centers of mass. Recall that for the center of mass of a system of particles, the 𝑥-coordinate is given by the sum of the products of the individual masses 𝑚 𝑖 multiplied by their 𝑥-coordinates 𝑥 𝑖 divided by the total mass, and similarly for the 𝑦-coordinate. We do not currently know the masses of the rods, but since they are uniform rods, they have uniform density. Therefore, their masses will be proportional to their lengths. Since the rod is uniform, its equilibrium position is independent of its density. So let us assume that the rod has a density of one kilogram per centimeter. The mass of the first rod 𝐴𝐵 is therefore 36 over five kilograms. And the mass of the second rod 𝐵𝐶 is 144 over five kilograms.
We now have everything we need to find the center of mass of the complete rod. For the 𝑥-coordinate of the center of mass, we have the mass of 𝐴𝐵, which is 36 over five kilograms, multiplied by its 𝑥-coordinate, which is zero, and then the mass of 𝐵𝐶, which is 144 over five kilograms, multiplied by its 𝑥-coordinate, 72 over five. We then divide it by the total mass, which is 36 over five plus 144 over five. This simplifies to 288 over 25, which we can place in our formula for tan 𝜃.
Now, we need to repeat the process for the 𝑦-coordinate. We take the mass of the first rod 𝐴𝐵, which is 36 over five, multiplied by the 𝑦-coordinate of its center of mass, 18 over five. Then, we take the mass of the second rod 𝐵𝐶, which is 144 over five, multiplied by the 𝑦-coordinate of its center of mass, which is 36 over five, and then divide by the total mass, which is 36 over five plus 144 over five. And this simplifies to 162 over 25. We can substitute this into our expression for tan 𝜃, which gives us 288 over 25 all over 162 over 25. The 25s will cancel giving us 288 over 162. And simplifying gives us the final answer, the tan of the angle that 𝐵𝐶 makes to the horizontal, 16 over nine.
