Whether in mathematical problems or physical applications, we often encounter similar problems: finding the center of mass (centroid) of the remaining part after a regular object has a regular part removed from (or added to) it.
The so-called center of gravity refers to the point where the gravitational forces acting on all parts of an object are concentrated. This point is called the center of gravity of the object. The center of mass, abbreviated as the centroid, refers to an imaginary point on a material system where the mass is considered to be concentrated. The concept of the center of mass is meaningful even in systems without a gravitational field, whereas the center of gravity is not. If the gravitational field is uniform, the center of mass and the center of gravity of the same material system coincide at the same imaginary point; in a non-uniform gravitational field, they usually do not.
The most direct method to find the center of mass is, of course, using multiple integrals. However, this method is not only complex but also difficult because the functions for arbitrarily given planes or objects are often not easily defined. Therefore, I have organized three methods below to deal with the aforementioned problems. These three methods start from different perspectives, each with its own ingenuity. Finally, a fourth method is provided for finding the center of gravity of irregular thin sheets.
I. Pappus’s Centroid Theorem
A finite plane moving in a direction everywhere perpendicular to it sweeps an area equal to the distance traveled by the centroid multiplied by the curve length, or the volume swept by a plane moving at a velocity perpendicular to it equals the distance traveled by the centroid multiplied by the area of the plane.
This sentence might be difficult to understand at first reading. In fact, you can focus on the second half: the direction of velocity is perpendicular to the plane. In this way, the movement mentioned above does not only refer to translation; in other words, the movement path is not necessarily a straight line, but can also be a rotation around a fixed axis. Here is a brief discussion of the proof of this theorem:
Let the area of the plane be S and the mass be M. Divide the plane into infinite small planes, each with an area s_i. The definition of the centroid is \vec{R}_{c}=\frac{\sum \vec{R}_{i}m_i}{M}. For a uniform plane, m_i \propto s_i, so the definition of the centroid can be rewritten as: \vec{R}_{c}=\frac{\sum \vec{R}_{i}s_i}{S} Differentiating both sides, we get: \begin{aligned}d\vec{R}_{c}=d\left(\frac{\sum \vec{R}_{i}s_i}{S}\right)=\frac{\sum d(\vec{R}_{i})s_i}{S} \\ S|d\vec{R}_{c}|=\sum |d(\vec{R}_{i})|s_i\end{aligned}
If the direction of motion is perpendicular to the plane, i.e., d(\vec{R}_{i}) is perpendicular, then the instantaneous swept area is s_i and the instantaneous swept length is |d(\vec{R}_{i})|, so the instantaneous swept volume is |d(\vec{R}_{i})|s_i. The volume swept over a period of time is: \int \left[\sum |d(\vec{R}_{i})|ds\right]=\int (S|d\vec{R}_{c}|)=S\int |d\vec{R}_{c}| The rightmost term is exactly the product of the distance traveled by the centroid and the area. The proof of Pappus’s theorem is complete.
The content above is just to help everyone understand Pappus’s theorem. If it is purely for application, you don’t need to worry about it. Now we can apply this theorem; we can use it to find volume or the centroid. For example, in the problem we proposed at the beginning: since this figure is symmetrical, the centroid must be on the axis of symmetry. Let the deficient circle rotate one revolution around the axis indicated by the red line in the figure below. We find the location of the centroid by using the "volume divided by area" approach.
According to Pappus’s theorem, the volume formed by the large circle rotating one revolution is \pi R^2 \cdot 2\pi R, and the volume formed by the small circle rotating one revolution is \pi r^2 \cdot 2\pi d. Thus, the volume of this solid of revolution is 2\pi^2(R^3-r^2 d). The area of the original figure is S=\pi(R^2-r^2). Therefore, the distance traveled by the centroid is \frac{2\pi^2(R^3-r^2 d)}{\pi(R^2-r^2)}=\frac{2\pi(R^3-r^2 d)}{R^2-r^2}.
The centroid is on AB, so the path it travels must also be a circle. We have 2\pi R_c=\frac{2\pi(R^3-r^2 d)}{R^2-r^2}, so the distance of the centroid from point P is R_c=\frac{R^3-r^2 d}{R^2-r^2}.
Pappus’s theorem is very useful for dealing with such problems and can conveniently find the volume of certain solids. However, it also has drawbacks, such as only being applicable to planar figures and sometimes involving cumbersome calculations.
II. Common Centroid of Centroids (Step-by-step Method, Incremental Method)
The common centroid of n particles is equal to the common centroid of the common centroid of a particles and the common centroid of the remaining (n-a) particles.
We already know the mass and centroid of the large circle before it was hollowed out and those of the small circle that was removed, so we can use the above method. For combinations of several solids, we can handle them very conveniently; for the case of being "hollowed out," the mass of the removed part needs to be recorded as a negative number. I call this method the step-by-step method or incremental method.
Continuing with the same problem, the distance between the centroids of the large and small circles is (R-d), and the ratio of the masses of the large and small circles is R^2:(-r^2). According to the definition of the centroid (similar to lever balance), we can find that the distance of the centroid from the center of the large circle is \frac{-r^2}{R^2+(-r^2)}(R-d) (this is a negative value, meaning it has moved away from the line segment between the two centers). Therefore, the distance of the centroid from point P is R-\frac{-r^2}{R^2+(-r^2)}(R-d)=\frac{R^3-r^2 d}{R^2-r^2}.
III. Gravitational Potential Energy Method
Dear readers, you did not read it wrong, and I did not type it wrong; it is indeed "gravitational potential energy." This is something BoJone suddenly thought of using to find the centroid while reading about gravitational potential energy in "Mechanics." Here, a physical law is applied:
The gravitational potential energy of an object is equal to the distance between the object’s centroid and the zero-potential surface, multiplied by the gravity acting on the object.
With this law, and according to the additivity of energy, we can easily find the location of the object’s centroid.
In the original problem, choose the red line part in the figure above as the zero-potential surface. Assuming the density of the circle is 1, the gravitational potential energy of the large circle is \pi R^2 \cdot g \cdot R, and the gravitational potential energy of the small circle is \pi r^2 \cdot g \cdot d. Therefore, the gravitational potential energy of the object is E_p=\pi R^2 \cdot g \cdot R - \pi r^2 \cdot g \cdot d. The mass of the object is M=\pi R^2 - \pi r^2, so the distance from the object to the zero-potential surface is \frac{E_p}{Mg}=\frac{R^3-r^2 d}{R^2-r^2}.
By now, I have basically finished explaining these three methods (exhausted...). These three methods start from different angles, including both mathematical and physical ones, and ultimately yield the same result. In fact, BoJone loves calculus but is not proficient in it, so I didn’t want to find the centroid from the perspective of multiple integrals and thus looked for the "shortcuts" above. In the process of looking for shortcuts, BoJone first considered using physical methods, initially thinking of using the law of universal gravitation, but then thought about the non-additivity of force and gave up. By chance, I saw "gravitational potential energy of a rigid body" and immediately realized! Pappus’s theorem was discovered during the process of searching for materials, and the second method was one BoJone had thought of before to find centroids, and now it has come in handy, haha!
Finally, I will introduce a "plumb line method" for finding the center of gravity of irregular thin plates. In fact, no matter how good your calculation skills are, it is unrealistic to use mathematical analysis to find the center of gravity in practice. We need to quickly determine the center of gravity in practice (even if it is not very precise). For a thin plate (not necessarily uniform), choose a fixed point (such as A and B) on it, tie a thin line to it, pull one end of the line, let the thin plate and the other end of the line hang naturally, and record the position of the line on the thin plate. Repeat the above operation with another fixed point. The intersection of the two lines drawn on the thin plate is the center of gravity of the thin plate.
When reposting, please include the original address of this article: https://kexue.fm/archives/759
For more detailed reposting matters, please refer to: Scientific Space FAQ