📚 Using Integration to Find Centre of Mass and Moment of Inertia | 用积分求质心与转动惯量
In the IB Mathematics curriculum, integration is not only a tool for finding areas and volumes; it also enables us to locate the centre of mass of a body and to compute its moment of inertia. These two quantities are central to mechanics and appear frequently in exam questions that blend calculus with physics.
在IB数学课程中，积分不仅是用来求面积和体积的工具，它还能帮助我们确定物体的质心位置并计算其转动惯量。这两个量是力学中的核心概念，也经常出现在结合微积分与物理的考题中。
1. What Are Centre of Mass and Moment of Inertia? | 什么是质心与转动惯量？
The centre of mass is the unique point at which the entire mass of an object can be considered to be concentrated for many dynamic purposes. For a system of discrete particles, the centre of mass coordinates are weighted averages of the particle positions, where the weights are the masses.
质心是这样一个特殊的点：在解决许多动力学问题时，可以把物体的全部质量看作集中在这一个点上。对于由离散质点组成的系统，质心坐标是各质点位置的加权平均值，权重就是各质点的质量。
The moment of inertia, on the other hand, measures how difficult it is to change the rotational motion of an object about a given axis. It depends not only on the total mass but also on how that mass is distributed relative to the axis.
而转动惯量衡量的是物体绕给定轴改变转动状态的难易程度。它不仅取决于总质量，还取决于质量相对于转轴的分布情况。
2. Coordinates of the Centre of Mass for Discrete Systems | 离散系统的质心坐标
For point masses m₁, m₂, …, mₙ located at coordinates (x₁, y₁), (x₂, y₂), …, (xₙ, yₙ), the centre of mass is given by:
对于位于坐标 (x₁, y₁), (x₂, y₂), …, (xₙ, yₙ) 的质点 m₁, m₂, …, mₙ，其质心坐标为：
x̄ = (Σ mᵢ xᵢ) / (Σ mᵢ), ȳ = (Σ mᵢ yᵢ) / (Σ mᵢ)
Here the Greek capital sigma Σ denotes summation over all particles. This weighted-average idea is the starting point for passing to continuous distributions by integration.
这里的希腊大写字母Σ表示对所有质点求和。这种加权平均思想是通过积分过渡到连续分布的出发点。
3. Centre of Mass of a One-Dimensional Continuous Rod | 一维连续细棒的质心
Suppose a thin rod lies along the x-axis from x = 0 to x = L. If its linear density is λ(x), then the element of mass between x and x + dx is dm = λ(x) dx. The centre of mass is therefore:
假设一根细棒沿x轴放置，从x = 0延伸到x = L。如果其线密度为λ(x)，则在x与x + dx之间的质量微元为dm = λ(x) dx。因此质心为：
x̄ = (∫₀ᴸ x λ(x) dx) / (∫₀ᴸ λ(x) dx)
If the density is constant, λ(x) = λ₀, then the numerator is λ₀ ∫₀ᴸ x dx = λ₀ L²/2, and the denominator is λ₀ L. Hence x̄ = L/2, as expected for a uniform rod.
如果密度恒定，即λ(x) = λ₀，则分子为λ₀ ∫₀ᴸ x dx = λ₀ L²/2，分母为λ₀ L，因此x̄ = L/2，这正是均匀细棒质心在正中间的预期结果。
A useful exam case is a rod with linearly increasing density λ(x) = kx. Then:
一个常见的考试情形是密度线性增加的细棒λ(x) = kx。此时：
x̄ = (∫₀ᴸ k x² dx) / (∫₀ᴸ k x dx) = (k L³/3) / (k L²/2) = 2L/3
Notice that the centre of mass shifts toward the denser end of the rod. This example illustrates how integration naturally handles non-uniform distributions.
注意质心会向密度较大的一端偏移。这个例子说明积分如何自然地处理非均匀分布。
4. Centre of Mass of a Two-Dimensional Lamina | 二维平面薄板的质心
For a lamina of area A with uniform surface density σ, the centre of mass coincides with the centroid of the area. If the lamina occupies a region R, then:
对于面积为A、面密度σ均匀的薄板，其质心与区域的形心一致。若薄板占据区域R，则：
x̄ = (1/A) ∫∫ x dA, ȳ = (1/A) ∫∫ y dA
When the region is bounded by y = f(x), the x-axis, and the lines x = a and x = b, these double integrals reduce to single integrals:
当区域由y = f(x)、x轴以及直线x = a和x = b围成时，这些二重积分可以化为单积分：
x̄ = (∫ₐᵇ x f(x) dx) / (∫ₐᵇ f(x) dx), ȳ = (∫ₐᵇ ½ [f(x)]² dx) / (∫ₐᵇ f(x) dx)
The second formula comes from the fact that the y-coordinate of the centre of the vertical strip is f(x)/2, and its area is f(x) dx. Multiplying by the strip’s y-coordinate gives ½ [f(x)]² dx.
第二个公式的来源是：竖条的质心纵坐标为f(x)/2，面积为f(x) dx。用竖条的质心纵坐标乘以面积，就得到½ [f(x)]² dx。
For example, the region under y = x² from x = 0 to x = 1 has area A = ∫₀¹ x² dx = 1/3. Then x̄ = ∫₀¹ x·x² dx / (1/3) = (1/4)/(1/3) = 3/4, and ȳ = ∫₀¹ ½ x⁴ dx / (1/3) = (1/10)/(1/3) = 3/10.
例如，由y = x²与x轴在x = 0到x = 1之间围成的区域，其面积A = ∫₀¹ x² dx = 1/3。于是x̄ = ∫₀¹ x·x² dx / (1/3) = (1/4)/(1/3) = 3/4，而ȳ = ∫₀¹ ½ x⁴ dx / (1/3) = (1/10)/(1/3) = 3/10。
5. Moment of Inertia: Definition | 转动惯量的定义
For a single particle of mass m at perpendicular distance r from an axis, the moment of inertia about that axis is defined as I = m r². For a continuous body, we sum infinitesimal contributions:
对于质量m、到转轴垂直距离为r的单个质点，其对该轴的转动惯量定义为I = m r²。对于连续物体，我们将所有微小贡献求和：
I = ∫ r² dm
The dimension of moment of inertia is mass × length², so its SI unit is kg·m². A smaller moment of inertia means that the object is easier to spin about that axis.
转动惯量的量纲是质量 × 长度²，因此其国际单位是kg·m²。转动惯量越小，说明物体绕该轴越容易转动。
6. Computing Moment of Inertia by Integration | 用积分计算转动惯量
In practice, we express dm in terms of the coordinate along the object. For a rod of length L and uniform linear density λ, rotating about one end:
在实际计算中，我们要把dm用物体上的坐标表示。对于长为L、线密度均匀为λ的细棒绕其一端转动：
I = ∫₀ᴸ x² λ dx = λ L³/3 = 1/3 M L²
For a uniform circular disc
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