📚 A-Level Mathematics: Determining and Solving Centroids | A-Level数学：质心的确定与求解
The concept of the centroid, or centre of mass, is a fundamental topic in A-Level mechanics and further mathematics. It is the point where the entire weight of a body may be considered to act, and mastering its calculation is essential for solving problems involving equilibrium, stability, and motion.
质心（或称质量中心）是A-Level力学和进阶数学中的一个基础概念。它是物体全部重量可以被认为作用的那一点，掌握其计算方法对于解决涉及平衡、稳定性和运动的问题至关重要。
1. Understanding the Centroid | 理解质心
For a system of particles or a rigid body, the centroid is the geometric centre of mass. In a uniform gravitational field, it coincides with the centre of gravity. For a lamina (a thin flat plate) of uniform density, the centroid depends only on the shape of the lamina, not on its material.
对于一个质点系统或刚体，质心是质量的几何中心。在均匀重力场中，质心与重心重合。对于均匀密度的薄板（二维平面薄片），质心仅取决于薄板的形状，而与其材料无关。
Mathematically, the position of the centroid of a set of discrete particles is the weighted average of their positions, where the weights are their masses. For a continuous body, this average is computed using integration over the volume, area, or length, depending on the dimensionality of the object.
数学上，一组离散质点的质心位置是其位置的质量加权平均值。对于连续体，这一平均值通过对物体的体积、面积或长度进行积分来计算，具体取决于物体的维度。
2. Centroids of Discrete Particle Systems | 离散质点系统的质心
Consider n particles with masses m₁, m₂, …, mₙ located at position vectors r₁, r₂, …, rₙ. The position vector of the centroid R is given by:
R = (m₁r₁ + m₂r₂ + … + mₙrₙ) / (m₁ + m₂ + … + mₙ)
In Cartesian coordinates, the x and y coordinates of the centroid are computed separately. For example, x-bar = Σ(mᵢxᵢ)/Σmᵢ and y-bar = Σ(mᵢyᵢ)/Σmᵢ.
在笛卡尔坐标系中，质心的x和y坐标分别计算。例如，x̄ = Σ(mᵢxᵢ)/Σmᵢ，ȳ = Σ(mᵢyᵢ)/Σmᵢ。
This is analogous to finding the weighted mean of a set of data points. The total mass M = Σmᵢ is used to normalise the weighted sum of moments.
这类似于求一组数据点的加权平均值。总质量M = Σmᵢ用于归一化力矩的加权和。
3. The Centroid of a Lamina by Integration | 薄板质心的积分求法
For a lamina occupying a region R in the xy-plane with uniform density ρ, the mass of an infinitesimal element is dA·ρ, where dA is the area element. The coordinates of the centroid are:
对于在xy平面上占据区域R、具有均匀密度ρ的薄板，无穷小元素的质量为ρdA，其中dA是面积元素。质心坐标为：
x̄ = (∫∫x dA) / (∫∫dA) , ȳ = (∫∫y dA) / (∫∫dA)
Here the denominator is simply the area A of the lamina. The numerator is the first moment of the area about the y-axis and x-axis respectively. For standard shapes, these integrals reduce to simple formulas.
这里分母就是薄板的面积A。分子分别是面积关于y轴和x轴的一次矩。对于标准形状，这些积分可简化为简单的公式。
A common A-Level technique is to use horizontal or vertical strips. If the lamina is bounded by y = f(x) and y = g(x), the area is A = ∫[g(x)−f(x)] dx, and the centroid coordinates are found by integrating the moments of each strip.
A-Level中常用的技巧是使用水平或垂直条带。如果薄板由y=f(x)和y=g(x)界定，则面积A = ∫[g(x)−f(x)] dx，质心坐标通过积分每个条带的力矩来找到。
4. Standard Results for Common Shapes | 常见形状的标准结果
Memorising the centroids of standard shapes saves considerable time in exams. The following table lists key results for uniform density objects.
记忆常见形状的质心可以在考试中节省大量时间。下表列出了均匀密度物体的关键结果。
| Shape | Centroid Location |
| Rectangle (width w, height h) | Centre: (w/2, h/2) |
| Triangle (base b, height h) | h/3 above the base, along the median |
| Circle (radius r) | Centre of the circle |
| Semicircle (radius r) | 4r/(3π) from the flat edge along the symmetry axis |
| Sector of a circle (radius r, angle 2α) | (2r sin α)/(3α) from the centre along the symmetry axis |
These formulas derive from setting up integrals over the respective regions. For example, a triangle’s centroid lies at the intersection of its medians, one-third of the way from the base to the apex.
这些公式来自于对各区域建立积分。例如，三角形的质心位于其三条中线的交点，即从底边到顶点的三分之一处。
5. Composite Bodies: The Principle of Decomposition | 组合体：分解原理
When a lamina or body consists of several simpler shapes, its centroid is found by treating each component as a point mass located at its own centroid. The total moment is the sum of the individual moments, and the total mass (or area) is the sum of the individual masses (or areas).
当一个薄板或物体由多个简单形状组成时，其质心通过将每个组件视为位于自身质心的质点来求解。总力矩是各分力矩之和，总质量（或面积）是各分质量（或面积）之和。
Mathematically, for a composite area made of regions A₁, A₂, …, Aₖ with centroids (x̄₁, ȳ₁), (x̄₂, ȳ₂), …, the overall centroid is:
数学上，对于由区域A₁, A₂, …, Aₖ（质心分别为(x̄₁, ȳ₁), (x̄₂, ȳ₂), …）组成的组合面积，整体质心为：
x̄ = Σ(Aᵢx̄ᵢ) / ΣAᵢ , ȳ = Σ(Aᵢȳᵢ) / ΣAᵢ
An important technical point: holes or cut-outs are treated as negative areas. If you remove a circle from a rectangle, you subtract its area and its contribution to the moment.
一个重要的技术要点：孔洞或切出部分被视为负面积。如果从矩形中去除一个圆，你需要减去其面积及其对力矩的贡献。
6. Symmetry and Simplification | 对称性与简化
Symmetry is a powerful tool. If a lamina is symmetric about a line, the centroid lies on that line. If it has two perpendicular axes of symmetry, the centroid is their intersection. For example, a uniform rectangle’s centroid is simply its centre.
对称性是一个有力的工具。如果薄板关于一条线对称，则质心位于该线上。如果有两条垂直的对称轴，则质心是它们的交点。例如，均匀矩形的质心就是其中心。
This property often reduces a two-dimensional problem to a one-dimensional one. When solving for the centroid of a semicircle, symmetry tells us that x̄ = 0 if the flat edge lies along the y-axis, leaving only ȳ to compute.
这一性质常将二维问题简化为一维问题。在求解半圆的质心时，如果直边沿y轴放置，对称性告诉我们x̄ = 0，只需计算ȳ。
In compound shapes, look for symmetry first before carrying out full integration. A shape with a line of symmetry allows you to ignore all moment calculations perpendicular to that line.
在组合形状中，先寻找对称性再进行完整积分。具有对称轴的形状可以忽略垂直于该轴的所有力矩计算。
7. Worked Example: Triangular Lamina | 例题解析：三角形薄板
Let the vertices of a triangle be at (0,0), (b,0), and (0,h). This is a right-angled triangle. The centroid has coordinates (b/3, h/3). Let us verify this by integration.
设三角形的顶点为(0,0)、(b,0)和(0,h)。这是一个直角三角形。其质心坐标为(b/3, h/3)。我们通过积分来验证这一点。
The hypotenuse has equation y = h − (h/b)x. Using vertical strips of width dx and height y(x), the area of each strip is y(x)dx. The total area A = ½bh, as expected.
斜边方程为y = h − (h/b)x。使用宽度为dx、高度为y(x)的垂直条带，每个条带的面积为y(x)dx。总面积A = ½bh，符合预期。
The first moment about the y-axis is ∫x·y(x) dx from 0 to b = ∫x(h − hx/b) dx = h[x²/2 − x³/(3b)] from 0 to b = hb²/6. Dividing by A = hb/2 gives x̄ = b/3.
关于y轴的一次矩为从0到b的∫x·y(x) dx = ∫x(h − hx/b) dx = h[x²/2 − x³/(3b)]，从0到b的值为hb²/6。除以A = hb/2得x̄ = b/3。
Similarly, the moment about the x-axis gives ȳ = h/3. Thus the centroid is indeed at (b/3, h/3), one-third of the way along each leg from the right-angle vertex.
类似地，关于x轴的矩给出ȳ = h/3。因此质心确实位于(b/3, h/3)，即从直角顶点沿每条直角边的三分之一处。
8. Worked Example: Composite Shape | 例题解析：组合形状
A lamina consists of a rectangle of width 4m and height 2m, with a semicircle of radius 1m attached to the top edge, centred above the midpoint of the top edge. The left vertex of the rectangle is at (0,0). Find the centroid.
一块薄板由一个宽4m、高2m的矩形和一个半径1m的半圆组成，半圆连接在矩形顶边中点上方。矩形左下角位于(0,0)。求质心。
The rectangle has area A₁ = 8 m², centroid at (2,1). The semicircle has area A₂ = ½π(1)² = π/2 m², and its centroid is located π/4 units above its straight edge. Since the straight edge of the semicircle lies at y = 2, its centroid is at (2, 2 + 4/(3π)).
矩形的面积为A₁ = 8 m²，质心在(2,1)。半圆的面积为A₂ = ½π(1)² = π/2 m²，其质心位于其直边上方4/(3π)处。由于半圆的直边在y = 2处，其质心在(2, 2 + 4/(3π))。
By symmetry, x̄ = 2. For ȳ, we use the composite formula:
根据对称性，x̄ = 2。对于ȳ，我们使用组合公式：
ȳ = [8(1) + (π/2)(2 + 4/(3π))] / [8 + π/2]
Simplifying the numerator: 8 + π + 2/3 = 8.667 + 3.142 + 0.667 ≈ 12.476. The denominator is 8 + 1.571 = 9.571. Thus ȳ ≈ 1.303 m. The centroid is approximately (2, 1.303).
简化分子：8 + π + 2/3 = 8.667 + 3.142 + 0.667 ≈ 12.476。分母为8 + 1.571 = 9.571。因此ȳ ≈ 1.303 m。质心约为(2, 1.303)。
9. Pappus’ Theorems | 帕普斯定理
Pappus’ first theorem states that the volume of a solid of revolution obtained by rotating a plane curve about an external axis in its plane is equal to the product of the initial area and the distance travelled by the centroid of the area.
帕普斯第一定理指出：将平面曲线绕其平面内的一条外部轴旋转所得的旋转体体积，等于初始面积与面积的质心所行经距离的乘积。
If area A is rotated through an angle θ about an axis, the volume is V = A·θ·d, where d is the distance from the centroid to the axis. For a full revolution, θ = 2π, so V = 2πAd.
若面积A绕轴旋转角度θ，则体积为V = A·θ·d，其中d是质心到轴的距离。对于完整旋转，θ = 2π，所以V = 2πAd。
Pappus’ second theorem applies to surface areas: the surface area generated by rotating a plane curve is S = θ·L·d, where L is the length of the curve. These theorems provide elegant shortcuts for finding centroids of unusual shapes.
帕普斯第二定理适用于表面积：旋转平面曲线所产生的表面积为S = θ·L·d，其中L是曲线长度。这些定理为求不规则形状的质心提供了简洁的捷径。
10. Centroids of Curved Wires | 曲线细线的质心
For a thin wire or rod bent into a curve, the centroid is found by integrating along the arc length. If the wire has uniform density, the centroid of a circular arc of radius r subtending angle 2α at the centre lies on the angle bisector at distance:
对于弯成曲线的细线或细杆，质心通过沿弧长积分来求解。若细线密度均匀，圆心角为2α、半径为r的圆弧的质心位于角平分线上，距圆心距离为：
d = r sin α / α
The derivation of this involves parametrising the arc and averaging x = rcosθ and y = rsinθ over the range −α to α. The arc length is 2rα, forming the denominator of the average.
该公式的推导涉及将弧参数化，并在−α到α范围内对x = rcosθ和y = rsinθ取平均。弧长为2rα，作为平均值的分母。
This formula is particularly useful in mechanics problems involving hanging wires or rotating links. Note the similarity to the sector formula: the sector’s centroid is further from the centre than the arc’s centroid for the same angle.
该公式在涉及悬垂细线或旋转连杆的力学问题中特别有用。注意与扇形公式的相似性：在相同角度下，扇形的质心比圆弧的质心离圆心更远。
11. Equilibrium and Stability Applications | 平衡与稳定性应用
When an object is suspended from a point, it comes to rest with its centroid directly below the point of suspension. This principle allows practical determination of centroids: suspend the object from two different points, and the centroid is where the vertical lines through the suspension points intersect.
当一个物体从一点悬挂时，它静止时质心位于悬挂点的正下方。这一原理允许实际确定质心：从两个不同点悬挂物体，通过悬挂点的铅垂线的交点就是质心。
For an object resting on a slope, stability requires that a vertical line through the centroid falls within the base of support. If it lies outside the base, the object topples over. This has direct applications in structural engineering and vehicle design.
对于放置在斜面上的物体，稳定性要求通过质心的铅垂线落在支撑基底内。如果铅垂线在基底外，物体将倾覆。这在结构工程和车辆设计中有直接应用。
In A-Level exams, students are often asked to find the maximum angle of a slope before a composite object topples. This requires locating the centroid and then setting the line through the centroid and the pivot point perpendicular to the slope.
在A-Level考试中，学生常被要求求组合物体在倾覆前斜面能够达到的最大角度。这需要定位质心，然后使通过质心和支点的连线垂直于斜面。
12. Exam Strategy and Common Pitfalls | 考试策略与常见陷阱
Students frequently forget that for cut-outs, the removed area must be treated as negative. Another common error is confusing the centroid of a sector with that of an arc. Always check the units: the centroid must lie inside the shape for convex figures.
学生经常忘记对于切出部分，移除的面积必须作为负来处理。另一个常见错误是混淆扇形质心与圆弧质心。始终检查单位：对于凸图形，质心必须位于形状内部。
When using composite formulas, ensure all component centroids are measured from the same origin. A diagram with a clearly marked origin is essential. Also, verify the physical plausibility of your final answer—if the centroid lies outside the lamina, re-examine your calculations.
使用组合公式时，确保所有分量质心都从同一原点测量。标记清晰原点的图至关重要。同时，验证最终答案的物理合理性——如果质心位于薄板之外，请重新检查计算。
Finally, remember that for uniform bodies, the centroid, centre of gravity, and centre of mass coincide. This equivalence is the reason we can use geometric area as a proxy for mass in all the formulas above.
最后，记住对于均匀物体，质心、重心和质量中心重合。这种等价性正是我们可以在上述所有公式中使用几何面积来代替质量的原因。
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