📚 Centre of Mass and Moment of Inertia in IB Mathematics | IB数学中的质心与转动惯量
In IB Mathematics, particularly in the Analysis and Approaches Higher Level and Applications and Interpretation Higher Level courses, the concepts of centre of mass and moment of inertia extend the fundamental ideas of integration into the realms of mechanics and physical reasoning. These topics require you to model mass distributions, apply definite integrals to compute first and second moments, and grasp the significance of symmetry and theorems such as parallel and perpendicular axes. A thorough understanding not only prepares you for exam questions on distributed mass systems but also deepens your appreciation of how calculus describes real-world rotational and translational behaviour.
在IB数学课程中，尤其是在分析与方法高级课程以及应用与解释高级课程里，质心和转动惯量的概念将积分的基本思想延伸到了力学和物理推理的领域。这些专题要求你建立质量分布模型，运用定积分计算一次矩和二次矩，并理解对称性以及平行轴定理和垂直轴定理的意义。透彻掌握这些内容不仅能让你从容应对关于分布质量系统的考试题目，还能加深你对微积分如何描述真实世界中转动与平动行为的理解。
1. Defining Centre of Mass and First Moment | 质心与一次矩的定义
ique point where the entire mass may be considered to be concentrated for the purpose of analysing translational motion. Mathematically, it is obtained by balancing the first moments of mass. For discrete particles, if particles of masses m₁, m₂, ... , mₙ are placed at positions x₁, x₂, ... , xₙ along a line, the centre of mass x̄ is given by x̄ ting, a differential element dm contributes a first moment x dm, and integration yields the total moment.
质心是一个系统的唯一一点，在分析平动时，可以认为全部质量都集中在该点。从数学上讲，它通过平衡质量的一次矩来求得。对于离散质点，如果质量为 m₁, m₂, ... , mₙ 的质点分别位于直线上的位置 x₁, x₂, ... , xₙ，则质心 x̄ 由公式 x̄ = (Σ mᵢ xᵢ) / (Σ mᵢ) 给出。在连续情形中，一个微元 dm 贡献一次矩 x dm，通过积分可求得总力矩。
2. Centre of Mass for a System of Particles in Two Dimensions | 二维质点系的质心
When particles lie in a plane, coordinates (x̄, ȳ d separately. For particles of masses mᵢ at (xᵢ, yᵢ), the centre of mass has coordinates x̄ = (Σ mᵢ xᵢ) / M and ȳ t is important to recognise that the centre of mass is a weighted average of the particles' positions, with masses serving as weights.
当质点在平面内时，坐标 (x̄, ȳ) 需要分别求得。对于位于 (xᵢ, yᵢ) 处、质量为 mᵢ 的质点系，质心坐标为 x̄ = (Σ mᵢ xᵢ) / M 及 ȳ = (Σ mᵢ yᵢ) / M，其中 M = Σ mᵢ。这些公式直接源于对 y 轴和 x 轴的一次矩的定义。需要认识到，质心是质点位置以质量为权重的加权平均值。
3. Centre of Mass of a Uniform Rod by Integration | 通过积分求均匀细杆的质心
h dx has mass dm = λ dx. Its first moment about the origin is x dm. The total moment is ∫₀ᴸ x λ dx = λ [x²/2]₀ᴸ = λ L²/2. Dividing by the total mass M = λL gives x̄ = L/2, as expected from symmetry.
对于一根放置在 x 轴上从 x = 0 到 x = L 的均匀细杆，线密度为常数 λ = M/L。长度为 dx 的小段具有质量 dm = λ dx。其对原点的一次矩为 x dm。总力矩为 ∫₀ᴸ x λ dx = λ [x²/2]₀ᴸ = λ L²/2。除以总质量 M = λL，得到 x̄ = L/2，这与对称性预期一致。
4. Centre of Mass of a Lamina with Variable Density | 变密度薄片的质心
ement is dm = σ dA. The coordinates of the centre of mass are given by double integrals: x̄ = (1/M) ∬ x σ dA, ȳ = (1/M) ∬ y σ dA, where M = ∬ amina is symmetric about a line and the density obeys the symmetry, the centre of mass lies on that line.
考虑占据 xy 平面内某个区域且具有面密度 σ(x,y) 的薄片。质量微元为 dm = σ dA。质心坐标由二重积分给出：x̄ = (1/M) ∬ x σ dA, ȳ = (1/M) ∬ y σ dA，其中 M = ∬ σ dA。在 IB 考题中，密度通常是常数或随某一坐标线性变化。对称性常常能简化计算：若薄片关于某条直线对称且密度服从该对称性，则质心位于该直线上。
5. Centre of Mass of a Solid of Revolution | 旋转体薄壳或实体的质心
id has uniform density, its centre of mass lies on the axis of rotation, so we only need x̄ s dV = π y² dx. The mass element is dm = ρ π y² dx. The first moment is x dm, leading to x̄ (∫ₐᵇ π y² dx), provided density cancels. The same principle extends to cylindrical shells.
当曲线 y = f(x) 在区间 [a,b] 上绕 x 轴旋转形成旋转体，且该实体具有均匀密度时，其质心位于旋转轴上，因此只需计算 x̄。体积微元是一个半径为 y、厚度为 dx 的圆盘，即 dV = π y² dx。质量微元为 dm = ρ π y² dx。一次矩为 x dm，因此 x̄ = (∫ₐᵇ x π y² dx) / (∫ₐᵇ π y² dx)，此时密度可以约去。同样的原理也适用于柱壳法。
6. Moment of Inertia – The Second Moment of Mass | 转动惯量——质量的二次矩
Moment of inertia I measures an object' here r is the perpendicular distance to the axis. The unit is kg·m². Just as mass is the inertia for translation, moment of inertia is the inertia for rotation.
转动惯量 I 衡量物体对绕给定轴转动加速度的抵抗能力。对于距离轴垂直距离为 r 的点质量 m，有 I = m r²。对于质点系，I = Σ mᵢ rᵢ²。在连续体中，它变为积分 I = ∫ r² dm，其中 r 是到轴的垂直距离。单位为 kg·m²。正如质量是平动的惯性，转动惯量是转动的惯性。
7. Calculating Moment of Inertia for a Thin Rod | 细杆转动惯量的计算
, axis at x=0. Let λ = M/L. An element dx at distance x has mass λ dx and r = x, so I = ∫₀ᴸ x² λ dx = λ [x³/3]₀ᴸ passes through the centre, the limits change to -L/2 to L/2, resulting in I = (1/12) M L².
计算质量为 M、长度为 L 的均匀细杆绕通过一端且垂直于杆的轴的转动惯量。将杆置于 x 轴上从 0 到 L，轴在 x=0 处。设 λ = M/L。距离为 x 的微元 dx 具有质量 λ dx，且 r = x，因此 I = ∫₀ᴸ x² λ dx = λ [x³/3]₀ᴸ = (λ L³)/3。代入 λ = M/L 得 I = (1/3) M L²。若轴通过中心，积分限变为 -L/2 到 L/2，结果为 I = (1/12) M L²。
8. Moment of Inertia of a Uniform Disc | 均匀圆盘的转动惯量
ment of moment of inertia is dI = r² dm = (2M/R²) r³ dr. Integrating from 0 to R yields I = (2M/R²) [r⁴/4]₀ᴿ = (1/2) M R². This classic result frequently appears in composite body problems.
对于质量为 M、半径为 R 的均匀薄圆盘，绕通过中心且垂直于盘面的轴的转动惯量可利用极坐标求得。考虑半径为 r、宽度为 dr 的薄圆环。其质量为 dm = (M / πR²) · 2πr dr = (2M/R²) r dr。到轴的距离为 r，因此转动惯量微元为 dI = r² dm = (2M/R²) r³ dr。从 0 积分到 R 得 I = (2M/R²) [r⁴/4]₀ᴿ = (1/2) M R²。这个经典结果经常出现在组合体问题中。
9. The Parallel Axis Theorem | 平行轴定理
If the moment of inertia of a body about an axis through its centre of mass, Icm, is known, the moment of inertia about any parallel axis a distance d away is I = Icm + M d². This theorem is immensely useful for shifting axes, for instance from the centre of a rod to its end. Verify: for a rod, Icentre = (1/12) M L², d = L/2; then Iend = (1/12) M L² + M (L/2)² = (1/12 + 1/4) M L² = (1/3) M L², matching direct integration.
如果已知物体对通过其质心的轴的转动惯量 Icm，则对与该轴平行且相距 d 的任意轴的转动惯量为 I = Icm + M d²。该定理在变换轴时非常有用，例如从杆的中心移到一端。验证：对细杆，I_中心 = (1/12) M L²，d = L/2；则 I_端点 = (1/12) M L² + M (L/2)² = (1/12 + 1/4) M L² = (1/3) M L²，与直接积分吻合。
10. The Perpendicular Axis Theorem for Laminae | 薄板的垂直轴定理
For a planar lamina lying in the xy-plane, the moment of inertia about the z-axis (perpendicular to the plane) is related to the moments about the x- and y- axes by Iz = Ix + Iy, provided the lamina is thin and the axes intersect at a common point. This theorem simplifies calculations for symmetric shapes such as rectangles and discs. For a uniform rectangular plate of sides a and b, Iz = (1/12) M (a² + b²) can be derived using Ix = (1/12) M b² and Iy = (1/12) M a².
对于位于 xy 平面内的平面薄板，若薄板很薄且各轴相交于同一点，则其对 z 轴（垂直于平面）的转动惯量与对 x 轴和 y 轴的转动惯量满足关系 Iz = Ix + Iy。该定理简化了诸如矩形和圆盘等对称形状的计算。对于边长为 a 和 b 的均匀矩形板，可利用 Ix = (1/12) M b² 和 Iy = (1/12) M a² 推导出 Iz = (1/12) M (a² + b²)。
11. Composite Bodies and Removal of Mass | 组合体与负质量法
adius r removed from its edge. The centre of mass of the remaining piece can be located by taking moments of the full disc and subtracting the moment of the removed disc, using their respective masses and centres. For moment of inertia, the same principle applies: I_remaining = Ifull he same axis. Practising these problems develops powerful problem-solving skills for the IB exam.
一个物体常常可以被看作简单形状的组合，或者是带有缺口的形状。由于质量和转动惯量对质量微元具有可加性，我们可以将缺失部分视为负质量贡献。假设从一个半径为 R 的圆盘边缘切除一个半径为 r 的小圆盘。剩余部分的质心可以通过取完整圆盘的力矩减去切除圆盘的力矩来定位，须使用各自的质量和质心位置。对于转动惯量，相同的原理适用：I_剩 = I_整 - I_切，前提是均针对同一轴线。练习这类问题能培养强有力的解题技巧，以应对 IB 考试。
12. Exam Strategy and Common Pitfalls | 考试策略与常见误区
In IB examination questions, you must carefully decide on the coordinate system, set up the correct mass element dm, and identify the distance r or lever arm. A common mistake is to use the wrong axis or to forget that the moment of inertia is a scalar with respect to a specific axis. Always check symmetry: if the problem involves a uniform solid of revolution, the centre of mass lies on the rotation axis. Remember the standard results for a rod (1/3 ML² about end), disc (1/2 MR²), and thin ring (MR²). Combine these with parallel and perpendicular axis theorems accurately. Show all integration steps clearly, and when a problem asks to 'show that' a formula holds, present your working logically with justification at each step. Practising with past paper questions will sharpen your ability to translate geometrical descriptions into mathematical integral set-ups.
在 IB 考试题目中，你必须仔细选定坐标系，建立正确的质量微元 dm，并明确距离 r 或力臂。常见错误包括使用错误的轴，或忘记转动惯量是相对于特定轴的标量。务必检查对称性：若题目涉及均匀旋转体，质心必在旋转轴上。牢记标准结果：细杆（绕端点 1/3 ML²）、圆盘（1/2 MR²）和薄圆环（MR²）。准确地结合平行轴和垂直轴定理。清晰展示所有积分步骤，当题目要求“证明”某公式成立时，逻辑严谨地呈现推导过程，并在每一步给出依据。通过练习历年真题，你将能更熟练地将几何描述转化为数学积分表达式。
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