Rigid Body Mechanics: Moment of Inertia, Equilibrium and Motion | 刚体力学：转动惯量、平衡与运动

📚 Rigid Body Mechanics: Moment of Inertia, Equilibrium and Motion | 刚体力学：转动惯量、平衡与运动

Rigid body mechanics is one of the most elegant and exam-relevant topics in IB Physics. It extends Newton’s laws from point particles to real objects, introducing rotational analogues of mass, force, momentum and kinetic energy. This article provides a comprehensive, step-by-step guide to moment of inertia, rotational equilibrium and rotational motion, complete with formulas, tables and problem-solving strategies tailored for the IB syllabus.

刚体力学是 IB 物理中最优雅、最贴近考点的话题之一。它将牛顿定律从质点推广到真实物体，引入了质量、力、动量和动能的转动对应量。本文提供关于转动惯量、转动平衡和转动运动的全面、循序渐进指导，包含公式、表格以及针对 IB 大纲量身定制的解题策略。

1. What Is a Rigid Body? | 什么是刚体？

A rigid body is an idealised object in which the distances between all pairs of constituent particles remain constant, regardless of the external forces or torques applied. In reality, every solid deforms slightly, but for most mechanics problems — from spinning wheels to balancing beams — the rigid body approximation is excellent. In rigid body mechanics, the motion is analysed in two parts: the translation of the centre of mass (CM) and the rotation about an axis through the CM (or any fixed axis).

刚体是一种理想化物体，其内部任意两个质点之间的距离在外力或外力矩作用下始终保持不变。事实上所有固体都会有微小形变，但对于大多数力学问题——从旋转的车轮到平衡的横梁——刚体近似都非常有效。在刚体力学中，运动分为两部分分析：质心的平动以及绕通过质心（或任一固定轴）的转动。

The translational behaviour of a rigid body is governed by the net external force: ΣF = Ma_cm, where M is the total mass and a_cm is the acceleration of the centre of mass. The rotational behaviour is governed by the net external torque. Understanding this separation is the first step to mastering the topic.

刚体的平动行为由合外力决定：ΣF = Ma_cm，其中 M 是总质量，a_cm 是质心加速度。转动行为则由合外力矩决定。理解这种划分是掌握本专题的第一步。

2. Moment of Inertia | 转动惯量

Just as mass quantifies a body’s resistance to linear acceleration, the moment of inertia I quantifies a body’s resistance to angular acceleration. For a collection of discrete particles, it is defined as the sum of each mass times the square of its perpendicular distance from the rotation axis:

正如质量量化物体对平动加速度的抵抗程度，转动惯量 I 量化物体对角加速度的抵抗程度。对于离散质点系，其定义为每个质点的质量乘以它到转轴垂直距离的平方之和：

I = Σ mᵢrᵢ²

For a continuous body, this sum becomes an integral: I = ∫ r² dm. The SI unit of moment of inertia is kg·m². Crucially, I depends not only on the total mass but also on how that mass is distributed relative to the axis. A mass far from the axis contributes much more to I than a mass close to it, which is exactly why a figure skater spins faster when pulling arms inward.

对于连续体，求和变为积分：I = ∫ r² dm。转动惯量的 SI 单位是 kg·m²。关键在于，I 不仅取决于总质量，还取决于质量相对于转轴的分布方式。离轴远的质量对 I 的贡献远大于离轴近的质量，这正是花样滑冰运动员收拢手臂时旋转更快的原因。

Another useful concept is the radius of gyration: by writing I = Mk², we define k as the distance from the axis at which the entire mass M could be concentrated to give the same moment of inertia. This is helpful in estimating I for irregular objects and is often tested in IB multiple-choice questions.

另一个有用的概念是回转半径：通过 I = Mk²，定义 k 为将全部质量 M 集中在此距离处可获得相同转动惯量的轴向距离。这对于估算不规则物体的 I 很有用，也常在 IB 选择题中出现。

3. Common Moments of Inertia | 常见转动惯量

The following table lists the moments of inertia for standard shapes about specified axes. These values are derived directly from I = ∫ r² dm and should be memorised or, at minimum, recognised in exam problems. In IB examinations, you may be expected to use these values without re-deriving them.

下表列出了标准形状绕指定轴的转动惯量。这些值直接从 I = ∫ r² dm 推导而来，应当熟记，至少要在考题中能识别它们。在 IB 考试中，你可能会被要求直接使用这些值而无需重新推导。

Object \| 物体	Axis \| 转轴	Moment of Inertia \| 转动惯量
Thin hoop / ring (环)	Through centre, ⊥ to plane	I = MR²
Solid cylinder / disc (圆柱/圆盘)	Central axis	I = ½MR²
Thin hollow cylinder (薄壁圆筒)	Central axis	I = MR²
Solid sphere (实心球)	Diameter	I = (2/5)MR²
Thin spherical shell (薄球壳)	Diameter	I = (2/3)MR²
Thin rod (细杆)	Through centre, ⊥ to length	I = (1/12)ML²
Thin rod (细杆)	Through one end, ⊥ to length	I = (1/3)ML²
Rectangular plate (矩形板)	Through centre, ⊥ to plane	I = (1/12)M(a² + b²)

Notice the pattern: for a given mass and radius, the moment of inertia is larger when mass is concentrated farther from the axis. The hoop (all mass at radius R) has I = MR², whereas the solid disc (mass distributed uniformly) has only half that value. This physical intuition helps with qualitative reasoning questions.

注意规律：对于给定的质量和半径，质量越集中在远离轴处，转动惯量越大。圆环（所有质量位于半径 R 处）的 I = MR²，而实心圆盘（质量均匀分布）仅为该值的一半。这种物理直觉有助于解答定性推理题。

4. The Parallel Axis Theorem | 平行轴定理

The parallel axis theorem allows us to calculate the moment of inertia about any axis parallel to one passing through the centre of mass. If I_cm is the moment of inertia about an axis through the CM, then about a parallel axis displaced by a perpendicular distance d, the moment of inertia is:

平行轴定理使我们能够计算绕任意与过质心轴平行的轴的转动惯量。若 I_cm 是绕通过质心的轴的转动惯量，则绕距离为 d 的平行轴的转动惯量为：

I = I_cm + Md²

Here M is the total mass of the body. This theorem is extremely useful. For example, the moment of inertia of a thin rod of mass M and length L about an axis through one end is obtained from I_cm = (1/12)ML² and d = L/2:

其中 M 是物体的总质量。这个定理极为有用。例如，质量为 M、长度为 L 的细杆绕通过一端的轴的转动惯量，可由 I_cm = (1/12)ML² 和 d = L/2 求得：

I = (1/12)ML² + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/3)ML²

This matches the table entry above. The theorem is a favourite in IB Paper 2 questions, especially when combined with energy conservation in pendulums or rolling objects.

这与上表条目一致。该定理在 IB Paper 2 中备受青睐，尤其常与能量守恒结合，用于摆或滚动物体的题目。

5. Torque and the Rotational Form of Newton’s Second Law | 力矩与牛顿第二定律的转动形式

Torque (or moment of force) is the rotational analogue of force. For a force F applied at a point whose position vector relative to the axis is r, the torque is:

力矩是力的转动对应量。对于作用点的位置矢量相对于转轴为 r 的力 F，力矩为：

τ = r × F, magnitude τ = rF sin θ

where θ is the angle between r and F. The perpendicular distance from the axis to the line of action of the force, r sin θ, is called the lever arm. The unit of torque is N·m (identical to the joule, but torque is never expressed in joules because it is not energy).

其中 θ 是 r 与 F 之间的夹角。从轴到力的作用线的垂直距离 r sin θ 称为力臂。力矩的单位是 N·m（与焦耳相同，但力矩从不以焦耳表示，因为它不是能量）。

The rotational equivalent of Newton’s second law states that the net torque acting on a rigid body equals the product of its moment of inertia and its angular acceleration:

牛顿第二定律的转动形式指出：作用在刚体上的合外力矩等于其转动惯量乘以角加速度：

Στ = Iα

The analogy is complete: F ↔ τ, m ↔ I, a ↔ α. This equation is the starting point for solving virtually every rigid-body dynamics problem. A common IB example involves a disc or pulley with a string wound around it: by applying Στ = Iα to the pulley and ΣF = Ma to the hanging mass, you can solve for the acceleration of the system.

对应关系完整：F ↔ τ、m ↔ I、a ↔ α。这个方程是解决几乎所有刚体动力学问题的出发点。一个常见的 IB 例题涉及绕绳的圆盘或滑轮：对滑轮应用 Στ = Iα，对悬挂质量应用 ΣF = Ma，即可求出系统的加速度。

6. Rotational Equilibrium | 转动平衡

A rigid body is in complete static equilibrium when it has neither translational nor rotational acceleration. The two conditions are:

当刚体既无平动加速度也无角加速度时，它处于完全静态平衡。两个条件为：

ΣF = 0 and Στ = 0 (about any axis)

The second condition must hold about any chosen axis; therefore, in solving problems you are free to select the pivot point that simplifies the mathematics most. Placing the pivot at the location of an unknown force eliminates that force from the torque equation, since its lever arm becomes zero.

第二个条件必须对任意选定轴都成立；因此，解题时可以自由选择最简化计算的支点。将支点放在某个未知力的作用位置，可使该力在力矩方程中消失，因为其力臂变为零。

Consider a classic IB problem: a uniform beam of mass m and length L rests on two supports at its ends. To find the reaction forces, take moments about one support. The weight mg acts at the centre (distance L/2 from either support), so:

考虑一个经典 IB 问题：质量为 m、长度为 L 的均匀横梁两端由两个支座支撑。为求支座反力，取关于其中一个支座的力矩。重力 mg 作用在中心（离任一支座距离 L/2），因此：

R₂ × L = mg × (L/2) → R₂ = mg/2

By symmetry or by ΣF = 0, the other reaction is also mg/2. When additional loads are placed on the beam, the same method — choose a pivot, write Στ = 0, then ΣF = 0 — always yields the unknowns. In IB, ladder problems and see-saw problems follow this exact pattern.

由对称性或 ΣF = 0，另一个反力也是 mg/2。当横梁上放置额外负载时，同样的方法——选择支点、列 Στ = 0、再列 ΣF = 0——总能解出未知量。在 IB 中，梯子问题和跷跷板问题都遵循这一模式。

7. Angular Momentum and Its Conservation | 角动量及其守恒

The angular momentum of a rigid body rotating about a fixed axis is:

绕固定轴转动的刚体的角动量为：

L = Iω

where I is the moment of inertia about that axis and ω is the angular speed. The SI unit is kg·m²/s. The rotational form of Newton’s second law can also be written as τ = dL/dt: the net external torque equals the rate of change of angular momentum.

其中 I 是绕该轴的转动惯量，ω 是角速度。SI 单位是 kg·m²/s。牛顿第二定律的转动形式也可写为 τ = dL/dt：合外力矩等于角动量对时间的变化率。

The conservation of angular momentum is a cornerstone principle: if the net external torque acting on a system is zero, then the total angular momentum of the system remains constant. Because L = Iω is conserved, a decrease in I leads to an increase in ω, and vice versa.

角动量守恒是基石级原理：如果作用于系统的合外力矩为零，则系统的总角动量保持不变。因为 L = Iω 守恒，I 减小导致 ω 增大，反之亦然。

The classic demonstrations — a spinning ice skater pulling in her arms, or a diver tucking into a somersault — are all consequences of this law. In IB Paper 1, you may face a question where a student stands on a rotating turntable holding weights; when the weights are pulled inward, I decreases and the angular speed increases. The key is recognising that no external torque acts about the vertical axis.

经典演示——旋转中的花样滑冰运动员收拢手臂，或跳水运动员团身翻腾——都是这一定律的体现。在 IB Paper 1 中，你可能会遇到这样的题目：学生站在旋转转台上手持重物；当重物向内收时，I 减小，角速度增大。关键是认识到竖直方向没有外力矩作用。

Equally important is the distinction between angular momentum and linear momentum: they are independent conserved quantities, each with its own conservation condition. A collision problem may simultaneously conserve both, imposing two independent equations.

同样重要的是区分角动量与线动量：它们是独立的守恒量，各有其守恒条件。碰撞问题可能同时守恒两者，提供两个独立方程。

8. Rotational Kinetic Energy and Work | 转动动能与功

A rotating rigid body possesses kinetic energy due to its rotation. The rotational kinetic energy is:

旋转的刚体因其转动而具有动能。转动动能为：

K_rot = ½Iω²

This expression is perfectly analogous to K_lin = ½mv². For an object that is both translating and rotating — such as a rolling ball — the total kinetic energy is the sum of the translational kinetic energy of the centre of mass and the rotational kinetic energy about the centre of mass:

此式与 K_lin = ½mv² 完全对应。对于既平动又转动的物体——如滚动的球——总动能为质心的平动动能与绕质心的转动动能之和：

K_total = ½Mv² + ½Iω²

The work done by a constant torque τ acting through an angular displacement Δθ is W = τΔθ, and the power delivered is P = τω. These relations mirror W = FΔx and P = Fv, completing the systematic analogy between linear and rotational quantities.

恒力矩 τ 作用通过角位移 Δθ 所做的功为 W = τΔθ，功率为 P = τω。这些关系与 W = FΔx 和 P = Fv 相互对应，完善了线量与转动量之间的系统类比。

This framework is invaluable for energy-conservation problems. For example, a sphere rolling down an inclined plane loses gravitational potential energy, which transforms into both translational and rotational kinetic energy. Without the rotational term, the final speed would be overestimated by up to 20% for a solid sphere.

这一框架对能量守恒问题极为重要。例如，一个球沿斜面滚下时损失的重力势能转化为平动动能和转动动能。如果没有转动项，实心球的最终速度会被高估至多 20%。

9. Rolling Without Slipping | 无滑动滚动

Rolling without slipping is a special and highly examinable motion. The condition connecting translational and angular motion is:

无滑动滚动是一种特殊且极具考查价值的运动。连接平动与角运动的条件为：

v_cm = ωR and a_cm = αR

where R is the radius of the rolling object. Under this condition, the point of contact is instantaneously at rest, which is why static friction (not kinetic friction) acts at the contact point — and static friction does no work, so mechanical energy is conserved.

其中 R 是滚动物体的半径。在此条件下，接触点瞬时静止，这就是为什么接触点处作用的是静摩擦力（而非动摩擦力）——而静摩擦力不做功，因此机械能守恒。

Consider the classic race between a hoop, a disc, a solid sphere and a hollow sphere released from rest at the top of an incline. Using energy conservation with the no-slip condition, the acceleration down the incline is:

考虑经典比赛：一个圆环、一个圆盘、一个实心球和一个薄球壳从斜面顶端由静止释放。利用能量守恒和无滑动条件，沿斜面向下的加速度为：

a = g sin θ / (1 + I/MR²)

The object with the smallest I/MR² ratio reaches the bottom first. Since I/MR² = 1 for a hoop, ½ for a disc, 2/5 for a solid sphere and 2/3 for a hollow sphere, the order of finish is: solid sphere, then disc, then hollow sphere, then hoop. Note that all objects travel the same distance, and the final speed depends only on the height of the incline, not on the mass or radius.

具有最小 I/MR² 比值的物体最先到达底部。由于圆环的 I/MR² = 1、圆盘为 ½、实心球为 2/5、薄球壳为 2/3，到达顺序为：实心球最先，然后是圆盘、薄球壳、圆环最后。注意所有物体行进相同距离，最终速度只取决于斜面高度，而与质量或半径无关。

This result is a favourite for IB Paper 2 extended-response questions, often combined with questions about friction direction. The friction force acts up the incline, providing the torque that causes rotation while reducing the translational acceleration.

这一结论是 IB Paper 2 扩展回答题的最爱，常结合摩擦力方向的问题。摩擦力沿斜面向上，提供产生转动的力矩，同时减小平动加速度。

10. Problem-Solving Strategies and Exam Tips | 解题策略与考试技巧

Success in rigid-body problems comes from a disciplined approach. Here are the essential strategies for IB examinations:

在刚体问题中取得成功的秘诀来自于有条理的方法。以下是 IB 考试必备的策略：

Draw a clear diagram showing the object, all forces, dimensions and the chosen axis of rotation. Label the pivot point explicitly. (画出清晰示意图，标出物体、所有力、尺寸和所选转轴。明确标出支点。)
Choose the axis strategically: place it where an unknown force acts, or where the torque expression is simplest. This often turns a three-equation problem into a two-equation problem. (策略性选择转轴：将其放在未知力作用处或力矩表达式最简处。这常将三方程问题化为两方程问题。)
Identify conserved quantities: if no external torque acts, angular momentum is conserved; if no external work (other than gravity) is done, mechanical energy is conserved. (识别守恒量：若无外力矩作用，角动量守恒；若除重力外无其他外力做功，机械能守恒。)
Verify dimensions and sign conventions: torque is a vector; choose clockwise or counterclockwise as positive and be consistent throughout. (核对量纲和符号约定：力矩是矢量；选定顺时针或逆时针为正，并全程保持一致。)
Do not confuse moment of inertia about CM with that about another axis: always check which axis the problem specifies, and apply the parallel axis theorem when necessary. (不要混淆绕质心的转动惯量与绕其他轴的转动惯量：始终检查题目指定的是哪个轴，必要时应用平行轴定理。)

Common student mistakes include using I = MR² for a disc when the problem involves a disc (should be ½MR²), forgetting the rotational kinetic energy term in rolling problems, and writing Στ = Iα about an accelerating point of contact instead of a fixed axis. Being vigilant about these pitfalls will immediately raise your accuracy.

学生常见错误包括：题目涉及圆盘时却使用 I = MR²（应为 ½MR²）、在滚动问题中遗漏转动动能项、以及绕加速的接触点列 Στ = Iα 而非绕固定轴列方程。警惕这些陷阱将立即提高你的正确率。

11. Conclusion | 结论

Rigid body mechanics unifies the translational and rotational worlds through a beautiful set of analogues: force ↔ torque, mass ↔ moment of inertia, linear momentum ↔ angular momentum, and kinetic energy forms ½mv² ↔ ½Iω². The three pillars — moment of inertia, equilibrium conditions and conservation laws — provide a complete framework for solving everything from balance beam problems to rolling races.

刚体力学通过一组优美的对应关系将平动世界与转动世界统一起来：力 ↔ 力矩、质量 ↔ 转动惯量、线动量 ↔ 角动量、动能形式 ½mv² ↔ ½Iω²。三大支柱——转动惯量、平衡条件和守恒定律——为解决从平衡梁问题到滚动比赛问题的一切提供了完整框架。

For IB success, memorise the standard moments of inertia, practise choosing optimal pivots, and always question whether rotation contributes to kinetic energy or angular momentum in the scenario. Work through past Paper 2 questions on rolling down inclines and rotating pulleys; these patterns recur frequently. With consistent practice, rigid body mechanics will become one of your most reliable scoring sections.

要在 IB 中取得好成绩，请熟记标准转动惯量、练习选择最优支点，并始终思考转动是否对该场景中的动能或角动量有贡献。认真做 Paper 2 中关于沿斜面滚动和旋转滑轮的历年真题；这些题型出现频率很高。通过持续练习，刚体力学将成为你最稳定的得分板块之一。

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Rigid Body Mechanics: Moment of Inertia, Equilibrium and Motion | 刚体力学：转动惯量、平衡与运动