IB Physics: Rigid Body Mechanics Explained | IB物理：刚体力学考点精讲

📚 IB Physics: Rigid Body Mechanics Explained | IB物理：刚体力学考点精讲

Rigid body mechanics is a key extension of classical mechanics in IB Physics, particularly for Higher Level students. In this article, we break down the core concepts, equations, and exam strategies for torque, rotational equilibrium, moment of inertia, angular momentum, and rotational kinetic energy.

刚体力学是IB物理中经典力学的重要延伸，尤其对高级别（HL）学生而言。本文将系统梳理力矩、转动平衡、转动惯量、角动量与转动动能等核心考点、公式及解题策略。

1. What Is a Rigid Body? | 什么是刚体

A rigid body is an idealised object in which the distance between any two particles remains constant, regardless of the forces applied. This means the body does not deform under load. In IB Physics, we treat objects like rods, discs, spheres, and pulleys as rigid bodies when analysing rotation.

刚体是一个理想化模型：无论受到怎样的外力，其内部任意两点之间的距离始终保持不变，即物体不发生形变。在IB物理中，我们通常将杆、圆盘、球体和滑轮等物体视为刚体来分析转动问题。

Two types of motion are possible for a rigid body:

刚体可以有两种运动类型：

Translational motion: every point moves with the same velocity, and the body’s orientation does not change.
Rotational motion: every point moves along a circular path about a fixed axis, and the body’s orientation changes.

平动：刚体上所有点具有相同的速度，物体的方向不变。
转动：刚体上所有点绕固定轴做圆周运动，物体的方向发生变化。

In many problems, the body performs both translation and rotation simultaneously, such as a rolling wheel. This is called general plane motion.

在许多问题中，刚体同时进行平动和转动，例如滚动的车轮。这种运动称为平面一般运动。

2. Torque and Rotational Force | 力矩与转动力

Torque (also called moment of force) measures the tendency of a force to rotate an object about an axis. It is defined as the product of the force and the perpendicular distance from the axis to the line of action of the force.

力矩（也称为力的矩）衡量力使物体绕轴转动的趋势。其定义是力与力的作用线到转轴的垂直距离的乘积。

τ = F × r ⊥ = F × r × sin θ

where τ is the torque (N·m), F is the magnitude of the force (N), r is the distance from the axis to the point of application (m), and θ is the angle between the force vector and the position vector.

其中 τ 是力矩（单位N·m），F 是力的大小（单位N），r 是转轴到力作用点的距离（单位m），θ 是力矢量与位置矢量之间的夹角。

方向	使物体逆时针转动为正
单位	N·m（牛顿·米）
矢量性	力矩是矢量，方向沿转轴（右手定则）

A common mistake in exams is using the full distance instead of the perpendicular component. Always identify the perpendicular lever arm: the shortest distance from the axis to the line of action of the force.

考试中常见的错误是使用完整距离而不是垂直分量。一定要找到垂直力臂：即转轴到力的作用线的最短距离。

3. Rotational Equilibrium | 转动平衡

A rigid body is in rotational equilibrium when the net external torque acting on it is zero. This means the body either remains at rest or rotates with constant angular velocity.

当刚体受到的合外力矩为零时，刚体处于转动平衡状态。这意味着刚体保持静止或者以恒定角速度转动。

Σ τ = 0

For a body to be in complete equilibrium (both translational and rotational), two conditions must be satisfied:

要使物体处于完全平衡状态（既平动平衡又转动平衡），必须满足两个条件：

Σ F = 0 (no net force, no linear acceleration)
Σ τ = 0 (no net torque, no angular acceleration)

Σ F = 0（合外力为零，无线性加速度）
Σ τ = 0（合外力矩为零，无角加速度）

When solving problems, choose a pivot point wisely. Often, choosing the point where an unknown force acts simplifies the calculation because that unknown force produces zero torque about that point.

解题时要合理选择支点。通常选择未知力作用点为支点可以简化计算，因为该未知力对支点的力矩为零。

4. Centre of Mass and Gravity | 质心与重心

The centre of mass of a rigid body is the point at which the entire mass of the body can be considered to act for translational motion. The centre of gravity is the point where the total weight of the body acts. In a uniform gravitational field, these two points coincide.

刚体的质心是物体全部质量可以视为集中作用的点，用于分析平动。重心是物体总重力作用点。在均匀引力场中，质心和重心重合。

For a system of particles, the position of the centre of mass is found using:

对于质点系，质心的位置由以下公式确定：

xₘ = (Σ mᵢxᵢ) / (Σ mᵢ)

where mᵢ is the mass of the i-th particle and xᵢ is its position. Similarly for y and z coordinates.

其中 mᵢ 是第 i 个质点的质量，xᵢ 是它的位置。y 和 z 坐标同理。

In rigid body rotation, the weight of the body acts through the centre of gravity, so when a body is suspended from a point, it will rotate until the centre of gravity lies directly below the suspension point.

在刚体转动中，重力作用于重心。因此，当物体从某点悬挂时，它会转动直到重心位于悬挂点正下方。

5. Moment of Inertia | 转动惯量

The moment of inertia (I) is the rotational equivalent of mass. It measures how difficult it is to change an object’s angular velocity. It depends on both the mass of the object and the distribution of that mass relative to the axis of rotation.

转动惯量（I）是转动中的“质量”，衡量改变物体角速度的难易程度。它既取决于物体的质量，也取决于质量相对于转轴的分布。

I = Σ mᵢrᵢ² = ∫ r² dm

where r is the perpendicular distance from the axis to each mass element. The unit of moment of inertia is kg·m².

其中 r 是转轴到每个质量微元的垂直距离。转动惯量的单位是 kg·m²。

Key point: the moment of inertia depends on the axis of rotation. A rod rotated about its centre has a different moment of inertia than the same rod rotated about its end.

关键点：转动惯量取决于转轴的位置。同一根杆绕中心转动和绕端点转动，其转动惯量不同。

Common moments of inertia you should remember for IB HL:

IB HL 需要记忆的常见转动惯量：

物体	转轴	转动惯量
细杆（长度 L，质量 M）	过中心且垂直于杆	I = (1/12)ML²
细杆（长度 L，质量 M）	过端点且垂直于杆	I = (1/3)ML²
实心圆柱/圆盘（半径 R）	中心轴	I = (1/2)MR²
实心球体（半径 R）	过球心	I = (2/5)MR²
薄圆环（半径 R）	中心轴	I = MR²

6. Parallel Axis Theorem | 平行轴定理

The parallel axis theorem relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the centre of mass.

平行轴定理将物体关于任意轴的转动惯量与关于通过质心的平行轴的转动惯量联系起来。

I = Iₘ + Mh²

where Iₘ is the moment of inertia about the centre-of-mass axis, M is the total mass, and h is the perpendicular distance between the two parallel axes.

其中 Iₘ 是通过质心轴的转动惯量，M 是物体的总质量，h 是两平行轴之间的距离。

Example: A rod of mass M and length L rotated about one end. The moment of inertia is (1/3)ML². Check: Iₘ = (1/12)ML² and h = L/2, so I = (1/12)ML² + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/3)ML². Correct.

示例：质量为 M、长度为 L 的细杆绕端点转动，其转动惯量为 (1/3)ML²。验证：Iₘ = (1/12)ML²，h = L/2，则 I = (1/12)ML² + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/3)ML²。正确。

This theorem is frequently tested in IB Paper 2 problems involving compound objects, such as a rod with a mass attached to its end.

这个定理在IB Paper 2中经常考查，尤其是涉及复合物体（如杆端连接一个质量块）的问题。

7. Newton’s Second Law for Rotation | 牛顿第二定律的转动形式

Just as force causes linear acceleration, torque causes angular acceleration. The rotational form of Newton’s second law is:

正如力产生线性加速度，力矩产生角加速度。牛顿第二定律的转动形式为：

τ_neτ = Iα

where τ_neτ is the net external torque, I is the moment of inertia, and α is the angular acceleration (rad/s²).

其中 τ_neτ 是合外力矩，I 是转动惯量，α 是角加速度（单位 rad/s²）。

This equation is directly analogous to F = ma. The angular acceleration is inversely proportional to the moment of inertia: for the same torque, a larger moment of inertia produces a smaller angular acceleration.

这个公式与 F = ma 直接对应。角加速度与转动惯量成反比：相同的力矩作用下，转动惯量越大，角加速度越小。

When solving dynamics problems involving pulleys or rolling objects, you often need to combine this equation with the linear version F = ma. For a pulley of mass M and radius R with a string, the tension forces on either side produce a net torque:

在求解涉及滑轮或滚动物体的动力学问题时，通常需要将这一方程与线性形式 F = ma 联立。对于质量为 M、半径为 R 的滑轮，两侧绳子的张力产生净力矩：

(T₂ − T₁)R = Iα = (1/2)MR² × (a/R)

Here the relationship a = Rα links linear and angular acceleration.

这里使用关系 a = Rα 将线性加速度与角加速度联系起来。

8. Angular Momentum | 角动量

Angular momentum is the rotational analogue of linear momentum. For a rigid body rotating about a fixed axis, the angular momentum is:

角动量是线性动量在转动中的对应量。对于绕固定轴转动的刚体，角动量为：

L = Iω

where L is the angular momentum (kg·m²/s), I is the moment of inertia, and ω is the angular velocity (rad/s).

其中 L 是角动量（单位 kg·m²/s），I 是转动惯量，ω 是角速度（单位 rad/s）。

The rate of change of angular momentum equals the net external torque:

角动量的变化率等于合外力矩：

τ_neτ = ΔL / Δt = dL / dt

This is another expression of Newton’s second law for rotation. If the net external torque is zero, angular momentum is conserved:

这是牛顿第二定律转动形式的另一种表达。如果合外力矩为零，角动量守恒：

I₁ω₁ = I₂ω₂

A classic demonstration: a spinning ice skater pulls her arms in, reducing her moment of inertia. Because angular momentum is conserved, her angular velocity increases dramatically.

经典例子：旋转中的花样滑冰运动员收回手臂，转动惯量减小。由于角动量守恒，她的角速度大幅增加。

9. Rotational Kinetic Energy | 转动动能

A rotating rigid body possesses kinetic energy due to the motion of its particles. The rotational kinetic energy is given by:

旋转的刚体因其各质点运动而具有动能。转动动能公式为：

E_k_rot = (1/2)Iω²

For an object that is both translating and rotating (e.g., a rolling ball), the total kinetic energy is the sum of translational and rotational parts:

对于既平动又转动的物体（如滚动的球），总动能为平动动能与转动动能之和：

E_k_total = (1/2)mv² + (1/2)Iω²

where v is the speed of the centre of mass. For rolling without slipping, v = Rω.

其中 v 是质心的速度。对于无滑动的滚动，有 v = Rω。

When solving energy problems, remember that friction may do work converting translational energy into rotational energy. For a body rolling down an inclined plane, gravitational potential energy is converted into both translational and rotational kinetic energy:

在解决能量问题时，注意摩擦力可能做功将平动动能转化为转动动能。对于沿斜面滚下的物体，重力势能转化为平动动能和转动动能：

mgh = (1/2)mv² + (1/2)Iω²

This equation is essential for comparing how quickly different shapes (solid sphere, hollow sphere, cylinder) roll down an incline.

这个方程在比较不同形状（实心球、空心球、圆柱体）沿斜面滚下快慢时至关重要。

10. Work and Power in Rotation | 转动中的功与功率

When a torque acts through an angular displacement, it does work. The work done by a constant torque is:

当力矩在角位移上做功时，其做功大小为。恒力矩做功的公式为：

W = τθ

where θ is the angular displacement in radians. The power delivered by a torque is:

其中 θ 是角位移（单位弧度）。力矩产生的功率为：

P = τω

These equations mirror the linear forms W = Fs and P = Fv. They are useful in problems involving motors, engines, and rotating machinery.

这些公式与线性形式 W = Fs 和 P = Fv 对应。它们在涉及电动机、发动机和旋转机械的问题中非常有用。

Remember that the work-energy theorem for rotation states: the net work done by torques equals the change in rotational kinetic energy.

请记住，转动的功能定理为：合外力矩所做的净功等于转动动能的变化量。

11. Rolling Motion Without Slipping | 无滑动滚动

Rolling without slipping is a special case of combined translation and rotation. The point of contact between the rolling object and the surface is instantaneously at rest.

无滑动滚动是平动与转动相结合的特殊情况。滚动物体与接触面的接触点瞬时静止。

For rolling without slipping, the following kinematic constraints hold:

对于无滑动滚动，以下运动学约束成立：

v = Rω, a = Rα

The static friction force between the object and the surface must be sufficient to prevent slipping. This condition affects both the linear and angular dynamics.

物体与表面之间的静摩擦力必须足够大以防止滑动。这一条件同时影响线性动力学和角动力学。

In an exam, you may be asked to compare the acceleration of different objects rolling down an incline. The general result is:

考试中可能会要求比较不同物体沿斜面滚下的加速度。一般结果为：

a = g sin θ / (1 + I/(mR²))

For a solid sphere, I = (2/5)mR², so a = (5/7)g sin θ. For a hollow sphere, I = (2/3)mR², so a = (3/5)g sin θ. Objects with smaller moments of inertia accelerate faster because their rotational energy requirement is lower.

对于实心球体，I = (2/5)mR²，所以 a = (5/7)g sin θ。对于空心球体，I = (2/3)mR²，所以 a = (3/5)g sin θ。转动惯量越小的物体加速越快，因为它需要的转动能量较少。

12. Common Problem Types and Exam Strategies | 常见题型与应试策略

In IB Physics examinations, rigid body mechanics typically appears in Paper 2 as structured extended-response questions. The following problem types are common:

在IB物理考试中，刚体力学通常以结构化扩展题的形式出现在Paper 2中。常见题型包括：

Torque and equilibrium problems: A ladder or rod supported at multiple points; solve using ΣF = 0 and Στ = 0.
Moment of inertia calculations: Use standard results or the parallel axis theorem for composite objects.
Pulley problems: Combine F = ma for hanging masses with τ = Iα for the pulley.
Energy conservation with rotation: Use total kinetic energy including rotational terms.
Angular momentum conservation: Apply when no external torque acts.

力矩与平衡问题：如梯子或多点支撑的杆；利用 ΣF = 0 和 Στ = 0 求解。
转动惯量计算：使用标准结果或对复合物体使用平行轴定理。
滑轮问题：将悬挂物的 F = ma 与滑轮的 τ = Iα 联立。
含转动的能量守恒：总动能包含转动项。
角动量守恒：当合外力矩为零时应用。

Key exam tips:

考试关键技巧：

Always state the sign convention for torques (positive for anticlockwise).
Draw a clear free-body diagram and label all forces at least once.
Clearly define the pivot point before taking torques.
When using energy methods, check whether friction does work. For pure rolling without slipping, static friction does no net work.
Units: torque in N·m, angular acceleration in rad/s², angular momentum in kg·m²/s.

明确写出力矩的正方向约定（逆时针为正）。
画出清晰的受力分析图，并至少标记一次所有力。
在计算力矩前明确指出支点位置。
使用能量法时，检查摩擦力是否做功。无滑动纯滚动时，静摩擦力不做净功。
注意单位：力矩 N·m，角加速度 rad/s²，角动量 kg·m²/s。

Practise converting between linear and angular quantities. The relationships s = rθ, v = rω, and a = rα are the bridge between translational and rotational kinematics. Mastering them will enable you to solve even the most complex rigid body problems systematically.

务必练习线性量与角量之间的转换。关系式 s = rθ、v = rω、a = rα 是平动运动学与转动运动学之间的桥梁。掌握它们，你将能够系统性地解决最复杂的刚体问题。

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IB Physics: Rigid Body Mechanics Explained | IB物理：刚体力学考点精讲