📚 IB Math: Circular Motion Key Points Explained | IB 数学：圆周运动 考点精讲
Circular motion is a topic that beautifully links trigonometry, parametric equations, and calculus in the IB Mathematics curriculum. Whether you are taking Analysis and Approaches (AA) or Applications and Interpretation (AI), understanding how to describe a particle moving along a circle using functions of time is a key skill. This article unpacks the essential concepts, from radian measure to differentiation of position vectors, and provides clear derivations of velocity and acceleration, with a special focus on centripetal acceleration. We will also highlight common pitfalls and demonstrate efficient problem-solving techniques that align with IB exam style.
在 IB 数学课程中，圆周运动将三角学、参数方程和微积分巧妙地联系在一起。无论你修读的是分析与方法 (AA) 还是应用与解释 (AI)，掌握如何用时间的函数描述一个沿圆周运动的质点都是一项关键技能。本文逐一拆解从弧度制到位置向量求导的核心概念，清晰推导速度与加速度，并重点剖析向心加速度。我们也会指出常见陷阱，并展示符合 IB 考试风格的高效解题技巧。
1. Parameterisation of Circular Motion | 圆周运动的参数表示
A particle moving counterclockwise on a circle of radius r, centred at the origin, can be described by the parametric equations x(t) = r cos θ(t), y(t) = r sin θ(t). If the motion is uniform, θ(t) = ωt + φ, where ω is constant angular speed and φ is the initial phase. This parameterisation is the starting point for all subsequent calculus.
沿逆时针方向在半径为 r、圆心在原点的圆上运动的质点，可以用参数方程 x(t) = r cos θ(t), y(t) = r sin θ(t) 描述。若为匀速圆周运动，则 θ(t) = ωt + φ，其中 ω 为恒定角速度，φ 为初相位。这一参数化是所有后续微积分运算的起点。
2. The Necessity of Radian Measure | 弧度制的必要性
In circular motion analysis, angles must be measured in radians. The radian is defined such that the arc length s = rθ. Consequently, the derivative of sin θ is cos θ only when θ is in radians. Using degrees would introduce annoying conversion factors and break the simplicity of kinematic links like v = rω and a = rω². Always switch your calculator to radian mode and express angular quantities in rad or rad s⁻¹.
分析圆周运动时，角度必须使用弧度制。弧度的定义使得弧长 s = rθ。因此，sin θ 的导数为 cos θ 这一关系仅在 θ 以弧度为单位时成立。使用角度制会引入繁琐的转换因子，并破坏 v = rω 和 a = rω² 等运动学关系的简洁性。请务必将计算器切换到弧度模式，并用弧度或 rad s⁻¹ 表达角量。
3. Angular Velocity, Period and Frequency | 角速度、周期和频率
For uniform circular motion, the angular velocity ω (in rad s⁻¹) is constant. The period T is the time for one full revolution: T = 2π/ω. The frequency f = 1/T = ω/(2π), measured in hertz (Hz). These relationships allow you to move easily between time domain and angular domain.
对于匀速圆周运动，角速度 ω（单位 rad s⁻¹）恒定。周期 T 是旋转一周所需的时间：T = 2π/ω。频率 f = 1/T = ω/(2π)，单位为赫兹 (Hz)。这些关系使得在时间域和角度域之间切换非常方便。
- ω = 2πf
- ω = 2π/T
角速度 ω = 2πf，ω = 2π/T。
4. Position Vector | 位置向量
The position vector of the particle as a function of time is r(t) = r cos(ωt) i + r sin(ωt) j (assuming motion starts at the positive x-axis). Its magnitude is constant: |r| = r. The direction of r always points radially outward from the centre.
质点位置关于时间的向量为 r(t) = r cos(ωt) i + r sin(ωt) j（假设从正 x 轴开始运动）。其大小恒定：|r| = r。r 的方向始终由圆心径向向外。
r(t) = r cos(ωt) i + r sin(ωt) j
5. Velocity Vector via Differentiation | 速度向量与参数求导
Velocity is the first derivative of position with respect to time. Differentiating each component yields v(t) = dr/dt = –rω sin(ωt) i + rω cos(ωt) j. The magnitude of velocity, or speed, is v = |v| = rω. The velocity vector is always tangent to the circle and perpendicular to the position vector (dot product r·v = 0).
速度是位置对时间的一阶导数。对每个分量求导得到 v(t) = dr/dt = –rω sin(ωt) i + rω cos(ωt) j。速度的大小，即线速率，为 v = |v| = rω。速度向量始终与圆相切，并与位置向量垂直（点积 r·v = 0）。
v(t) = –rω sin(ωt) i + rω cos(ωt) j ; v = rω
6. Speed and the Link v = rω | 线速率与 v = rω 的关联
The relationship v = rω is fundamental. It can be derived directly from arc length: s = rθ ⇒ ds/dt = r (dθ/dt) ⇒ v = rω. This formula holds for any circular motion, uniform or non-uniform, at an instant provided ω is the instantaneous angular speed. In exam questions, you often use it to find one quantity when the other two are known.
v = rω 是一个基本关系式。它可以从弧长直接导出：s = rθ ⇒ ds/dt = r (dθ/dt) ⇒ v = rω。无论是匀速还是非匀速圆周运动，该公式在任意瞬时都成立，只要 ω 取瞬时角速度。在考试题中，常根据已知的两项用此式求出第三项。
7. Acceleration Vector – Second Derivative | 加速度向量——二次求导
Acceleration is the derivative of velocity: a(t) = dv/dt = –rω² cos(ωt) i – rω² sin(ωt) j. Notice that a(t) = –ω² r(t). This shows that the acceleration vector points in exactly the opposite direction to the position vector, i.e., towards the centre of the circle. It is therefore called centripetal acceleration.
加速度是速度的导数：a(t) = dv/dt = –rω² cos(ωt) i – rω² sin(ωt) j。注意 a(t) = –ω² r(t)。这表明加速度向量正好与位置向量反向，即指向圆心。因此它被称为向心加速度。
a(t) = –ω² r(t) ; |a| = rω² = v²/r
8. Deriving Centripetal Acceleration Magnitude | 向心加速度大小的推导
Since a = –ω² r, the magnitude of centripetal acceleration is ac = ω² r. Using v = rω, you can also write ac = v² / r. This derivation requires only basic differentiation of trigonometric functions and the chain rule. Do not memorise it as an isolated physics formula; understand it as a consequence of the parametric position function and calculus.
由 a = –ω² r 可知，向心加速度的大小为 ac = ω² r。利用 v = rω，还可以写成 ac = v² / r。这一推导只用到三角函数的基本求导和链式法则。不要将其作为孤立的物理公式死记硬背，而应理解为参数位置函数与微积分结合的自然结果。
9. Tangential and Normal Components | 切向与法向分量
In non-uniform circular motion, the angular speed ω(t) is not constant. The acceleration then has two components: a tangential component at = rα (where α = dω/dt is angular acceleration) responsible for changing speed, and the centripetal (normal) component ac = rω² = v²/r responsible for changing direction. The net acceleration vector is the vector sum of these.
在非匀速圆周运动中，角速度 ω(t) 不是常数。此时加速度有两个分量：切向分量 at = rα（其中 α = dω/dt 为角加速度），负责改变速率；以及向心（法向）分量 ac = rω² = v²/r，负责改变方向。合加速度向量是两者的矢量和。
| Component / 分量 | Role / 作用 | Expression / 表达式 |
|---|---|---|
| Tangential at | Changes speed | rα = r dω/dt |
| Centripetal ac | Changes direction | rω² = v²/r |
切向分量改变速率，向心分量改变方向。
10. Worked Example 1 – Uniform Case | 典型例题 1 —— 匀速情形
Problem: A particle moves on a circle of radius 2 m with constant angular speed 3 rad s⁻¹. Find its speed, period, and centripetal acceleration.
题目：一质点沿半径 2 m 的圆以恒定角速度 3 rad s⁻¹ 运动。求其速率、周期和向心加速度。
Solution: v = rω = 2 × 3 = 6 m s⁻¹. T = 2π/ω = 2π/3 s. ac = rω² = 2 × 9 = 18 m s⁻². Also ac = v²/r = 36/2 = 18 m s⁻², confirming consistency.
解答：v = rω = 2 × 3 = 6 m s⁻¹。T = 2π/ω = 2π/3 s。ac = rω² = 2 × 9 = 18 m s⁻²。也可由 ac = v²/r = 36/2 = 18 m s⁻² 验证一致。
11. Worked Example 2 – Variable Angular Speed | 典型例题 2 —— 变角速度
Problem: The angular position of a particle is θ(t) = 4t² (rad). The radius is 1.5 m. Find the velocity and acceleration vectors at t = 2 s, and express their magnitudes.
题目：某质点的角位置为 θ(t) = 4t²（弧度），半径 1.5 m。求 t = 2 s 时的速度向量和加速度向量，并给出其大小。
Solution: ω(t) = dθ/dt = 8t. At t = 2 s, ω = 16 rad s⁻¹. α = dω/dt = 8 rad s⁻². Position: r = 1.5 cos(4t²) i + 1.5 sin(4t²) j. Velocity: v = 1.5 × 8t (–sin(4t²) i + cos(4t²) j). At t=2: v = 12 (–sin 16 i + cos 16 j) ≈ 12(0.2879 i – 0.9577 j) = (3.45 i – 11.49 j) m s⁻¹. Speed = rω = 1.5 × 16 = 24 m s⁻¹. Acceleration: a = –rω² r/r + rα θ (unit tangential). Magnitude a = √[(rω²)² + (rα)²] = √[(1.5×256)² + (1.5×8)²] = √(384² + 12²) ≈ 384.2 m s⁻².
解答：ω(t) = dθ/dt = 8t，t = 2 s 时 ω = 16 rad s⁻¹。α = dω/dt = 8 rad s⁻²。位置：r = 1.5 cos(4t²) i + 1.5 sin(4t²) j。速度：v = 1.5 × 8t (–sin(4t²) i + cos(4t²) j)。t=2 时：v = 12 (–sin 16 i + cos 16 j) ≈ 12(0.2879 i – 0.9577 j) = (3.45 i – 11.49 j) m s⁻¹。速率 = rω = 1.5 × 16 = 24 m s⁻¹。加速度：a = –rω² r/r + rα θ（切向单位向量）。大小 a = √[(rω²)² + (rα)²] = √[(1.5×256)² + (1.5×8)²] = √(384² + 12²) ≈ 384.2 m s⁻²。
12. Common Pitfalls and Tips | 常见陷阱与技巧
Degrees vs radians: Forgetting to use radian mode leads to wrong derivatives and incorrect v = rω. Always check your calculator settings at the start of an exam.
弧度与角度：忘记使用弧度模式会导致求导错误和 v = rω 计算错误。考试一开始就检查计算器设置。
Confusing centripetal and tangential components: In non-uniform motion, acceleration is not purely radial. Draw a vector diagram to separate ac and at.
混淆向心与切向分量：在非匀速运动中，加速度并非单纯径向。画出矢量图以区分 ac 和 at。
Sign errors in differentiation: When differentiating cos and sin, track the negative signs carefully. Writing the position vector clearly with i and j helps avoid sign slips.
求导时的符号错误：对 cos 和 sin 求导时，仔细处理负号。列出清晰的位置向量 i 和 j 分量有助于避免符号失误。
Memorising v²/r without proof: IB often asks for a calculus-based derivation. Be prepared to differentiate parametric equations from scratch.
死记 v²/r 而不知推导：IB 常要求基于微积分的推导。做好准备从参数方程入手逐步求导。
13. Summary | 总结
Circular motion in IB Maths bridges parametric equations, trigonometry, and calculus. Start with the parametric position r(t) = r cos(ωt) i + r sin(ωt) j. Differentiate to obtain velocity v(t) = –rω sin(ωt) i + rω cos(ωt) j with magnitude v = rω. Differentiate again to find centripetal acceleration a(t) = –ω² r(t) of magnitude ac = rω² = v²/r. For non-uniform motion, include the tangential component at = rα. Mastering these derivations and their applications will give you a strong advantage in both AA and AI papers.
IB 数学中的圆周运动连接了参数方程、三角学和微积分。从参数化的位置向量 r(t) = r cos(ωt) i + r sin(ωt) j 出发，一次求导得速度 v(t) = –rω sin(ωt) i + rω cos(ωt) j，大小为 v = rω。再次求导得到向心加速度 a(t) = –ω² r(t)，大小为 ac = rω² = v²/r。对于非匀速运动，还要加上切向分量 at = rα。掌握这些推导及其应用，将为你在 AA 和 AI 试卷中带来显著优势。
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