📚 9 Motion in Polar Coordinates | 极坐标中的运动
In many IB Higher Level and Further Mathematics problems, particles move along curved paths that are much easier to describe using a polar coordinate system rather than Cartesian coordinates. By focusing on the radial distance r and the angle θ, we can derive compact expressions for velocity and acceleration. This article unpacks the essential theory, derivations and practical applications of motion in polar coordinates, helping you master this topic for your examinations.
在许多 IB 高等级和进阶数学问题中，质点沿曲线运动，用极坐标系描述比直角坐标方便得多。通过关注径向距离 r 和角度 θ，我们可以推导出紧凑的速度和加速度表达式。本文将深入讲解极坐标中运动的核心理论、推导过程和实际应用，帮助你掌握这一考点，从容应对考试。
1. Polar Coordinates and the Position Vector | 极坐标与位置向量
tance from the origin (pole) and θ is the angle measured anticlockwise from the positive x‑ ions of time: r = r(t) and θ = θ(t). The position vector of the particle can be written as r = r er, where er is a unit vector pointing radially outward. This simple representation becomes the foundation for all subsequent kinematic quantities.
在极坐标系中，一个点 P 由有序对 (r, θ) 确定，其中 r 是到原点（极点）的距离，θ 是从正 x 轴逆时针测量的角度。随着质点运动，两个坐标都是时间的函数：r = r(t)，θ = θ(t)。质点的位置向量可写成 r = r er，其中 er 是指向径向外的单位向量。这种简洁的表示是后续所有运动量的基础。
2. Unit Vectors in Radial and Transverse Directions | 径向与横向单位向量
The radial unit vector er depends on θ and is expressed in Cartesian components as er = cosθ i + sinθ j. The transverse unit vector eθ is perpendicular to er and is obtained by a 90° anticlockwise rotation: eθ = – re not constant; their time derivatives are crucial for finding velocity and acceleration.
径向单位向量 er 依赖于 θ，在直角分量中表示为 er = cosθ i + sinθ j。横向单位向量 eθ 垂直于 er，可由 er 逆时针旋转90°得到：eθ = –sinθ i + cosθ j。与 i 和 j 不同，极坐标单位向量不是常向量；它们对时间的导数是求速度和加速度的关键。
der/dt = θ̇ eθ, deθ/dt = –θ̇ er
These relations arise because as θ changes, the direction of er rotates with angular speed θ̇, generating a transverse component. You must be able to derive these quickly using the chain rule and the Cartesian expressions above.
这些关系源自：当 θ 变化时，er 的方向以角速度 θ̇ 旋转，产生横向分量。你必须能利用链式法则和上述直角表达式快速推导出它们。
3. Velocity in Polar Coordinates | 极坐标中的速度
Differentiating the position vector r = r er with respect to time t gives the velocity v. Applying the product rule and the derivative of er yields:
将位置向量 r = r er 对时间 t 求导即可得到速度 v。使用乘法法则以及 er 的导数可得：
v = ṙ er + r θ̇ eθ
The radial component ṙ ents the rate of change of the distance from the origin, while the transverse component rθ̇ cular to the radius due to rotation. If the particle moves in a circle of constant radius, ṙ = 0 and the speed is simply r|θ̇|.
径向分量 ṙ 表示到原点的距离变化率，而横向分量 rθ̇ 表示因旋转而产生的垂直于半径的速度。如果质点沿恒定半径的圆运动，则 ṙ = 0，速率就只是 r|θ̇|。
4. Acceleration in Polar Coordinates | 极坐标中的加速度
To obtain acceleration, differentiate the velocity expression v = ṙ er + rθ̇ eθ. Careful use of the product rule and the derivatives of the unit vectors leads to the standard result:
为了得到加速度，对速度表达式 v = ṙ er + rθ̇ eθ 求导。仔细应用乘法法则和单位向量的导数，可以得到标准结果：
a = (r̈ – r θ̇²) er + (r θ̈ + 2 ṙ θ̇) eθ
The radial acceleration consists of two terms: r̈, the linear acceleration along the radius, and –rθ̇², the centripetal term directed towards the origin. The transverse acceleration combines rθ̈, which is the tangential acceleration due to changing angular speed, and 2ṙθ̇, the Coriolis‑type term that appears when both r and θ change.
径向加速度由两项构成：r̈ 是沿半径的线加速度，–rθ̇² 是指向原点的向心项。横向加速度则由 rθ̈（由于角速度变化引起的切向加速度）和 2ṙθ̇（当 r 和 θ 同时变化时出现的科里奥利型项）组合而成。
5. Deriving the Acceleration Components Step by Step | 逐步推导加速度分量
Start from v = ṙ er + rθ̇ eθ. Differentiating each term gives: d(ṙ er)/dt = r̈ er + ṙ θ̇ eθ. For the second term, d(rθ̇ eθ)/dt = ṙθ̇ eθ + rθ̈ eθ + rθ̇ (deθ/dt). Since deθ/dt = –θ̇ er, the contribution becomes –rθ̇² er. Collecting the er terms: r̈ – rθ̇². Collecting the eθ terms: ṙθ̇ + ṙθ̇ + rθ̈ = rθ̈ + 2ṙθ̇. This yields the concise acceleration formula.
从 v = ṙ er + rθ̇ eθ 出发。逐项求导：d(ṙ er)/dt = r̈ er + ṙ θ̇ eθ。对于第二项，d(rθ̇ eθ)/dt = ṙθ̇ eθ + rθ̈ eθ + rθ̇ (deθ/dt)。由于 deθ/dt = –θ̇ er，这一部分贡献了 –rθ̇² er。合并 er 项：r̈ – rθ̇²。合并 eθ 项：ṙθ̇ + ṙθ̇ + rθ̈ = rθ̈ + 2ṙθ̇。这样便得到了简洁的加速度公式。
Remembering this derivation makes it easier to reconstruct the formula under exam pressure. Always check that your units are consistent: ṙ and rθ̇ have units of velocity, and r̈, rθ̇², rθ̈, ṙθ̇ have units of acceleration.
记住这个推导过程有助于在考试压力下重新推导出公式。务必检查量纲是否一致：ṙ 和 rθ̇ 具有速度的量纲，而 r̈、rθ̇²、rθ̈ 和 ṙθ̇ 具有加速度的量纲。
6. Special Case: Uniform Circular Motion | 特例：匀速圆周运动
For uniform circular motion the radius is constant (r = R, ṙ = 0, r̈ = 0) and the angular speed is constant (θ̈ = 0). Let ω = θ̇. The velocity simplifies to v = Rω eθ, showing purely tangential motion. The acceleration becomes a = –Rω² er, which is directed towards the centre of the circle with magnitude Rω² or v²/R.
对于匀速圆周运动，半径保持不变（r = R，ṙ = 0，r̈ = 0），且角速度恒定（θ̈ = 0）。设 ω = θ̇。速度简化为 v = Rω eθ，表现为纯切向运动。加速度变为 a = –Rω² er，方向指向圆心，大小为 Rω² 或 v²/R。
sistency of our polar acceleration formula. Notice that the transverse acceleration term 2ṙθ̇ vanishes because ṙ = 0, and rθ̈ is zero because θ̈ = 0.
这一经典结果验证了极坐标加速度公式的一致性。注意横向加速度项 2ṙθ̇ 由于 ṙ = 0 而消失，而 rθ̈ 也因 θ̈ = 0 变为零。
7. Motion Along an Archimedean Spiral | 沿阿基米德螺线的运动
ar equation is r = aθ, where a is a positive constant. If the angular speed is constant, θ̇ = ω, then ṙ = aω and r̈ = 0. Substituting into the velocity and acceleration formulae:
考虑一个质点，其极坐标方程为 r = aθ，其中 a 为正常数。若角速度恒定，θ̇ = ω，则 ṙ = aω，r̈ = 0。代入速度和加速度公式：
v = aω er + aθ ω eθ
a = (–aθ ω²) er + (0 + 2 aω·ω) eθ = –aθ ω² er + 2aω² eθ
with θ, while the transverse acceleration is constant. At any instant the speed is v = aω√1 + θ². Such problems test your ability to interpret the terms in a physical context.
径向加速度指向内部且随 θ 增大而增大，而横向加速度保持不变。在任一时刻，速率为 v = aω√1 + θ²。这类问题考查你在物理情境中解释各项的能力。
8. Converting Between Cartesian and Polar Motion Forms | 直角坐标与极坐标运动形式的转换
ks for the polar components of velocity or acceleration. You can use the relationships r = √x² + y² and θ = arctan(y/x) to find ṙ and θ̇ by differentiation. Alternatively, once you have the Cartesian velocity vector v = ẋ i + ẏ j, the polar components can be found by projecting onto er and eθ: ṙ = v·er, rθ̇ = v·eθ. A similar projection works for acceleration.
有时题目给出 x(t) 和 y(t)，要求速度和加速度的极坐标分量。你可以利用关系式 r = √x² + y² 和 θ = arctan(y/x) 通过求导得出 ṙ 和 θ̇。另一种方法是，一旦得到直角速度向量 v = ẋ i + ẏ j，可将其投影到 er 和 eθ 上求得极坐标分量：ṙ = v·er，rθ̇ = v·eθ。加速度也可做类似投影。
Mastering both approaches gives you flexibility: direct differentiation of r(t) and θ(t) works well when motion is defined in polar form, while the projection method is powerful when the Cartesian path is known.
掌握两种方法可提供灵活性：当运动以极坐标形式定义时，直接对 r(t) 和 θ(t) 求导很有效；而当已知直角路径时，投影法十分强大。
9. Angular Momentum and the Transverse Component | 角动量与横向分量
In many mechanics problems, the transverse acceleration is zero, meaning rθ̈ + 2ṙθ̇ = 0. This can be rewritten as (1/r) d(r²θ̇)/dt = 0, so r²θ̇ round, mathematically recognising this constant helps solve differential equations of motion.
在许多力学问题中，横向加速度为零，即 rθ̈ + 2ṙθ̇ = 0。这可以改写为 (1/r) d(r²θ̇)/dt = 0，因此 r²θ̇ 为常数。这个量常被称为面积速度常数，与角动量守恒有关。即使没有物理背景，在数学上识别出这个常数也有助于求解运动微分方程。
r a central force, the transverse acceleration vanishes, and you can immediately write r²θ̇ = h (constant). This reduces the problem to analysing the radial equation alone.
例如，若质点在有心力作用下运动，横向加速度为零，你就可以立即写出 r²θ̇ = h（常数）。这样一来，问题便简化为只分析径向方程。
10. Common Mistakes and Examination Tips | 常见错误与考试提示
A very frequent error is forgetting to differentiate the unit vectors. Always remember er and eθ depend on time through θ. Another mistake is confusing the centripetal term –rθ̇² er with the radial acceleration r̈; the two coexist when r changes. Also, check the sign of the Coriolis‑type term: it is +2ṙθ̇, not minus.
一个十分常见的错误是忘记对单位向量求导。始终要记住 er 和 eθ 通过 θ 依赖于时间。另一个错误是将向心项 –rθ̇² er 与径向加速度 r̈ 混淆；当 r
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