📚 Circular Motion: Key Exam Points | 圆周运动 考点精讲
Circular motion is a core topic bridging IB and OCR Mathematics with mechanics, offering rich opportunities to test parametric equations, differentiation, vectors, and trigonometric modelling. Understanding how position, velocity, and acceleration vectors change in direction – not just magnitude – is the key to mastering centripetal acceleration and its applications. This revision guide unpacks every essential exam point, from first principles to problem-solving shortcuts.
圆周运动是连接 IB 与 OCR 数学和力学的重要主题，它将参数方程、求导、向量和三角函数建模结合在一起进行考察。理解位置、速度和加速度向量在方向上的变化——而不仅仅是大小变化——是掌握向心加速度及其应用的关键。这份考点精讲将从基本原理到解题技巧，逐一剖析每一个考试要点。
1. Parametric Equations of a Circle | 圆的参数方程
A point moving on a circle of radius r centred at the origin can be described by parametric equations x = r cos θ, y = r sin θ, where θ is the angular displacement measured from the positive x‑axis. In motion problems, θ is often expressed as a function of time t, typically θ = ωt for constant angular speed ω.
一个在半径为 r、圆心在原点的圆上运动的点，可用参数方程 x = r cos θ, y = r sin θ 描述，其中 θ 是从正 x 轴量起的角位移。在运动问题中，θ 通常表示为时间 t 的函数，对于恒定角速度 ω，一般为 θ = ωt。
If the centre is at (a, b), the equations become x = a + r cos ωt, y = b + r sin ωt. These forms appear frequently in both IB and OCR exam questions, especially when finding velocity and acceleration vectors.
如果圆心在 (a, b)，则方程变为 x = a + r cos ωt, y = b + r sin ωt。这类形式经常出现在 IB 和 OCR 的考题中，特别是在求速度和加速度向量时。
Always check the starting position: if the particle starts from the top of the circle, a phase shift is needed, such as x = r sin ωt, y = r cos ωt.
务必检查起始位置：如果质点从圆的顶端开始运动，则需要加入相位偏移，例如 x = r sin ωt, y = r cos ωt。
2. Angular Speed and Period | 角速度与周期
Angular speed ω (omega) is the rate of change of angular displacement with respect to time, measured in rad s⁻¹. The relationship ω = 2π/T links it to the period T, the time for one complete revolution. Frequency f = 1/T gives ω = 2πf.
角速度 ω 是角位移对时间的变化率，单位为 rad s⁻¹。关系式 ω = 2π/T 将其与周期 T（完成一整圈所需的时间）联系起来。频率 f = 1/T，因此 ω = 2πf。
In OCR questions, you may be given revolutions per minute (rpm) and need to convert to rad s⁻¹: multiply by 2π and divide by 60. IB often expects exact answers in terms of π.
在 OCR 题目中，可能会给出每分钟转数 (rpm)，需要将其转换为 rad s⁻¹：乘以 2π 再除以 60。IB 通常要求用 π 表示精确答案。
For non‑uniform circular motion, angular acceleration α = dω/dt is introduced, analogous to linear acceleration, but for uniform motion α = 0.
对于非匀速圆周运动，会引入角加速度 α = dω/dt，与线加速度类似；但对于匀速圆周运动，α = 0。
3. Position Vector and Displacement | 位置向量与位移
The position vector r(t) of a particle moving on a circle radius r, centre O, is given by r(t) = r cos ωt i + r sin ωt j, where i and j are unit vectors along the x‑ and y‑axes. Its magnitude |r| = r is constant.
在半径为 r、圆心为 O 的圆上运动的质点，其位置向量 r(t) 为 r(t) = r cos ωt i + r sin ωt j，其中 i 和 j 是沿 x 轴和 y 轴的单位向量。其模长 |r| = r 始终保持不变。
Displacement over a time interval is Δr = r(t₂) − r(t₁). While path length is an arc, displacement is the chord vector, and its magnitude is 2r sin(Δθ/2).
一段时间内的位移为 Δr = r(t₂) − r(t₁)。尽管运动路径是圆弧，位移却是弦向量，其大小为 2r sin(Δθ/2)。
Recognising that r is always perpendicular to the tangent at any point is fundamental to later vector proofs of acceleration direction.
认识到 r 始终垂直于任意点的切线，这对后续用向量证明加速度方向至关重要。
4. Velocity Vector and Linear Speed | 速度向量与线速度
Velocity v(t) is the derivative of position: v = dr/dt. Differentiating r cos ωt i + r sin ωt j yields v(t) = −rω sin ωt i + rω cos ωt j. Its magnitude is v = rω, the constant linear speed. The velocity vector is always tangent to the circle.
速度 v(t) 是位置对时间的导数：v = dr/dt。对 r cos ωt i + r sin ωt j 求导得 v(t) = −rω sin ωt i + rω cos ωt j。其模长 v = rω，即恒定的线速度。速度向量始终与圆相切。
Taking the dot product r · v = (r cos ωt)(−rω sin ωt) + (r sin ωt)(rω cos ωt) = 0 proves position and velocity are orthogonal. This purely mathematical step is a favourite in IB vector exams.
点积 r · v = (r cos ωt)(−rω sin ωt) + (r sin ωt)(rω cos ωt) = 0 证明位置与速度正交。这一纯数学推导是 IB 向量考试中的常见考点。
If the centre is not at the origin, differentiate the shifted parametric equations directly; the constant shift disappears upon differentiation, leaving the same velocity expression.
如果圆心不在原点，直接对平移后的参数方程求导即可；常数偏移在求导后消失，速度表达式保持不变。
5. Acceleration Vector and Centripetal Acceleration | 加速度向量与向心加速度
Acceleration a(t) is obtained by differentiating the velocity vector: a = dv/dt = −rω² cos ωt i − rω² sin ωt j. This simplifies to a = −ω² r, showing acceleration is directed opposite to the position vector, i.e. towards the centre.
对速度向量求导得到加速度 a(t)：a = dv/dt = −rω² cos ωt i − rω² sin ωt j。这可简化为 a = −ω² r，表明加速度方向与位置向量相反，即指向圆心。
The magnitude of this centripetal acceleration is a = rω². Using v = rω, it can also be written as a = v²/r. Both forms must be memorised; they are used interchangeably.
向心加速度的大小为 a = rω²。利用 v = rω，还可写为 a = v²/r。两种形式都须牢记；两者可相互转换使用。
In OCR mechanics questions, always identify whether v, ω, or r is given, and choose the appropriate expression. For IB, deriving a = v²/r from vector differentiation is often part of a longer proof question.
在 OCR 力学题中，要注意已知量是 v、ω 还是 r，并选择合适表达式。对于 IB，通过向量求导推导 a = v²/r 往往是长证明题的一部分。
6. Centripetal Force and Newton’s Second Law | 向心力与牛顿第二定律
Applying Newton’s second law in the radial direction gives the centripetal force F = ma = mrω² = mv²/r. This force points towards the centre and is required for any object moving in a circular path, regardless of whether the speed is constant.
在径向应用牛顿第二定律，得到向心力 F = ma = mrω² = mv²/r。无论速率是否恒定，任何做圆周运动的物体都需要这个指向圆心的力。
Common sources include tension (string), friction (car on a banked track), normal reaction (loop‑the‑loop), or gravity (orbits). Never invent a centrifugal force; in an inertial frame, only centripetal force exists.
常见的来源有：张力（绳）、摩擦力（倾斜弯道上的汽车）、法向反作用力（过山车回环）或引力（轨道运动）。切勿杜撰离心力；在惯性系中只存在向心力。
When solving problems, draw a clear free‑body diagram, resolve forces towards the centre, and set ΣF_radial = mv²/r or mrω². This is the single most reliable method.
解题时，画出清晰的受力图，将力沿径向分解，并列出 ΣF_径向 = mv²/r 或 mrω²。这是最可靠的方法。
7. Non‑Uniform Circular Motion | 非匀速圆周运动
When angular speed changes, angular acceleration α appears. The velocity magnitude is no longer constant; there is both a tangential acceleration a_t = rα and a radial centripetal acceleration a_r = rω² (or v²/r). The net acceleration vector has magnitude √(a_t² + a_r²).
当角速度变化时，会出现角加速度 α。速率不再恒定；既有切向加速度 a_t = rα，又有径向向心加速度 a_r = rω²（或 v²/r）。合加速度向量的模长为 √(a_t² + a_r²)。
In IB, non‑uniform circular motion is often explored through variable angular velocity functions like θ(t) = 2t³ − t. Students must differentiate to find ω and α, then apply rω² for radial acceleration.
在 IB 中，非匀速圆周运动常通过变角速度函数来考察，如 θ(t) = 2t³ − t。学生需通过求导得出 ω 和 α，再用 rω² 求径向加速度。
OCR may pose problems where forces must be resolved into tangential and normal components, for example a car accelerating around a bend. The friction vector can have both centripetal and tangential components.
OCR 可能会出一些需要将力分解为切向和法向分量的问题，例如汽车在弯道上加速行驶。此时摩擦力向量可能同时具有向心和切向分量。
8. Relationship with Simple Harmonic Motion | 与简谐运动的关系
The projection of uniform circular motion onto a diameter yields simple harmonic motion (SHM). If a particle moves as x = r cos ωt, y = r sin ωt, its x‑coordinate alone satisfies a_x = −ω² x, which is the SHM equation.
匀速圆周运动在一条直径上的投影会产生简谐运动 (SHM)。若质点运动方程为 x = r cos ωt, y = r sin ωt，仅其 x 坐标就满足 a_x = −ω² x，这正是简谐运动方程。
This connection is frequently tested in IB Mathematics Analysis & Approaches, where students must relate ω, period, and amplitude of circular motion to the corresponding SHM parameters.
这一联系在 IB 数学分析与方法课程中经常被考察，学生需将圆周运动的 ω、周期和振幅与相应的 SHM 参数关联起来。
This geometric link explains why SHM has sinusoidal solutions and why ω in SHM is called ‘angular frequency’ – it originates from the circular motion reference model.
这种几何联系解释了为什么 SHM 具有正弦形式的解，以及为什么 SHM 中的 ω 被称为“角频率”——它源自圆周运动参考模型。
9. Banking and Conical Pendulum | 斜面弯道与圆锥摆
A classic application is the banked curve without friction: the horizontal component of the normal reaction provides the centripetal force. For a bank angle θ, speed v, and radius r, tan θ = v²/(rg). This formula is derived by resolving normal force into vertical and horizontal components.
一个经典应用是无摩擦斜面弯道：法向反作用力的水平分量提供向心力。对于倾角 θ、速率 v 和半径 r，有 tan θ = v²/(rg)。该公式通过分解法向力的竖直和水平分量推导得出。
In a conical pendulum, a mass swings in a horizontal circle at the end of a string. The tension’s vertical component balances weight, while its horizontal component provides mv²/r. This yields relations like tan φ = v²/(rg), where φ is the half‑angle of the cone.
在圆锥摆中，一个质点在绳端做水平圆周运动。拉力的竖直分量与重力平衡，水平分量提供 mv²/r。由此可得 tan φ = v²/(rg)，其中 φ 是圆锥的半顶角。
Both situations require careful resolution of forces and are guaranteed to appear in either IB or OCR exams. Always identify the plane of the circle and resolve towards its centre.
这两种情况都需要仔细分解力，并且必定会出现在 IB 或 OCR 考试中。始终明确圆周所在的平面，然后沿指向中心的方向分解力。
10. Vertical Circular Motion and Energy | 竖直面内的圆周运动与能量
In vertical circular motion, speed is not constant because gravitational potential energy converts to kinetic energy. At the top of the loop, the minimum speed to maintain the circular path is v_min = √(gr), derived by setting the normal reaction to zero and using mg = mv²/r.
在竖直面内的圆周运动中，速率并不恒定，因为重力势能与动能相互转化。在回环顶部，保持圆周路径的最小速率为 v_min = √(gr)，这是通过令法向反作用力为零并利用 mg = mv²/r 推导得出的。
Energy conservation is key: ½ mv_top² + mg(2r) = ½ mv_bottom² (for a particle attached to a string). Combine this with radial force equations to solve for tensions and speeds at any point.
能量守恒是关键：对于系在细绳上的质点，有 ½ mv_顶部² + mg(2r) = ½ mv_底部²。将其与径向力方程结合，可求解任意点的拉力和速率。
OCR particularly likes questions where you must find the height at which a particle leaves a circular track (reaction becomes zero) or the speed required to complete a full loop.
OCR 特别喜欢考察以下类型：求质点脱离圆形轨道的高度（此时反作用力为零），或完成完整回环所需的速率。
Always check the condition for the string remaining taut or the object staying in contact: the radial net force must be at least the centripetal force needed.
务必检查保持细绳绷紧或物体保持接触的条件：径向合力必须至少等于所需的向心力。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
Misidentifying the radius is a top error – for a car on a circular track, the radius is the radius of the path, not necessarily the wheel radius or the car’s dimension. Always extract r from the geometry of the path.
错误识别半径是最常见的问题——对于在圆形轨道上的汽车，半径指的是路径的半径，而非车轮半径或汽车尺寸。务必从路径几何中提取 r。
Confusing frequency f (Hz) and angular frequency ω (rad s⁻¹) costs many marks. Remember ω = 2πf, and always convert to radians per second before using formulas like v = rω or a = rω².
混淆频率 f（Hz）与角频率 ω（rad s⁻¹）会导致大量失分。记住 ω = 2πf，并在使用 v = rω 或 a = rω² 等公式之前始终将其转换为弧度每秒。
Another frequent slip is using v²/r for tangential acceleration; v²/r is the radial (centripetal) acceleration only. Tangential acceleration is dv/dt or rα.
另一个常见错误是将 v²/r 用于切向加速度；v²/r 仅仅是径向（向心）加速度。切向加速度是 dv/dt 或 rα。
For IB proofs, structure your differentiation clearly: state r(t), then v(t) = r'(t), then a(t) = v'(t). Conclude with the negative scalar multiple of r(t) to show direction is towards centre.
对于 IB 证明题，清晰地组织求导过程：先写出 r(t)，然后 v(t) = r'(t)，再写出 a(t) = v'(t)。最后通过指出 a(t) 是 r(t) 的负标量倍数来表明方向指向中心。
In OCR mechanics, underline your consistent use of sign conventions: towards the centre is positive in the radial equation ΣF = ma. This avoids sign errors when dealing with tension and weight.
在 OCR 力学中，强调对符号约定的一致使用：在径向方程 ΣF = ma 中，指向中心为正。这能避免在处理拉力和重力时出现符号错误。
12. Summary of Key Formulas | 核心公式总结
| Relationship/Formula | Expression | Notes |
| Angular speed and period | ω = 2π/T | T is period for one revolution |
| Linear speed | v = rω | Always valid for rigid circular motion |
| Centripetal acceleration | a = rω² = v²/r | Directed radially inward |
| Centripetal force | F = mrω² = mv²/r | Resultant force towards centre |
| Tangential acceleration | a_t = rα = dv/dt | Only when ω is not constant |
| Banked curve (no friction) | tan θ = v²/(rg) | θ is bank angle to horizontal |
| Conical pendulum angle | tan φ = v²/(rg) | φ is half‑cone angle; r is circle radius |
Having these formulas at your fingertips, along with a strong grasp of vector differentiation and force resolution, gives you the confidence to tackle any circular motion problem on IB or OCR papers. Practise deriving them from first principles rather than just memorising, as this deepens understanding and prepares you for proof questions.
熟练掌握这些公式，再加上扎实的向量微分和受力分解能力，你就能自信地应对 IB 或 OCR 试卷中任何圆周运动问题。多从基本原理出发推导公式，而不是死记硬背，这能加深理解并为证明题做好准备。
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