📚 Circular Motion: WJEC A-Level Physics Key Points | A-Level WJEC 物理：圆周运动 考点精讲
Circular motion is a fundamental topic in A-Level Physics, particularly for the WJEC specification. It describes the motion of objects moving along a circular path at constant speed or varying speed. Understanding the concepts of angular displacement, centripetal acceleration and centripetal force is essential for tackling problems ranging from banking of roads to satellite orbits. This article provides a detailed, bilingual revision guide to help you master the key exam points.
圆周运动是A-Level物理的基础课题，尤其对WJEC考纲而言至关重要。它描述物体沿圆形路径以恒定或变化速度运动的情形。掌握角位移、向心加速度和向心力等概念，对于解决从倾斜弯道到卫星轨道等各类问题不可或缺。本文提供一份详细的中英双语复习指南，助你攻克考点。
1. Angular Displacement and Radian Measure | 角位移与弧度制
Angular displacement (θ) is the angle through which an object moves on a circular path. It is measured in radians (rad) rather than degrees, because radian measure simplifies the link between arc length and radius.
角位移(θ)是物体在圆形路径上转过的角度。它以弧度(rad)而非度来量度，因为弧度制简化了弧长与半径之间的关系。
One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius. Consequently, 2π rad = 360°.
一弧度定义为弧长等于半径的圆弧所对的圆心角。因此，2π rad = 360°。
s = rθ
The arc length s is given by s = rθ, where θ must be in radians. For conversion, θ(rad) = (π/180) × θ(deg).
弧长 s 由公式 s = rθ 给出，其中 θ 必须用弧度。换算公式为：弧度 = (π/180) × 角度。
2. Angular Velocity and Period | 角速度与周期
Angular velocity (ω) is the rate of change of angular displacement. For uniform circular motion, ω = Δθ/Δt. It is measured in rad s⁻¹.
角速度(ω)是角位移的变化率。对于匀速圆周运动，ω = Δθ/Δt，单位是 rad s⁻¹。
ω = 2π / T = 2πf
The period (T) is the time for one complete revolution. Since 2π radians are swept in that time, ω = 2π/T. Frequency f = 1/T, so ω = 2πf.
周期(T)是旋转一周所需的时间。由于一周对应 2π 弧度，因此 ω = 2π/T。频率 f = 1/T，所以 ω = 2πf。
WJEC questions may ask you to relate period, frequency and angular velocity for turntables, centrifuges or planets.
WJEC考题可能要求你针对转盘、离心机或行星，联系周期、频率与角速度。
3. Relation Between Linear and Angular Quantities | 线速度与角速度的关系
v = rω
The linear (tangential) speed v of a point on a rotating object is related to angular velocity by v = rω, where r is the radius. This relation holds provided ω is in rad s⁻¹.
旋转物体上一点的线（切向）速度 v 与角速度的关系为 v = rω，其中 r 是半径。只要 ω 的单位是 rad s⁻¹，该关系就成立。
Even if speed is constant, the velocity is not constant because the direction changes continuously. For non-uniform circular motion, a tangential acceleration at = rα exists, where α is the angular acceleration.
即使速率恒定，速度也不恒定，因为方向不断变化。对于非匀速圆周运动，还存在切向加速度 at = rα，α 是角加速度。
4. Centripetal Acceleration | 向心加速度
ac = v² / r = rω²
Even when speed is constant, an object in circular motion experiences a centre-pointing acceleration, called centripetal acceleration. Its magnitude is ac = v²/r = rω².
即使速率恒定，圆周运动中的物体也会经历指向圆心的加速度，即向心加速度。其大小为 ac = v²/r = rω²。
Using v = 2πr/T, you can also write ac = 4π²r / T². WJEC candidates should be able to derive the acceleration from a vector diagram of velocity change over a short time interval.
利用 v = 2πr/T，还可写成 ac = 4π²r / T²。WJEC考生应能通过短时间内的速度变化矢量图推导该加速度。
5. Centripetal Force | 向心力
F = mac = mv² / r = mrω²
From Newton’s second law, a net force must act towards the centre to produce centripetal acceleration: F = mv²/r = mrω².
根据牛顿第二定律，必须有净力指向圆心以产生向心加速度：F = mv²/r = mrω²。
Centripetal force is not a new kind of force; it is provided by familiar forces such as tension, friction, gravity or normal contact force. Always identify the physical source of the centripetal force in a free-body diagram.
向心力并非一种新型的力；它可以由张力、摩擦力、重力或支持力等常见力提供。务必在受力图中明确向心力的物理来源。
Typical WJEC prompts: “State what provides the centripetal force when a car rounds a bend.” Answer: the friction between tyres and road.
WJEC典型设问：“说明汽车转弯时是什么提供了向心力。”答案：轮胎与路面的摩擦力。
6. Conical Pendulum | 圆锥摆
A conical pendulum consists of a mass moving in a horizontal circle at the end of a string. The string traces a cone. Resolving vertically and horizontally yields two key equations.
圆锥摆由系在绳子末端的重物在水平面内做圆周运动构成，绳子划出一个圆锥面。竖直和水平分解得到两个关键方程。
T cosθ = mg (vertical)
T sinθ = mrω² (horizontal)
Here θ is the angle the string makes with the vertical, r = L sinθ (L = string length). Combining these gives the period T = 2π√(L cosθ / g).
其中 θ 是绳子与竖直方向的夹角，r = L sinθ（L为绳长）。联立可得周期 T = 2π√(L cosθ / g)。
Notice the period depends only on L and θ, not on the mass. This is a frequent WJEC calculation or multiple-choice question.
注意周期仅取决于绳长 L 和角度 θ，与质量无关。这是WJEC常考的计算题或选择题。
7. Banked Tracks and Vehicle Turning | 倾斜弯道与车辆转弯
For a vehicle turning on a flat road, the centripetal force is provided by static friction. The maximum safe speed before skidding is vmax = √(μs g r), where μs is the coefficient of static friction.
车辆在平坦路面转弯时，向心力由静摩擦力提供。打滑前的最大安全车速 vmax = √(μs g r)，μs 为静摩擦系数。
tanθ = v² / (rg) (ideal banking)
On a banked track, the normal reaction has a horizontal component that helps supply centripetal force, reducing reliance on friction. For a given speed v, the ideal banking angle satisfies tanθ = v²/(rg).
在倾斜弯道上，支持力的水平分量有助于提供向心力，减少对摩擦的依赖。给定车速 v，理想的倾斜角满足 tanθ = v²/(rg)。
When friction is also present, WJEC problems require resolving forces perpendicular and parallel to the slope, then using the condition that the net horizontal force equals mv²/r.
当同时存在摩擦时，WJEC题目要求沿斜面垂直和平行方向分解力，并利用净水平力等于 mv²/r 的条件。
8. Vertical Circular Motion: Critical Speeds | 竖直圆周运动：临界速度
Objects moving in a vertical circle (e.g., a roller coaster loop, a bucket of water swung overhead) experience varying speed and reaction forces. At the top, the speed must be sufficient to keep the object on the circular path.
在竖直面内做圆周运动的物体（如过山车回环、头顶甩动的水桶）速率和支持力不断变化。在最高点，速度必须足以维持圆周路径。
Top: mg + N = mv² / r
Bottom: N – mg = mv² / r
At the top, the normal reaction N and weight act downwards. For the object to just complete the loop, N ≥ 0, giving minimum speed vmin = √(gr).
在最高点，支持力 N 和重力均向下。物体恰好能完成回环时 N ≥ 0，得到最小速率 vmin = √(gr)。
At the bottom, N – mg = mv²/r, so the reaction is largest, which is critical for structural strength. Always draw separate free-body diagrams for top, bottom and side points.
在最低点，N – mg = mv²/r，支持力最大，这对结构强度至关重要。务必为最高点、最低点和侧点分别绘制受力图。
9. Energy Considerations in Circular Motion | 圆周运动中的能量分析
In vertical circles, mechanical energy is often conserved if no non-conservative forces do work. You can relate speed at different heights using mgh and ½ mv².
在竖直圆周运动中，若无非保守力做功，机械能往往守恒。可利用 mgh 和 ½ mv² 关联不同高度处的速率。
½ mvtop² + mg(2r) = ½ mvbottom²
For a complete loop, height change is 2r. Using energy conservation, if speed at the bottom is known, you can find speed at the top, then check if it exceeds √(gr).
对于完整回环，高度变化为 2r。利用能量守恒，若已知最低点速率，可求出最高点速率，进而检验是否超过 √(gr)。
For arcs like a pendulum, the vertical drop is r(1 – cosθ). WJEC may ask for speed at a given angle θ, requiring combined use of energy and circular motion equations.
对于单摆般的圆弧，竖直下落量为 r(1 – cosθ
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