A-Level CCEA Physics: Circular Motion – Key Points | A-Level CCEA 物理：圆周运动考点精讲

📚 A-Level CCEA Physics: Circular Motion – Key Points | A-Level CCEA 物理：圆周运动考点精讲

Circular motion appears throughout the CCEA A-Level Physics specification, from the motion of planets to the design of banked racetracks. Mastering the relationships between angular and linear quantities, the concept of centripetal force, and the application of free‑body diagrams to real‑world scenarios is essential for top marks. This revision guide breaks down every critical point, using straightforward explanations and worked‑style reasoning to help you build confidence for your exam.

圆周运动贯穿 CCEA A-Level 物理考纲，从行星运动到倾斜赛道的设计均有涉及。要取得高分，必须熟练掌握角量与线量之间的关系、向心力的概念，以及如何将受力分析应用于真实情境。本指南逐一拆解核心考点，配合清晰的解释与推导思路，帮助你巩固知识、从容应试。

1. Angular Displacement and the Radian | 角位移与弧度

Angular displacement θ is the angle through which an object moves on a circular path. In A‑Level Physics we always measure θ in radians (rad). One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius: when arc length s equals radius r, θ = 1 rad. The conversion between degrees and radians is 360° = 2π rad, so 1 rad ≈ 57.3°.

角位移 θ 是物体在圆周路径上转过的角度。A‑Level 阶段始终用弧度 (rad) 来度量 θ。当一段圆弧的长度 s 等于圆的半径 r 时，该圆弧所对的圆心角就是 1 弧度。度与弧度的换算关系为 360° = 2π rad，因此 1 rad ≈ 57.3°。

The general relationship between arc length s, radius r and angle θ in radians is s = rθ. This simple equation underpins almost every link between linear and angular motion, so it is crucial to be completely comfortable with it.

在弧度制下，弧长 s、半径 r 与圆心角 θ 之间满足 s = rθ 。这个简洁的公式是沟通线量与角量的基础，必须做到熟练运用。

2. Angular Velocity ω | 角速度 ω

Angular velocity ω is the rate of change of angular displacement. For uniform circular motion, where the object sweeps out equal angles in equal time intervals, the average angular velocity equals the instantaneous value:

ω = Δθ / Δt

The SI unit of angular velocity is rad s⁻¹. Because radians are dimensionless, ω can be treated as having dimensions of T⁻¹, but you must always quote the unit as rad s⁻¹ in numerical answers.

角速度 ω 表示角位移的快慢。对于匀速圆周运动，物体在相等时间内转过相等的角度，平均角速度就等于瞬时角速度。其定义式为 ω = Δθ / Δt ，国际单位是 rad s⁻¹。需要注意，弧度本身无量纲，因此 ω 的量纲可写为 T⁻¹，但在数值答案中必须带单位 rad s⁻¹。

In many problems ω is constant, and you can find it from the time taken to complete one full revolution. Since one revolution corresponds to an angular displacement of 2π rad, if the period is T, then ω = 2π / T. Equally, if you know the frequency f (number of revolutions per second), ω = 2π f.

许多题目中 ω 保持不变，此时可以通过转动一周所需的时间求出 ω。一周对应 2π rad，若周期为 T，则 ω = 2π / T；若已知频率 f（每秒转数），则 ω = 2π f。

3. Linking Linear Speed and Angular Velocity | 线速度与角速度的关联

Combining s = rθ with the definitions of speed and angular velocity gives the most frequently used relationship in circular motion:

v = r ω

where v is the instantaneous linear speed tangent to the circle. This equation tells you that for a fixed angular velocity, the linear speed increases with radius — a point on the rim of a spinning disc moves faster than a point near the centre.

将 s = rθ 与速度和角速度的定义结合，就得到圆周运动中最常用的关系式 v = r ω ，其中 v 是沿切线方向的瞬时速率。该式表明，在角速度相同时，半径越大线速度越大——旋转圆盘边缘处的点比靠近中心的点运动得更快。

If a problem gives you the diameter or radius and the RPM (revolutions per minute), convert RPM to rad s⁻¹ first: multiply by 2π and divide by 60. Then apply v = r ω to find the linear speed.

若题目给出直径或半径以及转速（RPM），应先将转速换算为 rad s⁻¹：乘以 2π 再除以 60，然后使用 v = r ω 计算线速度。

4. Period, Frequency and Their Link to ω | 周期、频率及其与 ω 的关系

The period T is the time for one complete revolution, measured in seconds. Frequency f is the number of revolutions per second, measured in hertz (Hz). For any repetitive circular motion:

T = 1 / f

As already noted, ω can be written in terms of T or f: ω = 2π / T, ω = 2π f. These equations are used constantly in CCEA examination papers, often as the first step in a calculation that then requires v = r ω or the centripetal acceleration formula.

周期 T 是完成一整圈所需的时间，单位为秒 (s)。频率 f 是每秒转动的圈数，单位为赫兹 (Hz)。二者满足 T = 1 / f 。如前所述，ω 也可用 T 或 f 表示：ω = 2π / T，ω = 2π f。这些公式在 CCEA 试卷中反复出现，通常作为后续代入 v = r ω 或向心加速度公式的第一步。

Be careful with unit conversions: a question might state “30 revolutions per minute”. This gives f = 30/60 = 0.5 Hz, T = 2 s, and ω = 2π × 0.5 = π rad s⁻¹. Always show these steps clearly.

注意单位换算：题目若给出“每分钟 30 转”，则 f = 30/60 = 0.5 Hz，T = 2 s，ω = 2π × 0.5 = π rad s⁻¹。答题时务必清晰展示这些换算过程。

5. Centripetal Acceleration | 向心加速度

Even when an object moves at constant speed in a circle, its velocity is continually changing direction, so it is accelerating. This acceleration is directed towards the centre of the circle and is called centripetal acceleration. Its magnitude is given by:

a = v² / r

Substituting v = r ω gives the alternative form:

a = r ω²

You must be able to choose the most convenient expression depending on the data provided. If you are given v and r, use a = v² / r; if you are given ω and r, use a = r ω².

即使物体以恒定速率做圆周运动，其速度方向也在不断改变，因此存在加速度。这个加速度始终指向圆心，称为向心加速度，其大小为 a = v² / r 。代入 v = r ω 可得另一常用形式 a = r ω² 。考试中需根据已知条件灵活选用：给出 v 和 r 时用 a = v² / r，给出 ω 和 r 时用 a = r ω²。

The direction of a is always radial and inward. In a diagram, draw the acceleration vector pointing from the object towards the centre. Do not confuse centripetal acceleration with a tangential acceleration; if the speed is constant, the tangential acceleration is zero.

向心加速度的方向总是沿半径指向圆心。作图时，应将加速度矢量画成从物体指向圆心。注意不要将向心加速度与切向加速度混淆；若速率恒定，切向加速度为零。

6. Centripetal Force | 向心力

According to Newton’s second law, a resultant force must act towards the centre to produce the centripetal acceleration. This resultant force is the centripetal force F_c:

F = m a = m v² / r = m r ω²

Centripetal force is not a new type of force; it is the name we give to the net radial force that keeps an object moving in a circle. Tension, friction, gravity or a normal reaction can all provide the centripetal force, depending on the context. In your free‑body diagram, identify the actual physical forces, then equate their resultant toward the centre to m v² / r or m r ω².

根据牛顿第二定律，必须有一个指向圆心的合力来产生向心加速度，这个合力就是向心力 F_c，表达式为 F = m a = m v² / r = m r ω² 。向心力并非一种新的力，而是对维持圆周运动的径向合力的称呼。根据具体情境，拉力、摩擦力、重力或法向反作用力都可以充当向心力。画受力图时，先识别所有实际存在的力，再将其指向圆心的合力与 m v² / r 或 m r ω² 建立等量关系。

A common misconception is to add a separate “centripetal force” arrow on the diagram. Examiners expect you to avoid this; instead, label the real forces and state that their resultant provides the centripetal force.

常见误区是在受力图上额外画一个“向心力”箭头。阅卷要求避免这种画法，应标出真实的力，并注明这些力的合力提供向心力。

7. Horizontal Circular Motion on a String | 水平面上的绳拉圆周运动

When a small object is whirled in a horizontal circle at the end of a string, the tension in the string supplies the centripetal force. If the motion is truly horizontal and the string is light and inextensible, resolving horizontally gives:

T = m v² / r

If the string makes an angle to the horizontal (as in a conical pendulum, discussed next), the horizontal component of tension provides the centripetal force, while the vertical component balances the weight.

当用细绳拉着一个小物体在水平面上做圆周运动时，绳的拉力提供向心力。若运动严格在水平面内，且细绳轻质不可伸长，水平方向的分量方程为 T = m v² / r 。如果细绳与水平方向有夹角（如下文所述的锥摆），则拉力的水平分量提供向心力，竖直分量与重力平衡。

For a perfectly horizontal circle, the string cannot be exactly horizontal unless some other vertical force (such as a smooth table) supports the weight. In practice, a slight dip is inevitable, but many simplified CCEA problems assume the tension acts horizontally. Always read the question carefully to see whether vertical forces need to be considered.

严格水平的圆周运动中，除非有其它竖直力（如光滑桌面）支撑重力，否则绳子不可能完全水平。实际情形中绳子会略微下垂，但许多 CCEA 简化题目假设拉力沿水平方向。解题时务必仔细读题，判断是否需要考虑竖直方向的力。

8. The Conical Pendulum | 锥摆

A conical pendulum consists of a mass tied to a string and swung in a horizontal circle so that the string traces out a cone. Here the string tension T has two perpendicular components:

Vertical equilibrium: T cos θ = m g
Horizontal centripetal force: T sin θ = m v² / r

where θ is the angle the string makes with the vertical. The radius r of the circular path is related to the string length L by r = L sin θ.

锥摆是将一个物体系在绳端，使其在水平面内做圆周运动，绳的轨迹形成圆锥面。此时绳的拉力 T 可沿竖直和水平方向分解：竖直方向平衡： T cos θ = m g；水平方向提供向心力： T sin θ = m v² / r。其中 θ 是绳与竖直方向的夹角，圆周半径 r 与绳长 L 的关系为 r = L sin θ。

Dividing the two equations eliminates T and gives tan θ = v² / (r g). Since v = r ω, this can also be written as tan θ = r ω² / g. These relations allow you to find ω directly from geometry:

ω = √(g tan θ / r)

This type of analysis is a classic CCEA question that tests your ability to resolve forces and combine kinematics.

两式相除可消去 T，得到 tan θ = v² / (r g)。代入 v = r ω 后得到 tan θ = r ω² / g，由此可直接从几何条件求出 ω： ω = √(g tan θ / r) 。该类分析是 CCEA 的经典考题，考查受力分解与运动学公式的综合运用能力。

9. Vertical Circular Motion | 竖直面内的圆周运动

When an object moves in a vertical circle, the speed often changes due to gravity, but at any instant the centripetal acceleration is still v² / r directed toward the centre. The net radial force equals m v² / r. An important skill is to apply this at the top and bottom of the circle.

物体在竖直面内做圆周运动时，速率常因重力而改变，但任意时刻向心加速度仍为 v² / r，方向指向圆心，且径向合力等于 m v² / r。考生需要重点掌握在圆周的最高点和最低点应用这一关系。

At the top: both weight mg and the normal reaction N (or tension) point downwards. The resultant radial force is mg + N = m v² / r. The minimum speed to maintain the circular path occurs when N = 0, giving v_min = √(g r).
At the bottom: the normal reaction N acts upwards and weight mg downwards, so N − mg = m v² / r. Hence N = mg + m v² / r, meaning the reaction is greater than the weight.

在最高点：重力 mg 和法向反作用力 N（或拉力）均向下，径向合力为 mg + N = m v² / r。维持圆周运动的最小速度出现在 N = 0 时，得 v_min = √(g r)。在最低点：N 向上，mg 向下，有 N – mg = m v² / r，因此 N = mg + m v² / r，即反作用力大于重力。

These expressions are commonly examined in the context of a bucket of water swung in a vertical circle, a roller‑coaster loop, or a mass on a string. Always draw a clear free‑body diagram and indicate the positive direction towards the centre.

这些表达式常见于“竖直面内水桶转动”、“过山车回环”或“绳端物体”等情境。务必画清受力图，并规定指向圆心的方向为正方向。

10. Vehicles on Flat and Banked Curves | 水平弯道与倾斜弯道上的车辆

When a car travels around a flat, unbanked bend, the friction between the tyres and the road provides the centripetal force. The maximum speed v_max before skidding is given by:

μ m g = m v_max² / r → v_max = √(μ g r)

where μ is the coefficient of static friction. This demonstrates that the maximum safe speed depends on μ and the radius of the bend.

汽车在水平无倾斜的弯道上行驶时，轮胎与路面间的摩擦力提供向心力。即将侧滑时的最大速度 v_max 满足 μ m g = m v_max² / r ，解得 v_max = √(μ g r) 。可见最高安全车速取决于静摩擦系数 μ 和弯道半径 r。

On a banked track, a component of the normal reaction helps to provide the centripetal force. For a frictionless banked curve at angle θ to the horizontal, the ideal speed v_ideal is given by:

tan θ = v_ideal² / (r g)

At this speed, no sideways frictional force is required. CCEA questions often ask you to derive this condition by resolving the normal reaction into horizontal and vertical components.

在倾斜弯道上，法向反作用力的水平分量帮助提供向心力。对于无摩擦且倾角为 θ（与水平面夹角）的理想弯道，理想车速 v_ideal 满足 tan θ = v_ideal² / (r g) 。以此速度过弯时，无需侧向摩擦力。CCEA 常要求考生通过对法向反作用力进行分解来推导这一条件。

11. Energy Considerations in Circular Motion | 圆周运动中的能量考量

While the centripetal force does no work (it is always perpendicular to the instantaneous velocity), energy methods can still be applied to circular motion problems, especially in vertical circles where speed changes. The work–energy principle or conservation of mechanical energy often helps to relate the speed at one point of a vertical circle to that at another.

虽然向心力始终与瞬时速度垂直而不做功，但在圆周运动问题中仍可使用能量方法，尤其是在竖直面内速率变化的场景。功能原理或机械能守恒常用于关联竖直圆周上不同位置的速度。

For example, a particle attached to a string and released from rest at the horizontal position will have a speed v at the lowest point given by:

m g r = ½ m v² → v = √(2 g r)

Combining this with the centripetal force equation at the bottom allows you to find the tension in the string. Such synoptic questions explicitly test the link between mechanics topics, a hallmark of A‑Level physics.

例如，一质点系于绳端从水平位置由静止释放，到达最低点时的速度 v 由机械能守恒给出： m g r = ½ m v² → v = √(2 g r) 。再结合最低点的向心力方程即可求出绳的拉力。这类综合性问题清晰体现了力学知识点的融会贯通，正是 A‑Level 物理的特色。

12. Exam Tips for CCEA Circular Motion Questions | CCEA 圆周运动考题答题技巧

Always identify the physical force(s) providing the centripetal force — never invent a “centripetal force”.
坚持先找出提供向心力的真实力，绝不虚构一个“向心力”。
Convert all units to SI: radians, metres, seconds. Do not forget to convert revolutions per minute to rad s⁻¹.
统一使用国际单位制：弧度、米、秒。切记将每分钟转数换算为 rad s⁻¹。
Show clearly any resolution of forces, often with a labelled diagram, and write the net radial force equation explicitly.
清晰地展示力的分解，最好配上受力分析图，并明确写出径向合力方程。
When a question involves two or more bodies (e.g., a mass sliding inside a hollow cylinder), apply Newton’s laws separately and link them through common accelerations or tensions.
涉及多个物体的问题（如滑块在空心圆筒内运动），要对各物体分别应用牛顿定律，再通过共同的加速度或拉力建立联系。
Check that your answer is physically reasonable: for instance, the tension at the bottom of a vertical circle should be larger than at the top.
检查答案的物理合理性：例如竖直圆周底部拉力应大于顶部。
Practice drawing vectors: velocity tangential, acceleration and net force radial inward.
多加练习矢量作图：速度沿切线方向，加速度和合力沿径向指向圆心。

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A-Level CCEA Physics: Circular Motion – Key Points | A-Level CCEA 物理：圆周运动 考点精讲