📚 AQA Maths: Mastering Circular Motion | AQA 数学：圆周运动考点精讲
Circular motion appears in the Mechanics section of AQA A-Level Mathematics, often in Paper 2 or 3. It builds on Newton’s laws and kinematics, requiring you to understand angular quantities, centripetal force, and how to resolve forces in a radial direction. This revision guide highlights the key formulas, typical exam scenarios, and common pitfalls, giving you a clear path to top marks.
圆周运动出现在 AQA A-Level 数学的力学部分（通常为 Paper 2 或 3）。它以牛顿定律和运动学为基础，要求你理解角量、向心力以及如何沿径向分解力。本文梳理核心公式、典型考题情景和常见失分点，帮助你高效备考。
1. Radians and Angular Displacement | 弧度与角位移
In circular motion, all angular measurements must be in radians. One radian is the angle subtended at the centre when the arc length equals the radius. Therefore, for a circle of radius r, the arc length s, angle θ (in radians), and radius are linked by s = rθ.
在圆周运动中，所有角度测量必须使用弧度。1 弧度的定义是弧长等于半径时的圆心角。因此，对于半径为 r 的圆，弧长 s、角度 θ（弧度）和半径的关系为 s = rθ。
Angular displacement is the change in the angular position of a particle moving along a circular path. When an object moves from point A to point B along the arc, the angle Δθ swept out is the angular displacement. Remember that 2π rad = 360°, so to convert degrees to radians multiply by π/180.
角位移是物体沿圆周路径运动时角位置的变化。当物体从 A 点运动到 B 点，扫过的角度 Δθ 就是角位移。记住 2π rad = 360°，所以从度数转为弧度需乘以 π/180。
2. Angular Velocity and Linear Speed | 角速度与线速度
Angular velocity ω (omega) measures how fast an object rotates. It is defined as the rate of change of angular displacement: ω = Δθ/Δt. Its units are rad s⁻¹. For uniform circular motion, ω is constant, and the object completes one full revolution in period T. Hence ω = 2π/T or ω = 2πf, where f is the frequency in Hz.
角速度 ω 衡量物体转动的快慢，定义为角位移的变化率：ω = Δθ/Δt，单位是 rad s⁻¹。在匀速圆周运动中 ω 恒定，物体在周期 T 内完成一整圈。因此 ω = 2π/T 或 ω = 2πf，其中 f 是频率（Hz）。
The linear speed v along the circular path is connected to angular velocity by the simple equation v = rω. This is one of the most used relationships in exams. Note that although the speed is constant in uniform circular motion, velocity is not constant because the direction changes continuously.
沿圆周的线速率 v 与角速度通过简单的方程关联： v = rω。这是考试中使用最频繁的关系式之一。注意，在匀速圆周运动中速率不变，但速度并不恒定，因为方向时刻变化。
v = rω , ω = 2π/T , v = 2πr/T
3. Centripetal Acceleration | 向心加速度
An object moving in a circle at constant speed is still accelerating because its direction changes. This acceleration is directed towards the centre of the circle and is called centripetal acceleration. Its magnitude can be expressed in two equivalent forms: a = v²/r or a = rω².
以恒定速率做圆周运动的物体仍然在加速，因为方向不断改变。该加速度指向圆心，称为向心加速度。其大小可用两种等价形式表示：a = v²/r 或 a = rω²。
You can derive a = v²/r by considering the change in velocity vector over a small time interval and the geometry of similar triangles. In AQA exams, you are usually given these formulas on the data sheet, but you must know how to apply them and when to use each version. Using a = rω² can save time when angular velocity is given directly.
可以通过考虑微小时间间隔内的速度矢量变化以及相似三角形来推导 a = v²/r。AQA 考试通常会在公式表给出这些公式，但你仍需知道如何应用以及何时使用哪个版本。当直接给出角速度时，使用 a = rω² 可以节省时间。
a = v²/r = rω²
4. Centripetal Force and Newton’s Second Law | 向心力与牛顿第二定律
Because there is a centripetal acceleration, there must be a net resultant force acting towards the centre according to Newton’s second law. This force is the centripetal force, provided by F = ma, so F = mv²/r = mrω². It is not a separate “new” force but the net force in the radial direction.
既然存在向心加速度，根据牛顿第二定律，必然有一个指向圆心的净合力。这个力就是向心力，由 F = ma 给出，即 F = mv²/r = mrω²。它并不是一个独立的“新”力，而是沿径向的合力。
A common misconception is to draw centripetal force as an outward force or to label it separately on a free-body diagram. In exams, you should identify the actual physical force (tension, friction, normal reaction, gravity component) that points towards the centre and equate it to mv²/r. Do not add a mysterious ‘centripetal force’ arrow.
一个常见的误解是把向心力画成向外的力，或在受力图上单独标注它。考试中应识别出指向圆心的实际力（如拉力、摩擦力、支持力、重力分量），并令其等于 mv²/r。不要添加一个神秘的“向心力”箭号。
F = mv²/r = mrω²
5. Identifying Centripetal Force in Different Contexts | 不同情境中向心力的识别
Exam questions love to test your ability to pick out the centripetal force from a scenario. For a mass on a string whirled horizontally, the tension provides the centripetal force. For a car rounding a flat bend, friction between tyres and road provides it. For a planet orbiting the Sun, gravitational attraction is the centripetal force.
考试喜欢考查你能否从具体情景中识别出向心力。对于用绳子水平旋转的物体，拉力提供向心力。对于在平直弯道上行驶的汽车，轮胎与路面的摩擦力提供。对于绕太阳运行的行星，万有引力就是向心力。
When an object moves in a vertical circle, the net force towards the centre at any point is a combination of tension/ normal reaction and a component of weight. You must resolve forces along the radius and set the net inward force equal to mv²/r. This is where many students lose marks by forgetting the weight component.
当物体在竖直面内做圆周运动时，任意一点指向圆心的合力由拉力/支持力与重力的分量共同构成。你必须沿径向分解力，并令向内的净力等于 mv²/r。很多学生因为忽略重力分量而失分。
| Scenario | Centripetal Force Provider |
|---|---|
| Car on flat curve | Friction |
| Car on banked track (no friction) | Horizontal component of normal reaction |
| Conical pendulum | Horizontal component of tension |
| Vertical circle (top) | Tension + weight (both towards centre) |
| Satellite orbit | Gravitational force |
6. The Conical Pendulum | 圆锥摆
A conical pendulum consists of a small mass attached to a string, moving in a horizontal circle at constant angular velocity. The string traces out a cone. Forces acting are tension T and weight mg. Resolving vertically: T cos θ = mg. Resolving radially: the horizontal component T sin θ provides the centripetal force, so T sin θ = mv²/r.
圆锥摆由系在绳子上的小质量块组成，它以恒定角速度在水平面内做圆周运动，绳子扫出一个圆锥面。受力有拉力 T 和重力 mg。竖直方向分解：T cos θ = mg。径向分解：水平分量 T sin θ 提供向心力，因此 T sin θ = mv²/r。
By combining these equations, you can relate the period, length of string, and angle. A favourite exam question is to show that the period depends only on the vertical depth h of the bob below the suspension point, and not on the mass or the length directly. The derived expression is T_period = 2π√(h/g).
结合这两个方程，可以建立周期、绳长和角度的关系。常考题型是证明周期仅取决于摆球在悬挂点下方的竖直深度 h，而与质量或绳长无直接关系。推导出的表达式为 T_period = 2π√(h/g)。
Period T = 2π√(L cos θ / g) = 2π√(h/g)
7. Vertical Circular Motion: Bucket of Water | 竖直圆周运动：水桶问题
When an object moves in a vertical circle, its speed usually changes depending on height due to conservation of energy. At the top of the circle, the centripetal force is provided by the sum of the tension T and weight mg both acting downwards. The equation is T + mg = mv²/r.
当物体在竖直面内做圆周运动时，由于能量守恒，速率通常会随高度改变。在圆周最高点，向心力由向下作用的拉力 T 和重力 mg 共同提供。方程为 T + mg = mv²/r。
For a bucket of water swung in a vertical circle, water does not fall out at the top if its speed is sufficient to keep it moving in a circle. The minimum speed occurs when the contact force (normal reaction or tension) falls to zero. Setting T = 0 gives mg = mv²/r, so v_min = √(gr). This is the critical speed at the top.
对于在竖直面内旋转的水桶，如果速度足够使水继续做圆周运动，水在顶部就不会洒出来。最小速度发生在接触力（支持力或拉力）减小为零时。令 T = 0，得 mg = mv²/r，因此 v_min = √(gr)。这就是最高点的临界速度。
At the bottom of the circle, tension and weight are opposite, with T upwards and weight downwards. The net centripetal force is T – mg = mv²/r. At intermediate positions, you must resolve weight radially and apply energy conservation to find speed.
在最低点，拉力和重力方向相反，T 向上，mg 向下。向心合力为 T – mg = mv²/r。在中间位置，必须沿径向分解重力，并利用能量守恒求出速度。
8. Vehicles on Curved Roads and Banked Tracks | 弯道上的车辆与倾斜轨道
For a car taking a flat, unbanked corner, the necessary centripetal force is provided entirely by static friction f between the tyres and the road. The maximum safe speed occurs when friction reaches its limiting value μR, where R is the normal reaction. Then μR = mv²/r, and since R = mg on a horizontal road, v_max = √(μgr).
对于在平坦无倾斜弯道上行驶的汽车，所需的向心力完全由轮胎与路面之间的静摩擦力 f 提供。最大安全速度发生在摩擦力达到极限值 μR 时，R 为支持力。水平路上 R = mg，因此 v_max = √(μgr)。
On a banked track with no side friction, the horizontal component of the normal reaction supplies the centripetal force. Resolving forces: R cos θ = mg vertically, R sin θ = mv²/r horizontally. Dividing gives tan θ = v²/(rg). This allows you to find the ideal banking angle for a given speed or the safe speed for a given angle.
在无侧向摩擦的倾斜轨道上，支持力的水平分量提供向心力。分解力：竖直方向 R cos θ = mg，水平方向 R sin θ = mv²/r。两式相除得 tan θ = v²/(rg)。由此可求给定速度的理想倾斜角，或给定角度下的安全速度。
Many AQA problems combine a banked track with friction, asking you to find the maximum and minimum speeds before slipping up or down the slope. Set up equations with friction acting up or down the plane and resolve parallel and perpendicular to the surface.
许多 AQA 题目结合倾斜轨道和摩擦力，要求你求出不向上或向下滑移的最大和最小速度。需设摩擦力沿斜面向上或向下，并沿平行和垂直表面分解力建立方程。
9. Critical Speed and Looping the Loop | 临界速度与轨道翻转
A particle moving on the inside of a circular track or a roller coaster loop must maintain contact at the top. The condition for contact is that the normal reaction N ≥ 0. At the top, if N = 0, the centripetal force is provided solely by gravity: mg = mv²/r ⇒ v = √(gr). This is the minimum speed at the top to stay on the track.
在圆形轨道内侧或过山车环圈上运动的物体，在顶部必须保持接触。接触条件为支持力 N ≥ 0。在顶部，若 N = 0，向心力仅由重力提供：mg = mv²/r ⇒ v = √(gr)。这是保持不脱离轨道所需的最小顶部速度。
If the speed at the top is less than √(gr), the particle loses contact earlier. At the bottom of the loop, the normal reaction is much larger because it must support the weight and provide centripetal force: N – mg = mv²/r. You can use conservation of energy to link speeds at different heights.
如果顶部速度小于 √(gr)，物体就会提前脱离轨道。在环圈底部，支持力非常大，因为它既要抵消重量又要提供向心力：N – mg = mv²/r。你可以利用能量守恒把不同高度的速度联系起来。
When tackling such problems, first define the radius of curvature and set up the radial force equation at the position of interest. Then use energy conservation (ΔKE + ΔPE = 0 or work-energy) to find the required speed. Always check whether the given initial speed satisfies the contact condition.
解决这类问题时，首先明确曲率半径并在所需位置建立径向力方程，然后利用能量守恒（动能变化 + 势能变化 = 0 或功能原理）求出速度。务必检查初始速度是否满足接触条件。
10. Exam Technique and Common Pitfalls | 答题技巧与常见易错点
Many students lose marks because they mix up tangential and radial directions. Remember: for uniform circular motion, tangential acceleration is zero; the net force is purely radial. In non-uniform vertical circles, radial force equation still holds at each instant, using the instantaneous speed.
许多学生因混淆切向和径向而失分。记住：对于匀速圆周运动，切向加速度为零；净力完全沿径向。在非匀速的竖直圆周运动中，径向力方程在每一瞬时仍然成立，使用瞬时速度即可。
Always draw a clear free-body diagram showing all real forces. Indicate the positive radial direction (usually towards the centre). Write the net inward force = mv²/r. If a question involves a banked track or a conical pendulum, resolve forces along perpendicular axes (vertical and horizontal, or along the plane) carefully.
一定要画清晰的受力图，标出所有实际力。标明正径向方向（通常指向圆心）。写出净向心力 = mv²/r。如果题目涉及倾斜轨道或圆锥摆，要仔细将力沿垂直轴（竖直和水平）或沿斜面分解。
Check your units: angular velocity must be in rad s⁻¹, not rpm or degrees per second unless converted. When using v = rω, r must be in metres and v in m s⁻¹. Also, don’t forget that frequency f = 1/T and ω = 2πf. Substituting directly can speed up your solution.
检查单位：角速度必须是 rad s⁻¹，不能直接使用 rpm 或度每秒，除非换算。使用 v = rω 时，r 以米为单位，v 以 m s⁻¹ 为单位。另外，不要忘记频率 f = 1/T，ω = 2πf。直接代入可以加快解题。
A final tip: if a particle is attached to a rod instead of a string, it can withstand compression, so at the top the rod can push outward. This changes the critical speed condition – a rod can have zero speed at the top and still maintain contact. Read the question carefully to distinguish between strings and rods.
最后提示：如果物体连接在杆上而非绳子，杆可以承受压力，所以在最高点杆可以向外推。这改变了临界速度条件——在顶部速度为零仍可保持接触。仔细读题，区分绳子和杆。
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