Circular Motion | 圆周运动考点精讲

📚 Circular Motion | 圆周运动考点精讲

Circular motion is a fundamental topic in the WJEC A-Level Mathematics mechanics syllabus. Mastering the relationships between angular and linear quantities, understanding centripetal force, and applying these concepts to horizontal and vertical circles are essential skills. This revision guide covers all the key points, common exam question types, and examiner tips to help you succeed.

圆周运动是WJEC A-Level数学力学部分的基础课题。掌握角量与线量之间的关系，理解向心力，并将这些概念应用于水平面和竖直面内的圆周运动，是必备的技能。这份考点精讲涵盖所有关键知识点、常见考题类型和阅卷技巧，助你顺利拿分。

1. Defining Circular Motion | 圆周运动的定义

An object moving in a circular path at constant speed is said to be in uniform circular motion. Despite the constant speed, the velocity continuously changes because its direction changes. This means the object experiences an acceleration directed towards the centre of the circle.

一个物体以恒定的速率沿圆形路径运动，就称为匀速圆周运动。尽管速率不变，但由于方向不断改变，速度矢量是变化的，因此物体具有指向圆心的加速度。

The time taken to complete one full revolution is the period T, measured in seconds. The number of revolutions per unit time is the frequency f, where f = 1/T.

完成一整圈所用的时间称为周期 T，单位为秒。单位时间内的转动圈数是频率 f，f = 1/T。

2. Angular Displacement and Radian Measure | 角位移与弧度制

Angular displacement θ is the angle through which a radius vector rotates. In circular motion problems, angles must be expressed in radians. One radian is the angle subtended at the centre when the arc length equals the radius.

角位移 θ 是半径矢量转过的角度。在圆周运动问题中，角度必须用弧度制表示。一弧度定义为弧长等于半径时所对圆心角的大小。

The conversion is: π rad = 180°. For an arc length s, we have s = rθ. This simple relationship underpins the link between linear and angular motion.

换算关系为：π 弧度 = 180°。弧长 s 满足 s = rθ。这个简洁的关系式是线量与角量联系的基础。

3. Linear and Angular Velocity | 线速度与角速度

Angular velocity ω is the rate of change of angular displacement: ω = θ/t (for constant ω). Its unit is rad s⁻¹. Linear speed v is the distance travelled along the circumference per unit time: v = s/t.

角速度 ω 是角位移的变化率：ω = θ/t（当 ω 恒定时），单位是 rad s⁻¹。线速度 v 是沿圆周移动的弧长除以时间：v = s/t。

Combining v = s/t with s = rθ gives the key relation:

结合 v = s/t 与 s = rθ 得到核心关系式：

v = rω

Since one revolution corresponds to an angle of 2π rad, the period is T = 2π/ω, and v = 2πr/T.

一整圈对应 2π 弧度，所以周期 T = 2π/ω，且 v = 2πr/T。

4. Centripetal Acceleration | 向心加速度

For an object moving uniformly in a circle, the acceleration is always directed towards the centre. This is the centripetal acceleration. Its magnitude is given by two equivalent formulas:

做匀速圆周运动的物体，其加速度始终指向圆心，称为向心加速度。其大小由两个等价的公式给出：

a = v²/r

a = ω²r

These expressions show that for a given radius, a higher speed requires a larger acceleration; for a fixed angular velocity, a larger radius results in greater acceleration. Even though speed is constant, a non-zero acceleration exists because the direction of velocity changes.

这些表达式表明，给定半径时，更大的线速度需要更大的加速度；角速度固定时，半径越大加速度越大。即使速率不变，由于速度方向变化，加速度也非零。

5. Centripetal Force | 向心力

Applying Newton’s second law in the radial direction, the net force towards the centre equals mass times centripetal acceleration. This resultant force is called centripetal force:

在径向应用牛顿第二定律，指向圆心的合力等于质量乘以向心加速度。这个合力就是向心力：

F = mv²/r = mω²r

It is crucial to understand that centripetal force is not a new type of force; it is simply the name for the resultant force directed towards the centre. In different situations, it may be provided by tension, friction, the normal reaction, or a component of weight.

关键是理解向心力并非一种新型的力，而是指向圆心的合力的称谓。不同情形下，它可由绳的张力、摩擦力、支持力或重力的分量提供。

When solving problems, always draw a clear free-body diagram, identify the forces acting, and set the net inward force equal to mv²/r or mω²r.

解题时，务必画出清晰的受力分析图，标出所有力，并令指向圆心的净力等于 mv²/r 或 mω²r。

6. Horizontal Circular Motion | 水平面内的圆周运动

Common horizontal circle scenarios include a mass on a smooth table rotating under tension, a vehicle turning on a flat road, or a particle on a rotating turntable. In each case, the vertical forces (weight and normal reaction) balance, and the horizontal resultant provides the centripetal force.

常见的水平圆周运动包括水平光滑桌面上的旋转物块、车辆在平路上的转弯、转盘上的物体等。每种情况下，竖直方向的力（重力和支持力）平衡，水平的合力充当向心力。

For a car turning on a flat road, the centripetal force is supplied by the friction between the tyres and the road. The maximum friction is μR = μmg, so the maximum safe speed is v_max = √(μgr). If the speed exceeds this, skidding occurs.

车辆在平路转弯时，向心力由轮胎与路面之间的摩擦力提供。最大静摩擦力为 μR = μmg，因此最大安全速度为 v_max = √(μgr)。超过此速度，车辆将发生侧滑。

7. Conical Pendulum and Banked Tracks | 圆锥摆与斜面转弯

A conical pendulum consists of a mass attached to a string revolving in a horizontal circle with the string tracing a cone. The forces are tension T and weight mg. Resolving vertically: T cos θ = mg. Horizontally: T sin θ = mω²r, where r = L sin θ, L being string length.

圆锥摆由一端系于固定点的轻绳和做水平圆周运动的小球构成，绳扫出圆锥面。对球受力分析：竖直方向 T cos θ = mg，水平方向 T sin θ = mω²r，且半径 r = L sin θ，L 为绳长。

Solving yields period T = 2π √(L cos θ / g), independent of the mass. This is a classic exam derivation question.

解出周期 T = 2π √(L cos θ / g)，与质量无关。这是典型的考试推导题。

For a banked track, the normal reaction and weight combine to provide the centripetal force. The ideal banking angle θ for a given speed v without relying on friction is tan θ = v²/(rg).

对于斜面弯道，支持力与重力的合力提供向心力。无需摩擦时，理想倾斜角 θ 满足 tan θ = v²/(rg)。

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

In a vertical circle, speed is not constant because gravity does work. Energy must be considered. A common setup is a particle attached to a light rod or string, moving in a vertical circle.

在竖直面内做圆周运动时，速率并不恒定，因为重力做功，必须考虑能量。常见模型是系于轻杆或轻绳的质点，在竖直圆上运动。

For a particle on a string, the tension varies. At the highest point, tension and weight both act downwards: T_top + mg = mv²/r. At the lowest point: T_bottom – mg = mv²/r. The minimum speed at the top for the string to remain taut is v_min = √(gr), giving T_top = 0.

对绳球模型，张力随位置变化。在最高点，张力与重力均向下：T_top + mg = mv²/r。在最低点：T_bottom – mg = mv²/r。绳子保持绷直的最高点最小速度为 v_min = √(gr)，此时 T_top = 0。

With a light rod, the particle can have zero speed at the top because the rod can provide a compressive force. The required minimum speed at the top for a rod is 0, but it still must complete the loop, often requiring energy analysis.

若为轻杆模型，质点可在最高点速度为零，因为杆可提供支撑力。杆约束下最高点最小速度可为零，但仍需机械能保证完整过圈，常需能量分析。

9. Energy in a Vertical Circle | 竖直圆中的能量

If no non-conservative forces act, mechanical energy is conserved. Taking the lowest point as zero potential energy, the speed at any angular position can be found from:

若无非保守力做功，机械能守恒。以最低点为零势能面，任意角度位置的速度可由下式得出：

½ mv²_top + mg(2r) = ½ mv²_bottom

For example, to find the speed at the top when given the speed at the bottom, use v²_top = v²_bottom – 4gr. This relation is frequently needed when calculating the tension or normal reaction at various points.

例如，已知最低点速度求最高点速度时，用 v²_top = v²_bottom – 4gr。在计算各点张力或支持力时，经常需要用到这个关系。

Combining energy and force equations allows us to determine whether an object will complete the circle or if the string goes slack mid-way.

将能量方程与受力方程结合，可判断物体能否完整过圈，或绳子是否会在中途松弛。

10. Summary of Key Formulas | 关键公式总结

The following table groups the essential equations you must recall for WJEC examinations. They are the foundation of nearly all circular motion problems.

下表汇总了WJEC考试中必须熟记的核心公式，它们是绝大多数圆周运动问题的基础。

Quantity 物理量	Formula 公式	Notes 备注
Angular velocity ω	ω = θ/t = 2π/T	Constant ω only 仅恒角速度
Linear speed v	v = rω = 2πr/T	Linked to angular speed
Centripetal acceleration a	a = v²/r = ω²r	Always towards centre 指向圆心
Centripetal force F	F = mv²/r = mω²r	Net inward force 净指向圆心的力
Conical pendulum period	T = 2π √(L cos θ / g)	Derivable from forces
Banked track ideal angle	tan θ = v²/(rg)	No friction required
Vertical circle top speed (string)	v_min_top = √(gr)	Tension ≥ 0
Energy conservation	½mv²_low = ½mv²_high + mg(2r)	Assuming zero friction

Always check the context to ensure you apply the correct form. For variable speed, energy methods often complement the force equations.

务必根据情境选用正确的公式形式。对于变速圆周运动，能量方法常与受力方程配合使用。

11. WJEC Exam Techniques | WJEC 考试技巧

WJEC mechanics questions on circular motion often combine kinematics, forces, and energy. Marks are awarded for clear diagrams, correct resolution of forces, and stated assumptions (e.g., light string, no air resistance).

WJEC 力学中关于圆周运动的题目常综合考查运动学、力和能量。清晰画图、正确分解力、明确假设（如轻绳、忽略空气阻力）都能得分。

Always write the general centripetal force equation first, then substitute the specific forces. For example, when a particle is at the top of a circle, write ‘mg + T = mv²/r’ rather than guessing the sign. This systematic approach reduces errors.

始终先写出向心力通用方程，再代入具体的力。例如，质点位于最高点时，先写 ‘mg + T = mv²/r’，而不是凭空写符号。这种系统性方法能减少错误。

If a problem involves a bead on a wire or a particle inside a circular track, be careful with the direction of the normal reaction. The reaction may press outwards or inwards depending on speed. Draw the force diagram for the specific instant and apply F = ma radially.

如果题目涉及穿在铁丝上的珠子或圆环轨道内的质点，注意支持力的方向可能向外或向内，取决于速度大小。针对那一瞬间画受力图，并从径向应用 F = ma。

When an object leaves the circular path (e.g., string goes slack or loses contact), the critical condition is that the normal reaction or tension becomes zero. Setting T=0 or R=0 in the radial equation gives the critical speed.

物体离开圆周路径时（如绳子松弛或脱离轨道），临界条件是支持力或张力为零。在径向方程中令 T=0 或 R=0，即可得到临界速度。

Practice past WJEC papers: questions frequently ask for derivations of v_min at the top, or the period of a conical pendulum. Ensure you can reproduce these derivations step by step, with explanations.

多练习 WJEC 历年真题：常考最高点最小速度的推导，或圆锥摆周期的推导。确保能一步步写出推导过程并加以解释。

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Circular Motion | 圆周运动 考点精讲