Circular Motion: Core Concepts for IB & WJEC Physics | 圆周运动：IB 与 WJEC 物理核心考点

📚 Circular Motion: Core Concepts for IB & WJEC Physics | 圆周运动：IB 与 WJEC 物理核心考点

Circular motion is a fundamental topic in both IB and WJEC physics syllabi, describing the motion of an object along a circular path. Mastering angular quantities, centripetal acceleration, and the forces that cause such motion is vital for solving problems involving loops, banked curves, satellites, and conical pendulums. This guide breaks down every key concept with clear explanations, derived equations using only Unicode symbols, and exam-focused tips to help you avoid common pitfalls.

圆周运动是 IB 和 WJEC 物理课程共同的基础内容，刻画沿圆形路径的运动。熟练掌握角量、向心加速度及其背后的力学原理，是攻克过山车、倾斜弯道、卫星轨道和圆锥摆等题型的必备能力。本文通过清晰的双语讲解、完全使用 Unicode 符号呈现的公式推导以及应试技巧，帮你避开典型失分点。

1. Defining Circular Motion | 圆周运动的定义

An object is in circular motion if its trajectory forms a circle. In uniform circular motion, the object's speed (magnitude of velocity) remains constant, yet the velocity vector changes continuously because its direction is always tangent to the circle. This continuous change in velocity implies the presence of an acceleration directed towards the centre, known as centripetal acceleration.

如果物体的运动轨迹是一个圆，就称其在做圆周运动。在匀速圆周运动中，物体的速率保持不变，但速度方向始终沿切线变化，因此速度矢量持续改变。这种速度的变化表明存在一个指向圆心的加速度，即向心加速度。

The motion can be described using linear quantities (displacement s, velocity v) along the circumference, or angular quantities (angular displacement Δθ, angular velocity ω) measured in radians. The two sets are linked by the radius r of the circle.

描述这种运动时，既可以用沿圆周的线量（弧长 s、线速度 v），也可以用使用弧度制表示的角量（角位移 Δθ、角速度 ω）。这两套物理量通过圆半径 r 相互联系。

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2. Angular Displacement, Velocity and Speed | 角位移、角速度与线速度

Angular displacement Δθ is the angle subtended at the centre when an object moves along an arc. It must be measured in radians for the fundamental relationships to be valid. One radian is the angle for which arc length equals the radius.

角位移 Δθ 是物体沿圆弧运动时在圆心所张的角度。必须使用弧度（rad）作单位，后续关键公式才能成立。1 弧度定义为弧长等于半径时所对应的圆心角。

displacement: ω = Δθ/Δt. The linear speed v of the object is related to ω by the equation:

v = rω

角速度 ω 是角位移随时间的变化率：ω = Δθ/Δt。物体的线速率 v 与角速度 ω 满足关系：v = rω。

(with θ in radians), and differentiating this with respect to time gives v = rω. This connection is essential for converting between linear and angular descriptions in all circular motion problems.

因为 v 是切线方向速度的大小，它的方向始终与半径垂直。在弧度制下，弧长 s = rθ，两边对时间求导即得 v = rω。这一联系是所有圆周运动问题中在角度描述与线量描述之间转换的核心。

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3. Period and Frequency | 周期与频率

revolutions per second, so f = 1/T. Angular velocity can be expressed in terms of T and f:

ω = 2π/T = 2πf

周期 T 是物体完成一整圈运动所需的时间。频率 f 是每秒转动的圈数，满足 f = 1/T。角速度可以用 T 和 f 表示为：ω = 2π/T = 2πf。

Substituting these into v = rω gives the useful relationships:

v = 2πr/T = 2πrf

将其代入 v = rω，得到常用关系式：v = 2πr/T = 2πrf。

When solving problems, always check whether the system rotates with a constant period (e.g. a satellite) or whether the period varies. Many exam questions ask for the velocity or acceleration at a specific point in the motion, requiring you to link f or T to the linear quantities.

解题时要分清系统是否以恒定周期旋转（例如卫星），还是周期在变化。许多考题会要求你计算某一点的线速度或加速度，这时就要将 f 或 T 与线量关联起来。

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4. Centripetal Acceleration – Magnitude and Direction | 向心加速度的大小和方向

For an object moving with constant speed v in a circle of radius r, the magnitude of the centripetal acceleration aₙ is:

aₙ = v²/r = ω²r

物体以恒定速率 v 在半径为 r 的圆周上运动时，向心加速度 aₙ 的大小为：aₙ = v²/r = ω²r。

The direction of aₙ is always towards the centre of the circle, perpendicular to the instantaneous velocity. Since the acceleration is perpendicular to velocity, it changes the direction of the velocity without changing its magnitude. This is why uniform circular motion still requires a resultant centripetal force.

向心加速度的方向始终指向圆心，与瞬时速度垂直。正是因为加速度与速度垂直，它只能改变速度的方向而不改变大小，这也就是匀速圆周运动需要合外力提供向心力的原因。

form is particularly helpful when analysing planetary motion or objects with a known rotation period.

利用周期 T 可将向心加速度写为 aₙ = 4π²r/T²。在分析行星运动或已知旋转周期的问题时，这种形式特别有用。

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5. Derivation of Centripetal Acceleration | 向心加速度的推导

he object moves from point A to B, sweeping out an angle Δθ. The change in velocity Δv = v•B – v• the triangle formed by velocities and the triangle formed by the radii are similar, giving |Δv|/v = Δs/r, where Δs is the chord length. For small Δt, chord length ≈ arc length vΔt.

设想物体以恒定速率 v 做圆周运动。在很短的时间 Δt 内，物体从 A 点运动到 B 点，转过的角度为 Δθ。速度的变化量 Δv = v_B – v_A 指向圆心。由几何关系，速度矢量三角形与半径三角形相似，因此有 |Δv|/v = Δs/r，其中 Δs 是弦长。Δt 很小时，弦长可近似为弧长 vΔt。

Using the approximate relation Δs ≈ vΔt, we obtain |Δv| ≈ (v × vΔt)/r. Dividing by Δt gives the magnitude of acceleration:

aₙ = |Δv|/Δt ≈ v²/r

将 Δs ≈ vΔt 代入，得到 |Δv| ≈ v²Δt/r。两边除以 Δt 即得加速度大小近似为 v²/r。取 Δt → 0 极限就是严格的瞬时向心加速度。

s aₙ = ω²r. Both forms appear regularly in IB and WJEC exam questions, so understanding the underlying geometry helps you remember which formula to use.

这一推导解释了加速度为何指向圆心，以及大小为何是 v²/r。改用角速度 ω 表述，则得到 aₙ = ω²r。这两个形式在 IB 和 WJEC 试题中都会反复出现，理解其中的几何关系有助于你在考场上准确选用。

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6. Centripetal Force: It's Not a New Force | 向心力：并非新的力

Newton's second law applied to the centripetal acceleration gives the centripetal force Fₙ required to keep an object moving in a circle:

Fₙ = m aₙ = mv²/r = mω²r

将牛顿第二定律用于向心加速度，即得到维持圆周运动所需的向心力 Fₙ：Fₙ = m v²/r = m ω²r。

Many students mistakenly treat centripetal force as a distinct "new" force, like gravity or tension. In reality, centripetal force is simply the name given to the resultant force directed towards the centre of the circle. It is provided by one or more real forces – tension in a string, friction on a flat track, the normal component on a banked curve, or gravitational attraction for a satellite.

很多同学错误地把向心力当作一种独立的、像重力或绳拉力那样的“新”力。实际上，向心力只是对指向圆心的合外力的称呼。它由一个或多个真实的力提供：绳子拉力、水平面的摩擦力、倾斜面上的支持力分量、卫星受到的万有引力等。

When drawing free-body diagrams, never include a separate arrow labelled "centripetal force". Instead, identify the physical forces present and then equate their radial component to mv²/r or mω²r. This approach will keep you safe from the common "centrifugal force" distraction in exams.

在画受力分析图时，千万不要单独画一个标着“向心力”的箭头。正确做法是先标出所有真实力，然后将指向圆心的合力分量设为 mv²/r 或 mω²r。这一习惯能帮你避免陷入“离心力”的考试陷阱。

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7. Horizontal Circular Motion Problems | 水平面内的圆周运动问题

ng in a horizontal circle with negligible air resistance; the tension T acts radially inward, so T = mv²/r.

物体在水平面内做圆周运动时，指向圆心的径向合力提供向心力。一个典型例子是忽略空气阻力时绳子拴着的小球在水平面旋转：绳子拉力 T 指向圆心，因此 T = mv²/r。

al force up to its maximum value fₘₐₓ = μₛ m g. The condition for the mass not to slip is:

mv²/r ≤ μₛ m g ⇒ v ≤ √μₛ g r

另一种常见情况是粗糙转盘上的物体。静摩擦力 f 提供向心力，其最大值为 f_max = μₛ m g。物体不滑动的条件为：mv²/r ≤ μₛ m g，即 v ≤ √μₛ g r。

In such problems, always identify the force that acts towards the centre. If friction acts perpendicular to the instantaneous velocity, it does no work, so speed can stay constant even without a motor. Exam questions often ask for the minimum coefficient of friction or the maximum angular velocity before slipping occurs.

处理这类问题时，务必要明确指向圆心的力是哪一个。摩擦力若始终与瞬时速度垂直，就不做功，因此即使没有驱动装置，速率也可保持不变。考题常要求计算不滑动所需的最小摩擦系数或最大角速度。

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8. Vertical Circular Motion – Loops and Hills | 竖直面内的圆周运动 – 过山车与拱桥

For an object attached to a string and whirled in a vertical circle, tension varies with position. At the top, both tension and weight point downwards, contributing to the centripetal force:

T_top + mg = mv²_top / r

物体用绳子系住在竖直面内旋转时，绳子拉力随位置变化。在最高点，拉力与重力都向下，共同提供向心力：T_top + mg = mv²_top / r。

At the bottom, tension acts upward while weight acts downward, so:

T_bottom – mg = mv²_bottom / r

在最低点，拉力向上而重力向下，有：T_bottom – mg = mv²_bottom / r。

For a string to remain taut at the top, the speed must be large enough that the required centripetal force is at least mg. The critical minimum speed at the top is v_min = √gr. If v_top < √gr, the string goes slack and the object leaves the circular path. For a rigid rod, however, the rod can exert a compressive force, so the speed at the top can be as low as zero.

要使绳子在最高点保持拉直，速度必须足够大，使得所需向心力至少等于 mg。最高点的临界最小速度是 v_min = √gr。如果 v_top < √gr，绳子将松弛，物体脱离圆形轨道。若是轻杆连接，杆可提供向外的支持力，因此最高点速度可以小至零。

The same principles apply to a car moving over a convex hill. At the top of the hill, the normal reaction N is less than the weight:

mg – N = mv²/r

汽车过拱桥顶部也是类似情形。在拱顶，支持力 N 小于重力：mg – N = mv²/r。速度越大，N 越小，当 v = √gr 时 N = 0，车辆将失去路面支撑。

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9. Banked Tracks and Conical Pendulum | 倾斜轨道与圆锥摆

A banked curve allows a vehicle to negotiate a corner at a certain speed without relying on friction. The horizontal component of the normal reaction provides the centripetal force. Resolving forces for a bank angle θ and neglecting friction gives:

N sinθ = mv²/r

N cosθ = mg

Dividing the two equations yields the ideal banking angle condition:

tanθ = v²/(rg)

倾斜弯道能令车辆在特定速度下不用借助摩擦力即可转弯。支持力的水平分量提供向心力。忽略摩擦时，设斜坡倾角为 θ，则有：N sinθ = mv²/r，N cosθ = mg。两式相除得到理想倾斜角条件：tanθ = v²/(rg)。

. The same force resolution applies, with the radius r = L sinθ, where L is the string length. The period of rotation is independent of the mass and is given by T = 2π√L cosθ/g, a classic result often examined.

圆锥摆是通过绳子将小球系住，小球在水平面内做匀速圆周运动，悬线扫出圆锥面的形状。受力分解与倾斜轨道相同，半径满足 r = L sinθ（L 为绳长）。旋转周期与摆球质量无关，T = 2π√L cosθ/g，这是一个经典考点。

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10. Gravitational Force as Centripetal – Orbits and Satellites | 万有引力提供向心力 – 轨道与卫星

For a satellite of mass m in a circular orbit about a planet of mass M, the gravitational attraction supplies the centripetal force:

G M m / r² = mv²/r

质量为 m 的卫星在质量 M 的行星周边做圆周轨道运动时，万有引力提供向心力：G M m/r² = mv²/r。

Cancelling m gives the orbital speed:

v = √GM/r

这表明轨道速度与卫星质量无关：v = √GM/r。

Using v = 2πr/T, the orbital period T satisfies Kepler's third law:

T² = (4π²/GM) r³

The period increases with orbital radius. Geostationary satellites have a period of 24 hours and an orbital radius of about 42,200 km from the Earth's centre. Questions often require you to derive the orbital radius from T or vice versa, and to compare the speeds and periods of satellites at different altitudes.

利用 v

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