A-Level OCR Maths: Circular Motion Revision Essentials | A-Level OCR 数学：圆周运动考点精讲

📚 A-Level OCR Maths: Circular Motion Revision Essentials | A-Level OCR 数学：圆周运动考点精讲

Circular motion is a cornerstone of the OCR A-Level Mathematics Mechanics module. It builds on kinematics and Newton’s laws to describe objects moving along curved paths at constant speed. Mastering radian measure, angular velocity, centripetal acceleration and the forces that cause it is essential for both straightforward calculations and the more challenging modelling of conical pendulums, banked curves and vertical circles.

圆周运动是 OCR A-Level 数学力学模块的基石。它建立在运动学和牛顿定律的基础上，描述物体以恒定速率沿曲线路径运动。掌握弧度制、角速度、向心加速度以及产生向心加速度的力，对于直接计算和更具挑战性的锥摆、倾斜弯道和竖直圆环建模都至关重要。

1. Radian Measure and Angular Speed | 弧度制与角速度

An angle of 1 radian is subtended at the centre of a circle by an arc equal in length to the radius. This means there are exactly 2π radians in a full circle, so 360° = 2π rad, and the conversion factor is 1 rad ≈ 57.3°.

1 弧度是圆心处由长度等于半径的弧所对的角。这意味着整个圆周角恰好为 2π 弧度，因此 360° = 2π rad，转换因子为 1 rad ≈ 57.3°。

Angular speed ω (omega) measures how quickly the angular displacement θ changes with time. The average angular speed is defined as ω = θ / t, with SI units of rad s⁻¹.

角速度 ω 衡量角位移 θ 随时间变化的快慢。平均角速度定义为 ω = θ / t，国际单位为 rad s⁻¹。

For uniform circular motion, the time taken to complete one full revolution is the period T, and the number of revolutions per second is the frequency f. They are linked by T = 2π / ω and f = 1 / T.

对匀速圆周运动，完成一整周所需的时间是周期 T，每秒转动的圈数是频率 f。它们的关系为 T = 2π / ω 和 f = 1 / T。

2. Relation Between Linear and Angular Velocity | 线速度与角速度的关系

When an object moves along a circular arc, the distance travelled s equals rθ, where θ is measured in radians. Differentiating both sides with respect to time gives the instantaneous speed v as v = rω, since dθ/dt = ω and r is constant.

当物体沿圆弧运动时，经过的路程 s 等于 rθ，其中 θ 以弧度为单位。两边对时间求导即得瞬时速度 v：因为 dθ/dt = ω 且 r 为常数，所以 v = rω。

v = rω

This relationship holds for every particle in a rigid rotating body, meaning that although all particles share the same angular speed, those farther from the centre have a larger linear speed.

这一关系对刚体内每一个质点都成立，这意味着虽然所有质点具有相同的角速度，但离中心越远的质点线速度越大。

3. Centripetal Acceleration | 向心加速度

Even if the speed of an object is constant, the direction of its velocity is continuously changing as it moves around the circle. This change of velocity implies an acceleration that points towards the centre of the circle. It is called centripetal acceleration, often denoted a.

即使物体的速率恒定，当它绕圆运动时速度方向也在不断变化。这种速度变化意味着存在一个指向圆心的加速度，称为向心加速度，通常记作 a。

The magnitude of centripetal acceleration can be expressed in two equivalent forms: a = v² / r and a = rω². Substituting v = rω into either form confirms their equivalence.

向心加速度的大小有两种等价表达式：a = v² / r 和 a = rω²。将 v = rω 代入即可验证它们等价。

a = v² / r = rω²

In vertical circle problems, the centripetal acceleration is still given by v² / r using the instantaneous speed at that point, and its direction is always towards the centre.

在竖直圆周问题中，向心加速度依然用该点的瞬时速率按 v² / r 计算，其方向始终指向圆心。

4. Centripetal Force | 向心力

From Newton’s second law, whenever there is an acceleration there must be a net force in the same direction. The net force directed towards the centre of a circular path is called the centripetal force. It is not a new type of force, but the resultant of real forces like tension, friction, or gravity.

根据牛顿第二定律，只要有加速度就必然存在同方向的净力。指向圆周路径中心的净力称为向心力。它不是一种新的力，而是张力、摩擦力或重力等真实力的合力。

The magnitude of centripetal force needed to keep an object of mass m moving in a circle of radius r with speed v is F = mv² / r. Equivalently, F = mrω².

为了使质量为 m 的物体以速率 v 沿半径 r 的圆周运动，所需的向心力大小为 F = mv² / r。这等价于 F = mrω²。

F = mv² / r = mrω²

In exam questions, you must resolve forces towards the centre and set the net inward force equal to mv² / r or mrω². Always draw a free-body diagram and clearly show the direction of the positive radial axis (usually towards the centre).

在考题中，你必须将力沿径向分解，并令指向中心的净力等于 mv² / r 或 mrω²。一定要画出受力图，并清楚地标出正径向坐标轴的方向（通常指向圆心）。

5. Horizontal Circular Motion: Conical Pendulum | 水平圆周运动：锥摆

A conical pendulum consists of a mass attached to a string that traces a horizontal circle. The string sweeps out a cone. The two forces acting are tension T in the string and weight mg. The vertical component of tension balances the weight, while its horizontal component provides the centripetal force.

锥摆由一个连接细绳的质量块组成，它在水平面内做圆周运动，细绳扫出一个圆锥。物体受到两个力：绳的张力 T 和重力 mg。张力的竖直分力平衡重力，水平分力提供向心力。

Resolving vertically gives Tcosθ = mg, where θ is the angle the string makes with the vertical. Resolving horizontally gives Tsinθ = mrω² (or mv²/r). The radius r = l sinθ, where l is the string length.

竖直方向分解得 Tcosθ = mg，其中 θ 是绳与竖直方向的夹角。水平方向分解得 Tsinθ = mrω²（或 mv²/r）。半径 r = l sinθ，l 为绳长。

Combining these leads to several useful results: the period T_p = 2π √(l cosθ / g) and the angular speed ω = √(g / (l cosθ)). These allow you to relate the angle of the cone to the speed of rotation.

联立上述方程可得几个有用结论：周期 T_p = 2π √(l cosθ / g) 以及角速度 ω = √(g / (l cosθ))。利用这些关系可以将圆锥的角度与旋转速率联系起来。

6. Horizontal Circular Motion: Car Rounding a Bend | 水平圆周运动：汽车转弯

When a car turns on a flat road, the centripetal force is supplied by friction between the tyres and the road. The maximum friction force is μR, where R is the normal reaction (equal to mg on a level road). Thus, the maximum safe speed without slipping satisfies μmg = mv² / r, giving v_max = √(μgr).

当汽车在水平路面上转弯时，向心力由轮胎与路面之间的摩擦力提供。最大摩擦力为 μR，其中 R 是法向反作用力（在水平路面上等于 mg）。因此，不侧滑的最大安全速度满足 μmg = mv² / r，得到 v_max = √(μgr)。

For a banked road, the normal reaction itself has a horizontal component that can help to provide the centripetal force. On a perfectly banked bend designed for a speed v, there is no reliance on friction, and tanθ = v² / (rg).

对于倾斜弯道，法向反作用力本身具有水平分量，可以帮助提供向心力。在为一个速度 v 设计的完全倾斜弯道上，无需依赖摩擦力，此时 tanθ = v² / (rg)。

v = √(rg tanθ)

In exam scenarios, you may be asked to calculate the ideal banking angle, the friction needed if the car travels faster or slower than the design speed, or the minimum coefficient of friction to prevent skidding.

在考试情景中，可能会要求计算理想倾斜角、当汽车行驶速度快于或慢于设计速度时所需的摩擦力，或者防止侧滑的最小摩擦系数。

7. Vertical Circular Motion: Speed and Forces | 竖直圆周运动：速度与力

Objects moving in a vertical circle, such as a mass on a string or a bead on a wire, experience changing speed because gravity does work. The centripetal acceleration at any point is still given by v² / r, where v is the instantaneous speed, but the tension or normal reaction varies significantly.

在竖直圆环内运动的物体（例如绳上的质量块或穿在铁丝上的珠子）由于重力做功，速率会变化。任意点的向心加速度仍由 v² / r 给出，其中 v 为瞬时速率，但张力或法向反作用力变化很大。

At the lowest point, the net inward force is T – mg, so the force equation is T – mg = mv² / r. At the highest point, for a string (or a light rod where the mass is ‘above’ the pivot), both weight and tension act downwards, giving T + mg = mv² / r.

在最低点，指向中心的净力为 T – mg，因此力方程为 T – mg = mv² / r。在最高点，对于细绳（或质量块位于支点之上的轻杆），重力和张力均向下，故 T + mg = mv² / r。

For an object attached to a string or rod, the string can only pull, so it must remain taut. The minimum speed at the top to keep the string taut is when T = 0, giving v_min = √(gr). For a stiff rod, the mass can reach the top even with zero speed because the rod can push, but tension in the rod will then be a thrust rather than a pull.

对于系在绳或杆上的物体，绳只能拉，因此必须保持绷紧。保持绳绷紧的最高点最小速率发生在 T = 0 时，解得 v_min = √(gr)。对于刚性杆，因为杆可以推，即使速率为零也能到达最高点，但此时杆中的力将是推力而非拉力。

8. Vertical Circular Motion: Energy Considerations | 竖直圆周运动：能量分析

In problems where friction is negligible, mechanical energy is conserved. Relating the speed at the lowest point to the speed at any other height often involves setting up an energy equation. For a mass moving from the bottom (speed u) to a point at height h above the bottom, ½ mu² = ½ mv² + mgh.

在忽略摩擦的问题中，机械能守恒。建立最低点速率与另一高度点速率的关系通常需要写出能量方程。对于从底部（速率为 u）运动到底部上方高度为 h 的一点的质量块：½ mu² = ½ mv² + mgh。

For the standard vertical circle of radius r, the height difference between the top and bottom is 2r. If the speed at the bottom is u, then at the top the speed v satisfies ½ mu² = ½ mv² + 2mgr. This is frequently used together with the force equation to find the tension or speed at crucial points.

对于半径为 r 的标准竖直圆，顶部与底部的高度差为 2r。若底部速率为 u，则顶部速率 v 满足 ½ mu² = ½ mv² + 2mgr。这个关系常常与力方程联立，用于求关键点的张力或速率。

Exam questions may also ask for the energy lost when friction is present, or for the minimum initial speed required to just complete a full vertical circle under a string. The rigorous condition is that the mass must have at least √(gr) at the top, which leads to a minimum speed at the bottom of √(5gr) via energy conservation.

考题也可能涉及存在摩擦时损失的能量，或刚好能完成一整圈竖直圆周运动所需的最小初始速率。严格条件是质量块在顶部必须至少具有 √(gr) 的速率，利用能量守恒可导出底部最小速率为 √(5gr)。

9. Common Misconceptions and Exam Tips | 常见误区与应试技巧

Always switch your calculator to radian mode when working with angular quantities. Using degrees for ω or θ in formulas like s = rθ or v = rω will give completely wrong answers.

在处理角度相关的物理量时，务必将计算器切换为弧度模式。在 s = rθ 或 v = rω 等公式中若使用度而非弧度，将得出完全错误的答案。

Remember that centripetal force is not a separate ‘extra’ force; it is just the name given to the resultant of real forces pointed towards the centre. Never add a ‘centripetal force’ to your free-body diagram alongside tension or gravity.

牢记向心力并非一种独立的“额外”力，它只是指向圆心的真实力合力的称谓。千万不要在受力图中将“向心力”与重力、张力等真实力并列。

Be meticulous with the direction of the positive radial axis. In horizontal circles, point it towards the centre; in vertical circles, the direction is still towards the centre, meaning that at the top the positive direction is downward. This will determine the signs of mg and T in your equations.

处理正径向坐标轴的方向时要一丝不苟。在水平圆周中，将正方向指向圆心；在竖直圆周中，正方向仍指向圆心，这意味着在最高点正方向是向下的。这将决定方程中 mg 和 T 的正负号。

For vertical circles, do not assume the speed is constant. Use energy conservation to find speeds at different positions, and then apply the radial force equation. A common error is to use v = rω with a constant ω throughout a vertical loop.

对于竖直圆周，切勿假设速率恒定。利用能量守恒求出不同位置的速率，然后再应用径向力方程。一个常见错误是在整个竖直圆环中使用常量 ω 套用 v = rω。

Finally, always check the physical constraints of the problem: can the string go slack? Is friction limiting? Does the road banking angle assume no friction? Identifying these subtleties is what distinguishes high-level answers.

最后，务必检查问题的物理约束：绳子是否会松弛？摩擦力是否达到极限？路面倾斜角是否假设无摩擦？辨别这些细微之处正是高分答案的体现。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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A-Level OCR Maths: Circular Motion Revision Essentials | A-Level OCR 数学：圆周运动 考点精讲