📚 A-Level Maths: Circular Motion Revision | A-Level 数学：圆周运动 考点精讲
In A-Level Mathematics (mechanics), circular motion appears as a core topic where particles move along circular paths at constant speed or with acceleration. You must understand the relationships between angular and linear quantities, the concept of centripetal force, and how to resolve forces when objects travel in horizontal or vertical circles. This article revisits every crucial formula, common modelling scenarios, and typical pitfalls to help you master the topic.
在 A-Level 数学（力学）中，圆周运动是一个核心主题，涉及粒子沿圆形路径匀速或变速运动。你需要掌握角量与线量的关系、向心力的概念，以及在水平面或竖直面内运动时如何进行受力分析。本文带你回顾所有关键公式、常见建模情景和典型误区，帮助你彻底掌握这个考点。
1. Angular Velocity and Period | 角速度与周期
Angular velocity ω (rad s⁻¹) measures the rate of change of angle. For a particle moving around a circle of radius r, if the period (time for one complete revolution) is T, then ω = 2π / T. Frequency f (in Hz) is defined as f = 1/T, so ω = 2π f. These relationships are fundamental and often appear in multi-step questions.
角速度 ω（单位 rad s⁻¹）衡量角度的变化率。对于绕半径为 r 的圆周运动的质点，若周期（完成一圈的时间）为 T，则 ω = 2π / T。频率 f（单位 Hz）定义为 f = 1/T，因此 ω = 2π f。这些关系是基础，经常出现在多步计算题中。
Make sure you work in radians. If a question gives angular speed in revolutions per minute (rpm), convert to rad s⁻¹ by multiplying by 2π/60. A common mistake is to use degrees instead of radians.
务必使用弧度制。如果题目给出转速（例如每分钟转数 rpm），需要乘以 2π/60 转换成 rad s⁻¹。常见错误是误用角度制而不是弧度制进行计算。
2. Linear Velocity | 线速度
The linear speed v of a particle moving along a circular path is tangent to the circle and given by v = r ω. This formula links angular and linear worlds. If the particle moves with constant speed, the magnitude of v is constant, but its direction changes continuously.
质点沿圆周运动的线速度 v 沿切线方向，且满足 v = r ω。这个公式将角量世界和线量世界连接起来。如果质点匀速运动，v 的大小不变，但其方向时刻改变。
In some problems, you might be given v and r, then need to find ω or T. Simply rearrange: ω = v / r, T = 2πr / v. For objects like a wheel or a pulley, the point on the rim moves with linear speed v while the whole object rotates with angular speed ω.
在某些问题中，你可能已知 v 和 r，需要求 ω 或 T。只需重新排列公式：ω = v / r，T = 2πr / v。对于轮子或滑轮等物体，边缘上一点以线速度 v 运动，而整个物体以角速度 ω 旋转。
v = r ω
3. Centripetal Acceleration | 向心加速度
Even if speed is constant, the direction change implies acceleration towards the centre of the circle. The centripetal acceleration a has magnitude a = v² / r = r ω². Derivation can appear in exams, but more often you apply these expressions directly in Newton’s second law.
即使速度大小不变，方向的改变意味着存在指向圆心的加速度。向心加速度 a 的大小为 a = v² / r = r ω²。推导可能出现在考试中，但更常见的是直接将这些表达式用于牛顿第二定律。
The vector always points to the centre. Do not confuse with a tangential acceleration, which only exists if the speed changes. In uniform circular motion, tangential acceleration is zero.
该矢量始终指向圆心。不要与切向加速度混淆，切向加速度仅在速度大小变化时才存在。在匀速圆周运动中，切向加速度为零。
a = v² / r = r ω²
4. Centripetal Force | 向心力
By Newton’s second law, the net force causing centripetal acceleration is F = m a = m v² / r = m r ω². This force is not a new type of force; it is provided by real forces such as tension, friction, normal reaction, or gravity. Always start by sketching a free-body diagram and then equate the resultant force towards the centre to m v²/r.
根据牛顿第二定律，产生向心加速度的合力为 F = m a = m v² / r = m r ω²。这个力不是某种新型力，而是由真实力提供，如拉力、摩擦力、法向反作用力或重力。始终从绘制受力图开始，然后将指向圆心的合力等于 m v²/r。
Many students mistakenly label a separate ‘centripetal force’ on diagrams. Do not add it as an extra arrow. Identify the actual force(s) directed towards the centre, and set their sum equal to m v²/r.
很多学生错误地在受力图中单独标出“向心力”。不要把它画成一个额外的箭头。找出指向圆心的实际力（一个或多个），并将它们的合力设为 m v²/r。
5. Applying Newton’s Second Law | 牛顿第二定律的应用
For uniform horizontal circular motion, resolve forces radially. The sum of components towards the centre equals m v²/r, and perpendicular to the plane, forces balance. Example: a particle tied to a string and whirled in a horizontal circle – the string tension provides the centripetal force, and weight is balanced by no vertical acceleration if the circle is perfectly horizontal.
对于水平面内的匀速圆周运动，沿径向分解力。指向圆心的合力等于 m v²/r，垂直于运动平面的方向上受力平衡。例如：用绳子系住一个小球使其在水平面内做圆周运动——绳子的拉力提供向心力，竖直方向重力若与某个力平衡，则无竖直加速度。
If the string is not horizontal (e.g., conical pendulum), you must consider both vertical and horizontal equilibrium. Write two equations: T cos θ = mg (vertical) and T sin θ = m v² / r (radial). Combining them can eliminate T and give relationships involving speed and angle.
如果绳子不水平（如圆锥摆），必须同时考虑竖直和水平方向的平衡。写出两个方程：T cos θ = mg（竖直方向）和 T sin θ = m v² / r（径向）。联立可以消去 T，得到速度与角度之间的关系。
6. Horizontal Circular Motion: Conical Pendulum | 水平圆周运动：圆锥摆
A conical pendulum consists of a mass rotating in a horizontal circle at the end of a light string. The string traces a cone. The radius r of the circle is L sin θ, where L is the string length. From the radial equation T sin θ = m r ω² and vertical equation T cos θ = mg, you can derive ω² = g / (L cos θ) or tan θ = v² / (r g).
圆锥摆由一个系在轻绳末端的质点绕水平圆周旋转构成。绳子扫出一个圆锥。圆周半径 r = L sin θ，其中 L 为绳长。由径向方程 T sin θ = m r ω² 和竖直方程 T cos θ = mg，可推导出 ω² = g / (L cos θ) 或 tan θ = v² / (r g)。
Questions may ask for the period T in terms of L, θ, or g. Substituting ω = 2π/T gives T = 2π √(L cos θ / g). This result is independent of mass, as expected. Many learners fail to connect that as the angular speed increases, θ increases, and the mass rises.
题目可能要求用 L、θ 或 g 表示周期 T。代入 ω = 2π/T 可得 T = 2π √(L cos θ / g)。该结果与质量无关，符合预期。许多学生未能理解：当角速度增大时，θ 增大，质点上升。
7. Horizontal Circular Motion: Vehicle Cornering | 车辆转弯
When a car travels around a curved horizontal road, friction between tyres and road provides the centripetal force. For a flat bend, the limiting speed without skidding is when friction reaches maximum f = μ N = μ mg. Thus μ mg = m v² / r, giving v_max = √(μ g r).
汽车在水平弯道上行驶时，轮胎与路面之间的摩擦力提供向心力。对于平直弯道，不侧滑的极限速度出现在摩擦力达到最大值 f = μ N = μ mg 时。因此 μ mg = m v² / r，得出 v_max = √(μ g r)。
On a banked track, the normal reaction has a horizontal component contributing to centripetal force, reducing reliance on friction. You resolve normal force N into components: N sin θ = m v² / r horizontally, N cos θ = mg vertically. The design speed for no sideways friction is v = √(r g tan θ). Exam questions may involve finding the angle of banking required for a given speed or the friction needed beyond that speed.
在倾斜彎道上，法向反作用力有水平分量提供向心力，减少对摩擦力的依赖。分解支持力 N：水平方向 N sin θ = m v² / r，竖直方向 N cos θ = mg。无侧向摩擦的理想设计速度为 v = √(r g tan θ)。考试题可能要求计算给定速度所需的倾斜角，或超出该速度时需提供的摩擦力。
8. Vertical Circular Motion: Top and Bottom | 竖直圆周运动：最高点与最低点
Vertical circles are more complex because gravity plays a role. At the top of a loop or circle, the net force towards the centre is the sum of the normal reaction (or tension) and weight, unless the object is moving so fast that the normal reaction direction changes. For a particle on a string moving in a vertical circle, at the highest point, mg + T = m v² / r. At the lowest point, T − mg = m v² / r.
竖直面内的圆周运动更为复杂，因为重力参与了作用。在轨道的最高点，指向圆心的合力是法向反作用力（或拉力）与重力之和，除非物体速度太大导致方向改变。对于系在绳子上在竖直面内做圆周运动的质点，在最高点有 mg + T = m v² / r。在最低点有 T − mg = m v² / r。
For an object to just complete a full loop (critical condition), the tension at the top can be zero. Then mg = m v² / r, giving minimum speed v_min = √(g r). In a roller coaster loop, the normal force from the track must be at least zero at the top for passengers to not fall out.
要使物体刚好完成整个圆环（临界条件），最高点的拉力可以为零。此时 mg = m v² / r，得到最小速度 v_min = √(g r)。在过山车圆环中，轨道对乘客的支持力在最高点至少为零，乘客才不会掉下来。
9. Energy Considerations in Vertical Circles | 竖直圆周运动中的能量分析
If speed is not constant, energy conservation may be required. For a particle started from rest or from a known height, use conservation of mechanical energy to find speed at any angular position. For a pendulum or a bead on a wire, ΔGPE + ΔKE = 0 (assuming no external work). Then substitute v into radial force equations.
如果速度不恒定，可能需要能量守恒。对于从静止或已知高度释放的质点，利用机械能守恒求任意角度位置的速度。对于摆球或穿在金属丝上的珠子，Δ重力势能 + Δ动能 = 0（假设无外力做功）。然后将 v 代入径向力方程。
A typical problem: A particle on a light rod is given just enough speed at the bottom to reach the top. At the bottom, KE = ½ m v², at the top height 2r, GPE increase = 2mgr. So v_bottom_min² = 4gr + v_top_min². Since v_top_min = √(gr), then v_bottom_min = √(5gr). This is a classic result for a rod (not a string) as a rod can support the particle without going slack.
一个典型问题：用轻杆固定在底部的质点获得刚好足够速度到达顶部。在底部，动能 = ½ m v²，在顶部高度 2r，重力势能增加 = 2mgr。所以 v_底部_min² = 4gr + v_顶部_min²。由于 v_顶部_min = √(gr)，则 v_底部_min = √(5gr)。这是轻杆（而非绳子）情况下的经典结果，因为杆可以支撑质点而不会松弛。
10. Common Pitfalls | 常见误区
Do not confuse centripetal force with centrifugal force. Centrifugal force is a fictitious force in a rotating reference frame and is not part of the A-Level syllabus. Always analyse from an inertial frame using real forces towards the centre.
不要混淆向心力与离心力。离心力是旋转参考系中的虚拟力，不属于 A-Level 课程内容。始终从惯性系出发，使用指向圆心的真实力进行分析。
Avoid using the same symbol for different radii when a particle moves on a horizontal circle attached to a string at an angle; the radius of the circle is not the string length. Remember to resolve components correctly: T sin θ goes radially, T cos θ vertically.
当质点系在倾斜绳子上做水平圆周运动时，不要把绳长和半径用同一个符号。记住正确分解分量：T sin θ 沿径向，T cos θ 沿竖直方向。
When speed changes in a vertical circle, the centripetal acceleration is still v²/r at that instant, but you must use the instantaneous speed. Combine energy methods to find v and then apply Newton’s law at critical points.
在竖直圆周运动中速度变化时，向心加速度在该瞬间仍然是 v²/r，但你必须使用该瞬间的速度。结合能量方法求出 v，然后在关键点应用牛顿定律。
Check units consistently: angular speed in rad s⁻¹, radius in metres, velocity in m s⁻¹, force in newtons. Be especially careful when converting from revolutions per minute or kilometres per hour.
确保单位统一：角速度用 rad s⁻¹，半径用米，速度用 m s⁻¹，力用牛顿。从每分钟转数（rpm）或千米每小时（km/h）转换时要格外小心。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导