📚 AS Mathematics: Circular Motion Key Points | AS 数学：圆周运动 考点精讲
Circular motion is a cornerstone topic in AS Mathematics Mechanics that appears frequently in exams. Understanding the relationships between angular and linear quantities, centripetal force, and the critical conditions in vertical circles can make the difference between an average grade and a top score. Mastery of these concepts also lays the groundwork for more advanced mechanics.
圆周运动是 AS 数学力学部分的重要考点，经常出现在各类考试中。理解角量与线量之间的关系、向心力的本质以及竖直圆周中的临界条件，是取得高分的关键。扎实掌握这些概念也为后续学习更高阶的力学内容打下坚实基础。
1. Uniform Circular Motion | 匀速圆周运动
An object travelling in a circle at constant speed performs uniform circular motion. Even though the speed is unchanging, the direction of the velocity vector is continuously altering, so the object is accelerating. This acceleration always points towards the centre of the circle.
一个物体以恒定速率做圆周运动，称为匀速圆周运动。尽管速率不变，但速度向量的方向时刻改变，因此物体具有加速度，且该加速度始终指向圆心。
The time taken to complete one full revolution is called the period, T. The motion is periodic and the object returns to its starting position after each period, covering an angular displacement of 2π radians.
完成一整圈所需的时间称为周期 T。运动具有周期性，物体每经过一个周期，角位移为 2π 弧度，回到起始位置。
2. Angular Velocity and Linear Speed | 角速度与线速度
Angular velocity ω (omega) measures how fast the angle changes: ω = Δθ/Δt. It is expressed in radians per second (rad s⁻¹). For one revolution, Δθ = 2π, so ω = 2π/T.
角速度 ω 描述角度变化的快慢：ω = Δθ/Δt，单位为弧度每秒（rad s⁻¹）。一整圈的 Δθ = 2π，因此 ω = 2π/T。
Linear speed v along the circumference is linked to angular velocity by v = rω, where r is the radius. The direction of the linear velocity is tangent to the circle at any instant.
圆周上的线速度 v 与角速度的关系为 v = rω，其中 r 为半径。瞬时线速度的方向始终沿圆的切线方向。
Period T and frequency f are reciprocals: T = 2π/ω and f = 1/T. Doubling the angular speed halves the period while doubling the radius (for fixed ω) doubles the linear speed.
周期 T 与频率 f 互为倒数：T = 2π/ω，f = 1/T。若角速度加倍，周期减半；若角速度不变而半径加倍，线速度变为原来的两倍。
3. Centripetal Acceleration | 向心加速度
Any object in uniform circular motion experiences an acceleration towards the centre, called centripetal acceleration. Its magnitude is given by a = v²/r or, using v = rω, a = rω². Both forms are essential for solving problems.
做匀速圆周运动的物体会受到指向圆心的加速度，称为向心加速度。其大小可由 a = v²/r 或 a = rω² 表示，两种形式在解题中都十分常用。
The derivation of a = v²/r comes from considering the change in the velocity vector over a tiny time interval. The triangular construction shows that the acceleration is perpendicular to velocity and has magnitude v²/r.
a = v²/r 的推导来自对极短时间间隔内速度向量变化的分析。通过几何构造可知加速度垂直于速度，其大小为 v²/r。
Though the speed is constant, the velocity changes because its direction changes. Hence, there is acceleration even with no change in the speedometer reading.
虽然速率不变，但由于速度方向持续改变，因此存在加速度，这与日常经验中“速度数值不变就没有加速”的直觉不同。
4. Centripetal Force | 向心力
Newton’s second law tells us that a net force must act towards the centre to produce centripetal acceleration: F = ma, so F = mv²/r = mrω². This force is called centripetal force, but it is not a new type of force—it is the result of real forces like tension, gravity, friction or the normal reaction acting towards the centre.
牛顿第二定律指出，必须有指向圆心的合力来产生向心加速度：F = ma，因此 F = mv²/r = mrω²。这个力名为向心力，但它并非一种新型的力，而是由真实的力（如张力、重力、摩擦力或支持力）指向圆心的分量共同提供。
In horizontal circular motion, the centripetal force is often supplied by the horizontal component of tension in a string or by static friction between tyres and the road. If the required centripetal force exceeds the maximum available force, skidding or breakage occurs.
在水平圆周运动中，向心力通常由绳子张力的水平分量或轮胎与路面间的静摩擦力提供。若所需向心力大于实际能提供的最大力，就会发生打滑或绳子断裂。
Always identify the forces acting towards the centre and write Fnet = mv²/r. Never invent a separate ‘centripetal force’ arrow on the diagram; it is the net inward force.
解题时，应标记所有指向圆心的实际作用力，并建立方程 Fnet = mv²/r。不要在受力分析图上凭空添加一个“向心力”箭头，它是内指的合力。
5. Horizontal Circular Motion Examples | 水平圆周运动实例
A common exam model is the conical pendulum: a mass attached to a string moves in a horizontal circle with the string tracing a cone. The vertical component of tension balances weight, T cosθ = mg, and the horizontal component provides centripetal force, T sinθ = mrω². Dividing gives tanθ = rω²/g = v²/(rg).
锥摆是常考模型：一个系在绳子上的物体在水平面内做圆周运动，绳子扫出一个圆锥面。张力的竖直分量平衡重力：T cosθ = mg，水平分量提供向心力：T sinθ = mrω²。两式相除可得 tanθ = rω²/g = v²/(rg)。
For a vehicle on a banked curve without friction, the horizontal component of the normal reaction supplies the centripetal force. This leads to the ideal banking angle: tanθ = v²/(rg). With friction, the maximum safe speed can be increased.
对于无摩擦的倾斜弯道，支持力的水平分量提供向心力，由此得出理想倾斜角公式 tanθ = v²/(rg)。当有摩擦时，可提高最大安全车速。
On a flat curve, the centripetal force is entirely static friction: f = μsN = μsmg = mv²/r, giving a maximum speed vmax = √(μsgr). Exceeding this speed leads to skidding.
在平坦弯道上，向心力完全由静摩擦力提供：f = μsN = μsmg = mv²/r，得出最大安全车速 vmax = √(μsgr)。超过此速度便会打滑。
6. Vertical Circular Motion | 竖直圆周运动
When an object moves in a vertical circle, its speed changes because gravity does work. The tension in the string or the normal reaction varies with position. At the top, both gravity and tension act downwards, contributing to the centripetal force; at the bottom, tension acts upwards while gravity acts downwards.
当物体在竖直平面内做圆周运动时，由于重力做功，速率会发生变化。绳中张力或支持力随位置改变：在最高点，重力和张力同向下，共同提供向心力；在最低点，张力向上而重力向下。
At the highest point, we have T + mg = mv²/r. For the string to remain taut, T ≥ 0, which imposes a critical minimum speed: vmin = √(gr). If the speed drops below this, the string goes slack and the object can no longer follow the circle.
在最高点，有 T + mg = mv²/r。为使绳子保持绷直，需满足 T ≥ 0，由此得到临界最小速率：vmin = √(gr)。若速率低于此值，绳子将松弛，物体无法继续维持圆周路径。
At the bottom of the circle, T – mg = mv²/r, so tension is greatest here. The difference between top and bottom tension gives valuable insights: Tbottom – Ttop = 6mg for an object moving with just enough speed at the top.
在最低点，T – mg = mv²/r，此处张力最大。对于以最小速率刚好能过顶点的运动，最低点与最高点张力差为 Tbottom – Ttop = 6mg，这一关系可快速检验计算。
7. String vs. Rod Models | 绳模型与杆模型
It is crucial to distinguish between a light string and a light rod. A string can only pull (tension), so there is a minimum speed at the top to keep it taut. A rod, however, can either push or pull; it can exert a force away from the centre (upwards at the top), allowing the object to pass over the top with a speed that can even be zero in theory.
必须清楚地区分轻绳和轻杆模型。轻绳只能提供拉力，因此在最高点需要保持最小速率以确保绳子绷直。而轻杆既可以拉也可以推，在最高点可以提供背离圆心的力（向上支持），理论上物体在顶点速率可以为零。
Therefore, the critical condition at the top for a particle on a rod or a bead on a wire is simply that the normal reaction adjusts to provide the required centripetal force, so the top speed can be zero. For a string, the minimum speed is √(gr).
因此，对于杆或穿过圆环的珠子，最高点的临界条件是支持力可以自动调整以满足向心力需求，最高点速度可以为零；而对于绳子，最小速度必须为 √(gr)。
Exam questions often ask: “Find the minimum speed at the top so that the particle completes a full circle.” If it is a string, use v = √(gr); if it is a rod, the answer is v = 0 (provided the rod can push). Never confuse the two.
考试常问：“求刚好能完成整个圆周运动时最高点的最小速度。”若是绳子模型，答案为 v = √(gr)；若是杆模型，理论上 v = 0（前提是杆可提供推力）。务必分清模型。
8. Working with Angular and Linear Quantities | 角量与线量的转换
Many problems involve converting between linear and angular variables. Always check whether the given data uses radius, angular speed or period, and select the appropriate form of centripetal acceleration: a = v²/r or a = rω². For instance, if you know the revolutions per minute (rpm), first convert to rad s⁻¹.
很多题目需要在角量和线量之间转换。注意观察已知条件给出的是半径、角速度还是周期，然后选用合适的向心加速度表达式：a = v²/r 或 a = rω²。例如，若已知每分钟转数 (rpm)，应首先转换为 rad s⁻¹。
Always set up Newton’s second law towards the centre: ΣFtowards centre = mv²/r or mrω². Be careful to use the velocity at that specific point for non-uniform circular motion.
始终沿径向列出牛顿第二定律方程：ΣF指向圆心 = mv²/r 或 mrω²。对于非匀速圆周运动，务必使用该特定位置的瞬时线速度。
Do not forget that the speed may change in a vertical circle due to conservation of energy. You can link speeds at different heights using ½mv₁² + mgh₁ = ½mv₂² + mgh₂, taking a reference level for height.
不要忘记，竖直圆周运动中的速率会因能量守恒而改变，可以利用 ½mv₁² + mgh₁ = ½mv₂² + mgh₂ 联系不同高度处的速率，选取合适的零势能面。
9. Step‑by‑Step Problem Solving | 解题步骤拆解
Begin by drawing a clear free-body diagram at the position of interest, showing all real forces. Resolve forces along the radial direction (towards the centre is positive) and set up the net force equation. For vertical circles, consider using energy conservation to find speed if not given.
解题时，先画出所选位置清晰的受力分析图，展示所有真实力。将力沿径向分解（以指向圆心为正），建立合力方程。对于竖直圆周，若速度未直接给出，可考虑利用能量守恒来求解。
Check for special conditions: at the top of a vertical circle, T + mg = mv²/r; at the bottom, T – mg = mv²/r. Identify whether the string/rod can go slack or how a bead on a wire behaves.
检查特殊条件：在竖直圆周最高点，有 T + mg = mv²/r；在最低点，T – mg = mv²/r。判断绳子或杆是否会松弛，或考虑珠子在圆环上的行为。
Always substitute numerical values only after deriving the symbolic formula. This minimises rounding errors and makes it easier to check units. Use g = 9.8 m s⁻² or 10 m s⁻² as instructed by the question.
坚持先推导符号表达式再代入数值，以减少舍入误差并便于检查单位。根据题目要求，重力加速度 g 通常取 9.8 m s⁻² 或 10 m s⁻²。
10. Common Mistakes and How to Avoid Them | 常见错误与避坑指南
Mistake 1: Forgetting that centripetal force is a resultant, not a separate real force. Always sum the real forces pointing towards the centre.
错误一：把向心力当作独立的实在力。务必记住它是所有指向圆心的实际力的合力。
Mistake 2: Using a constant speed assumption in vertical circles when it clearly changes. Apply energy principles to find speeds at different points.
错误二：错误地假设竖直圆周运动速率不变。应正确使用能量原理找出不同位置的速度。
Mistake 3: Confusing the radius of the circle with the length of the string when the string is at an angle, as in a conical pendulum. The relevant radius is r = L sinθ, where L is the length of the string.
错误三：在锥摆等模型中混淆圆半径与绳长。圆半径应为 r = L sinθ，其中 L 为绳长，不是绳长本身。
Mistake 4: Applying the minimum speed formula √(gr) to a rod model. The rod can support the mass, so the condition is different.
错误四：将绳模型的最小速率公式 √(gr) 套用到杆模型上。杆能提供支持力，临界条件完全不同。
11. Practice Example with Solution | 典型例题与解答
Example: A particle of mass 0.4 kg is attached to a light string of length 0.5 m and whirled in a vertical circle. Find the minimum speed at the top needed to keep the string taut, and the tension at the bottom if the speed there is 6 m s⁻¹.
例题：质量 0.4 kg 的物体系在长 0.5 m 的轻绳一端，在竖直平面内做圆周运动。求保持绳子绷直时最高点的最小速率，以及当最低点速率为 6 m s⁻¹ 时该处的张力。
Solution: At the top, T + mg = mv²/r, with T ≥ 0 gives vmin = √(gr) = √(9.8 × 0.5) ≈ 2.21 m s⁻¹. At the bottom, T – mg = mv²/r ⇒ T = m(g + v²/r) = 0.4(9.8 + 36/0.5) = 0.4(9.8 + 72) = 32.72 N.
解答：最高点 T + mg = mv²/r，令 T ≥ 0 得 vmin = √(gr) = √(9.8 × 0.5) ≈ 2.21 m s⁻¹。最低点 T – mg = mv²/r ⇒ T = m(g + v²/r) = 0.4(9.8 + 36/0.5) = 0.4(9.8 + 72) = 32.72 N。
This simple example illustrates the two key equations. Always draw a diagram, mark the forces, and write the net radial force equal to mv²/r.
这个简单例子展示了两个关键方程的应用。解题时一定要绘制受力图，标出所有力，并列出径向合力等于 mv²/r。
12. Summary and Quick Reference | 总结速查表
Memorise these essential formulas:
Angular speed: ω = 2π/T
Linear speed: v = rω
Centripetal acceleration: a = v²/r = rω²
Centripetal force: F = mv²/r = mrω²
Critical speed for string at top: v = √(gr)
Conical pendulum: tanθ = v²/(rg)
牢记以下核心公式：
角速度：ω = 2π/T
线速度：v = rω
向心加速度：a = v²/r = rω²
向心力：F = mv²/r = mrω²
绳模型最高点临界速度：v = √(gr)
锥摆关系：tanθ = v²/(rg)
Approach every problem systematically: draw forces, resolve towards centre, apply F = mv²/r or energy conservation as needed, and check for special conditions like slack strings or rod support.
解题时系统性地执行：画受力图、向圆心分解、根据需要应用 F = mv²/r 或能量守恒，并检查如绳子松弛或杆支撑等特殊条件。
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