📚 Deriving Centripetal Acceleration: a = v²/r – Examiner’s Insight | 向心加速度 a = v²/r 推导及考官报告剖析
The June 2018 A-Level Physics examiner’s report (Paper 4) highlighted that many candidates struggled when asked to derive the formula for centripetal acceleration, a = v²/r. Although students could often state the final equation, they lost critical marks because their derivations lacked clear vector diagrams, omitted the small-angle approximation, or failed to connect average acceleration to instantaneous acceleration. This article reconstructs that examiner feedback into a step-by-step derivation, showing exactly what examiners expect to see. Mastering this derivation not only secures marks in structured questions but also deepens your understanding of circular motion, a topic that bridges kinematics, dynamics, and fields.
2018年6月A-Level物理考官报告（试卷4）指出，许多考生在被要求对向心加速度公式 a = v²/r 进行推导时遇到困难。虽然学生往往能够直接写出最终公式，但由于缺少清晰的矢量图、忽略了小角度近似，或者未能将平均加速度与瞬时加速度联系起来，他们丢失了关键的分数。本文将该考官反馈重构为一个循序渐进的推导过程，精确展示考官所期望的内容。掌握这一推导不仅能在结构化问题中确保得分，还能加深你对圆周运动的理解，而圆周运动正是衔接运动学、动力学与场的桥梁。
1. Circular Motion Basics | 圆周运动基础
An object moving in a circle at constant speed still experiences acceleration because its velocity vector is constantly changing direction. Velocity is a vector, so even if the magnitude (speed) stays the same, a change in direction produces a non-zero acceleration. This acceleration points towards the centre of the circle and is called centripetal (centre‑seeking) acceleration.
即使速率恒定，做圆周运动的物体仍然具有加速度，因为其速度矢量的方向在不断改变。速度是矢量，因此即便大小（速率）不变，方向的改变也会产生非零的加速度。这个加速度指向圆心，称为向心加速度。
In a time interval Δt, the object moves from point A to point B, subtending an angle Δθ at the centre. The radius r is constant. The speed v is given by v = (arc length AB) / Δt = rΔθ / Δt, so Δθ = v Δt / r. This relation will be used shortly.
在一个时间间隔 Δt 内，物体从点 A 运动到点 B，在圆心处张角为 Δθ。半径 r 是恒定的。速率 v 由 v = (弧长 AB)/Δt = rΔθ/Δt 给出，因此 Δθ = v Δt / r。这个关系稍后会被用到。
2. Velocity Vectors and Direction Change | 速度矢量与方向变化
At point A the velocity vA is tangential to the circle, pointing in one direction. At point B the velocity vB is also tangential but has been rotated through the same angle Δθ. Because the speed is constant, |vA| = |vB| = v. To find the change in velocity, we place the two vectors tail‑to‑tail: Δv = vB − vA.
在点 A，速度 vA 沿圆的切线方向指向某一方向。在点 B，速度 vB 同样沿切线方向，但已经转过相同角度 Δθ。由于速率恒定，|vA| = |vB| = v。为了求速度的变化，我们将两个矢量尾尾相连：Δv = vB − vA。
The vector Δv is directed roughly towards the centre of the circle. As Δt becomes very small, Δv points exactly toward the centre. Constructing a vector triangle with sides of length v and v enclosing the angle Δθ shows that the magnitude of Δv is given by the chord of the sector, not the arc. Using the isosceles triangle, the chord length is 2v sin(Δθ/2).
矢量 Δv 大致指向圆心。当 Δt 变得非常小，Δv 正好指向圆心。构建一个两边长为 v、夹角为 Δθ 的矢量三角形，可以看出 Δv 的大小由扇形的弦长给出，而非弧长。利用等腰三角形，弦长为 2v sin(Δθ/2)。
3. Small-Angle Approximation | 小角度近似
For very small Δt, the angle Δθ becomes tiny. Examiners expect you to use the small‑angle approximation sin θ ≈ θ (where θ is in radians). Therefore, Δv ≈ 2v (Δθ/2) = v Δθ. This step is essential; omitting it or not stating the approximation explicitly often costs a mark.
对于非常小的 Δt，角度 Δθ 变得极小。考官期望你使用小角度近似 sin θ ≈ θ（其中 θ 以弧度为单位）。因此，Δv ≈ 2v (Δθ/2) = v Δθ。这一步至关重要；如果省略这一步，或者没有明确说明该近似，往往会丢分。
Substituting the earlier expression for Δθ gives Δv ≈ v (v Δt / r) = v² Δt / r. Thus the magnitude of the change in velocity is directly proportional to the time interval, with the factor v²/r multiplying Δt.
代入之前 Δθ 的表达式，得到 Δv ≈ v (v Δt / r) = v² Δt / r。因此，速度变化的大小与时间间隔成正比，比例因子为 v²/r 乘以 Δt。
4. Average Acceleration to Instantaneous | 从平均加速度到瞬时值
The average acceleration during the interval Δt is aavg = Δv / Δt. Using the result above, aavg ≈ (v² Δt / r) / Δt = v²/r. Notice that the time Δt cancels out, so the average acceleration magnitude is independent of the length of the interval, provided the small‑angle approximation holds.
在时间间隔 Δt 内的平均加速度为 aavg = Δv / Δt。利用上述结果，aavg ≈ (v² Δt / r) / Δt = v²/r。注意到时间 Δt 被消去，因此只要小角度近似成立，平均加速度的大小与时间间隔长度无关。
As Δt → 0, the average acceleration becomes the instantaneous acceleration, and the direction of Δv becomes exactly radial (toward the centre). Consequently, the instantaneous centripetal acceleration is a = v²/r, directed toward the centre of the circle. This is the key derivation.
当 Δt → 0，平均加速度变为瞬时加速度，而 Δv 的方向恰好沿径向指向圆心。因此，瞬时向心加速度为 a = v²/r，方向指向圆心。这就是关键的推导过程。
5. Alternative Form Using Angular Speed | 使用角速度的替代形式
Since v = ωr, where ω is the angular speed in rad s⁻¹, we can rewrite the centripetal acceleration as a = (ωr)²/r = ω²r. This form appears frequently in rotational dynamics problems, especially those involving orbits or rotating platforms. The examiner’s report noted that students who could flexibly switch between a = v²/r and a = ω²r scored higher in multi-step calculations.
由于 v = ωr，其中 ω 为角速度（单位 rad s⁻¹），我们可将向心加速度改写为 a = (ωr)²/r = ω²r。这种形式在旋转动力学问题中经常出现，尤其是在涉及轨道或旋转平台的题目中。考官报告指出，能够在 a = v²/r 和 a = ω²r 之间灵活转换的学生，在多步计算中得分更高。
Both expressions are equivalent, but choosing the one that avoids an intermediate calculation of linear speed often reduces arithmetic errors. For example, if a question gives the period T, use ω = 2π/T and then a = ω²r directly.
这两种表达式等价，但选择那个能够避免线速度中间计算的公式，往往可以减少算术错误。例如，如果题目给出周期 T，可利用 ω = 2π/T，然后直接使用 a = ω²r。
6. Force Aspect: Centripetal Force | 力的方面：向心力
From Newton’s second law, the net force causing this acceleration is F = ma = mv²/r, also directed toward the centre. This centripetal force is not a new type of force; it is the resultant of real forces such as tension, gravity, friction, or the normal reaction. Many candidates in the 2018 exam incorrectly labelled ‘centripetal force’ as an additional force on free-body diagrams, leading to lost marks.
根据牛顿第二定律，引起该加速度的净力为 F = ma = mv²/r，同样指向圆心。这个向心力并不是一种新型的力；它是真实力（如张力、重力、摩擦力或法向反作用力）的合力。2018年考试中，许多考生在受力图中错误地将“向心力”标注为一种额外的力，从而导致失分。
Always identify the physical source of the centripetal force. For a car rounding a bend, it is the friction between tyres and road; for a planet orbiting a star, it is gravity; for a stone whirled on a string, it is tension. Stating this clearly satisfies the examiners’ requirement for rigorous explanation.
务必指出提供向心力的物理来源。对于转弯的汽车，它是轮胎与路面之间的摩擦力；对于绕恒星运行的行星，它是引力；对于用绳子旋转的石块，它是绳子的张力。清晰地说明这一点，可以满足考官的严格解释要求。
7. Examiner’s Common Pitfalls | 考官指出的常见错误
The June 2018 examiner’s report (Paper 4) catalogued several recurring errors in derivation questions:
2018年6月考官报告（试卷4）列举了推导题中的若干常见错误：
- Missing vector diagram: Examiners expect a clear sketch showing vA, vB and Δv. Without it, the geometry of the derivation is impossible to follow. (缺少矢量图：考官期望清晰画出 vA、vB 和 Δv。没有矢量图，推导的几何关系就无从理解。)
- Confusing chord length with arc length: Some students wrote Δv = vΔθ immediately without explaining that it arises from the chord of the vector triangle, losing a mark for the geometric justification. (混淆弦长与弧长：有些学生直接写出 Δv = vΔθ，而没有解释它来源于矢量三角形的弦长，从而丢失了几何论证的分数。)
- Not stating the small-angle approximation: The step Δv = vΔθ relies on sin(Δθ/2) ≈ Δθ/2. Omitting this logical connection breaks the derivation. (没有说明小角度近似：Δv = vΔθ 这一步依赖于 sin(Δθ/2) ≈ Δθ/2。省略这个逻辑联系会打断整个推导。)
- Failing to link average to instantaneous acceleration: Simply writing a = Δv/Δt = v²/r is incomplete; you must mention that as Δt → 0, the direction becomes radial and the acceleration becomes instantaneous. (未能将平均加速度与瞬时加速度联系起来：仅仅写出 a = Δv/Δt = v²/r 是不完整的；必须提及当 Δt → 0 时，方向变为径向，加速度变为瞬时值。)
8. Worked Example: Derivation in an Exam Context | 实例演练：考试情境下的推导
Let us reconstruct a model answer that would gain full marks in a typical A-Level structured question: “By considering the change in velocity over a short time interval, derive a = v²/r for an object moving in a circle of radius r at constant speed v.”
我们来重构一份在典型 A-Level 结构化问题中能够获得满分的模范答案：“通过考虑短时间间隔内的速度变化，推导出沿半径为 r、速率恒为 v 的圆周运动的物体的加速度 a = v²/r。”
Step 1: Draw a large, neat circle with centre O. Mark points A and B close together, radii OA and OB, angle Δθ. Draw tangential velocity vectors vA and vB of equal length. Draw the vector triangle for vB – vA showing Δv. Step 2: Note that |vA| = |vB| = v and the angle between them equals the angle moved, Δθ. Step 3: From the isosceles triangle, |Δv| = 2v sin(Δθ/2). Step 4: For small Δθ (as Δt → 0), sin(Δθ/2) ≈ Δθ/2, so |Δv| ≈ vΔθ. Step 5: Arc length AB = rΔθ = vΔt, hence Δθ = vΔt/r. Step 6: Substitute to get |Δv| ≈ v²Δt/r. Step 7: Average acceleration magnitude = |Δv|/Δt = v²/r. As Δt→0, this becomes the instantaneous acceleration, directed toward O.
步骤1：画出一个大而清晰的以 O 为圆心的圆，标出靠得很近的点 A 和 B，半径 OA 与 OB，角度 Δθ。画出等长的切向速度矢量 vA 和 vB。画出 vB – vA 的矢量三角形，显示 Δv。步骤2：注意到 |vA| = |vB| = v，两者之间的夹角等于物体移动的角度 Δθ。步骤3：由等腰三角形可得 |Δv| = 2v sin(Δθ/2)。步骤4：对于很小的 Δθ（当 Δt → 0），sin(Δθ/2) ≈ Δθ/2，因此 |Δv| ≈ vΔθ。步骤5：弧长 AB = rΔθ = vΔt，因此 Δθ = vΔt/r。步骤6：代入得到 |Δv| ≈ v²Δt/r。步骤7：平均加速度的大小 = |Δv|/Δt = v²/r。当 Δt→0，此即瞬时加速度，方向指向 O。
This answer includes every mark-scoring step: diagram, small-angle approximation explicitly stated, substitution of Δθ, cancellation of Δt, and the limit as Δt→0 to justify instantaneous acceleration.
这份答案包含了每个得分步骤：图示、明确写出的小角度近似、Δθ 的代入、Δt 的约分，以及用 Δt→0 的极限来证明瞬时加速度。
9. Why Derivation Matters in A-Level Physics | 为何推导在 A-Level 物理中如此重要
The 2018 examiner’s report stressed that derivation questions assess AO2 (application of knowledge) and AO3 (analysis and evaluation). Simply memorising the final formula a = v²/r is not enough; you must demonstrate understanding of the underlying vector mechanics. This is especially true in synoptic questions that link circular motion with fields, such as deriving the radius of an electron’s orbit in a magnetic field.
2018 年考官报告强调，推导题考查的是 AO2（知识应用）和 AO3（分析与评价）。仅仅记住最终公式 a = v²/r 是不够的；你必须展现出对底层矢量力学的理解。这在将圆周运动与场联系起来的综合性题目中尤为如此，例如推导电子在磁场中的轨道半径。
Furthermore, derivations build problem-solving intuition. When you encounter a non-standard circular motion scenario — like a charged particle in a magnetic field — you can adapt the same logical steps to find the required acceleration or force without relying on rote learning.
此外，推导过程能培养解题直觉。当你遇到非标准的圆周运动场景（例如带电粒子在磁场中），你可以应用相同的逻辑步骤来求出所需的加速度或力，而无需依赖于死记硬背。
10. Practice and Self-Check | 练习与自检
To consolidate the derivation, close your notes and re-derive a = v²/r from scratch. Afterwards, ask yourself:
为了巩固推导过程，请合上笔记，从零开始重新推导 a = v²/r。完成后，请自问：
- Did I draw a clear vector diagram with labelled angles? (我是否画出了标注角度的清晰矢量图？)
- Did I write sin(Δθ/2) ≈ Δθ/2 and say why it is valid? (我是否写明了 sin(Δθ/2) ≈ Δθ/2 并说明其合理性？)
- Did I explicitly let Δt → 0 to obtain instantaneous acceleration? (我是否明确令 Δt → 0 以得到瞬时加速度？)
- Could I express the final answer in terms of ω as a = ω²r? (我能否将最终结果用 ω 表示为 a = ω²r？)
If you missed any of these, repeat the derivation until the steps become natural. Examiner reports repeatedly note that fluent derivations differentiate A-grade students from the rest.
如果你遗漏了任何一项，请重复推导直到这些步骤变得自然。考官报告一再强调，流畅的推导是将 A 等级学生与其他学生区分开来的关键。
11. Linking to Other Topics | 与其它知识点的联系
Once a = v²/r is established, it seamlessly connects to gravitational orbits: for a satellite, mg = mv²/r and thus v = √(gr). In electromagnetism, a charged particle moving perpendicularly to a uniform magnetic field experiences a centripetal force Bqv = mv²/r, giving r = mv/(Bq). These applications are high-frequency A-Level problems, and starting from a solid derivation prevents conceptual mistakes.
一旦确立了 a = v²/r，它就可以无缝连接到引力轨道：对于卫星，有 mg = mv²/r，因此 v = √(gr)。在电磁学中，一个垂直于匀强磁场运动的带电粒子受到向心力 Bqv = mv²/r，从而得到 r = mv/(Bq)。这些应用是 A-Level 高频题目，从扎实的推导入手可以避免概念性的错误。
The June 2018 report specifically praised candidates who linked the centripetal acceleration derivation to the concept of uniform circular motion in magnetic fields, showing deep synoptic understanding. Including such links in your exam answers demonstrates a top-level grasp of physics.
2018年6月的报告特别表扬了那些将向心加速度推导与磁场中匀速圆周运动概念联系起来的考生，这展示了深入的综合性理解。在考试答案中包含此类联系，可以显示出对物理学的顶级掌握。
12. Final Advice from the Examiner | 考官的最后建议
The examiner concluded that derivation questions are not about reproducing a textbook proof word for word but about communicating a logical, physically sound argument. Always start with a diagram, define your symbols, state assumptions (constant speed, small angle, etc.), and finish with a concluding sentence that ties the mathematical result back to the physical direction (toward the centre).
考官总结道，推导题的要点并不是一字不差地复现教材上的证明，而是传达一个逻辑上合理、物理上可靠的论证。务必从图示开始，定义所有符号，说明假设（速率恒定、小角度等），最后用一句结论将数学结果与物理方向（指向圆心）联系起来。
By internalising the derivation of a = v²/r through the lens of the 2018 examiner’s report, you are not only preparing for potential derivation questions but also reinforcing the vector reasoning that underpins all of mechanics. Practise it, understand it, and you will find circular motion problems become far more approachable.
通过从 2018 年考官报告的视角内化 a = v²/r 的推导，你不仅是在为可能出现的推导题做准备，更是在强化支撑整个力学的矢量推理能力。勤加练习、透彻理解，你会发现圆周运动问题变得容易得多。
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