📚 Deriving Centripetal Acceleration: A-Level Physics Jun 18 Paper 1 Formula Derivation | 向心加速度公式推导：2018年6月A-Level物理卷1公式解析
In the June 2018 A-Level Physics Paper 1, a structured question required candidates to derive the formula for centripetal acceleration for an object moving in a circle at constant speed. This task is a classic assessment of vector manipulation, geometric reasoning, and the application of limits. The question typically asks: ‘Using a diagram, show that the acceleration of a particle moving with constant speed v in a circular path of radius r is directed towards the centre and has magnitude a = v²/r.’ A confident step‑by‑step derivation not only fetches method marks but also deepens understanding of circular motion. In this revision article we unpack the full derivation, mirroring the logic expected in the exam.
在2018年6月的A-Level物理卷1中，一道结构化试题要求考生推导匀速圆周运动中物体的向心加速度公式。这一任务经典地考查了矢量操作、几何推理和极限思想的应用。题目通常表述为：“利用图示，证明一个以恒定速率v在半径为r的圆周上运动的粒子，其加速度指向圆心且大小为a = v²/r。”逐步清晰的推导不仅能获得方法分，还能加深对圆周运动的理解。在这篇复习文章中，我们将完整拆解推导过程，重现考试要求的逻辑。
1. Statement of the Problem | 问题陈述
The exam question from June 2018 Paper 1 directly targets a core skill: translating a physical situation into a vector‑geometry argument to obtain a = v²/r. Candidates are given constant speed v, circle radius r, and must show both the magnitude and the centripetal direction of the acceleration. The marking scheme rewards clear diagrams, labelled angles, similarity statements, small‑angle approximations, and a limiting argument to make the derivation rigorous. Understanding what the examiners look for helps you structure your answer perfectly.
2018年6月卷1的这道题直接针对一项核心技能：将物理情景转化为矢量‑几何论证以得出a = v²/r。题目给出恒定速率v、圆周半径r，要求同时证明加速度的大小和向心方向。评分方案奖励清晰的图示、标注的角度、相似性陈述、小角度近似以及使推导严谨的极限论证。了解考官的期待有助于你完美地组织答案。
The derivation links uniform circular motion to the definition of acceleration as the rate of change of velocity. Although the speed is constant, the velocity vector changes direction continuously, meaning a non‑zero acceleration must be present. The challenge is to calculate this acceleration using only basic geometry and the concepts of average and instantaneous acceleration.
这一推导将匀速圆周运动与加速度作为速度变化率的定义联系起来。尽管速率恒定，但速度矢量的方向不断变化，这意味着必定存在非零加速度。挑战在于仅用基本几何和平均与瞬时加速度的概念来计算这个加速度。
2. Uniform Circular Motion Basics | 匀速圆周运动基础
In uniform circular motion, the particle travels around a circle with a steady speed v. Because velocity is a vector, any change in direction constitutes a change in velocity. Therefore, the particle is accelerating even though its speed never changes. This acceleration is called centripetal acceleration. The derivation starts by considering the velocity vectors at two adjacent points on the circle. The velocity at any point is tangential to the circle and perpendicular to the radius at that point.
在匀速圆周运动中，粒子以恒定速率v沿圆周运动。由于速度是矢量，方向的任何变化都构成速度的变化。因此，尽管速率不变，粒子也在加速。这个加速度叫作向心加速度。推导从考察圆周上两个相邻点的速度矢量开始。任意一点的速度沿该点切线方向，并与该点半径垂直。
A common mistake is to assume acceleration is zero because speed is constant. Examiners specifically test the idea that a force and acceleration are required to change the direction of motion. Remember Newton’s first law: a body continues in a straight line unless acted upon by a resultant force. Circular motion is therefore a continuously forced motion, and the related acceleration must be directed towards the centre of the circle.
一个常见错误是认为速率不变则加速度为零。考官特地考察改变运动方向需要力和加速度这一概念。记住牛顿第一定律：物体将沿直线运动，除非受到合力作用。因此，圆周运动是持续受迫的运动，相关的加速度必须指向圆心。
3. Velocity Vectors at Two Instants | 两个时刻的速度向量
Let the particle be at point P at time t, moving with velocity v₁ of magnitude v. After a short time interval Δt, it arrives at point Q, with velocity v₂, also of magnitude v. Draw the circle of radius r with centre O. The angle subtended by the arc PQ at O is a small angle Δθ. Since v₁ is perpendicular to OP and v₂ is perpendicular to OQ, the angle between the velocity vectors v₁ and v₂ is also Δθ. This relationship is the geometric heart of the derivation.
设粒子在t时刻位于P点，以大小为v的速度v₁运动。经过微小时间间隔Δt后，它到达Q点，速度为v₂，大小也为v。画出以O为圆心、半径为r的圆。弧PQ在O点所张的角是一个小角度Δθ。因为v₁垂直于OP且v₂垂直于OQ，所以速度矢量v₁与v₂之间的夹角也是Δθ。这一几何关系是推导的核心。
Place the velocity vectors tail‑to‑tail to visualise their difference. The vector Δv = v₂ – v₁ represents the change in velocity over Δt. For small Δθ, the magnitude of Δv is roughly the base of an isosceles triangle with equal sides of length v and included angle Δθ. The direction of Δv is such that it points roughly toward the centre of the circle. In the limit Δt → 0, the approximation becomes exact.
将速度矢量尾对尾放置以直观展示它们的差。矢量Δv = v₂ – v₁代表在Δt内的速度变化量。对于微小的Δθ，Δv的大小近似等于一个等腰三角形的底边，该三角形的两个等边长度为v，夹角为Δθ。Δv的方向大致指向圆心。在极限Δt → 0下，近似转变为精确结果。
4. Change in Velocity and Geometry | 速度变化与几何关系
Consider the triangle formed by the two radial lines OP and OQ and the chord PQ. Compare this with the velocity vector triangle formed by v₁, v₂ and Δv. Because the velocity vectors are each perpendicular to the corresponding radii, the two triangles are similar (both are isosceles with the same apex angle Δθ). The similarity tells us that the ratio of the sides is the same: |Δv| / v = (chord PQ) / r.
考虑由两条径向线OP、OQ和弦PQ构成的三角形。将其与由v₁、v₂和Δv构成的速度矢量三角形进行比较。因为速度矢量各自垂直于相应的半径，这两个三角形相似（两者都是顶角为Δθ的等腰三角形）。相似性告诉我们边长的比例相同：|Δv| / v = (弦PQ) / r。
For a very small time interval Δt, the length of the chord PQ is almost equal to the length of the arc PQ. The arc length is simply vΔt because the particle travels a distance equal to speed multiplied by time. Thus, chord PQ ≈ vΔt. Substituting into the ratio gives |Δv| / v ≈ (vΔt) / r. Rearranging, the average acceleration magnitude aav = |Δv| / Δt ≈ v² / r.
对于非常小的时间间隔Δt，弦PQ的长度几乎等于弧PQ的长度。弧长就是vΔt，因为粒子走过的路程等于速率乘以时间。因此，弦PQ ≈ vΔt。代入比例式得到|Δv| / v ≈ (vΔt) / r。重新整理，平均加速度大小 aav = |Δv| / Δt ≈ v² / r。
5. Deriving the Magnitude of Acceleration | 加速度大小推导
Acceleration is defined as the instantaneous rate of change of velocity. By taking the limit as Δt → 0, both the approximation of chord length to arc length and the average acceleration become exact. The small angle Δθ also tends to zero. We obtain the exact expression:
加速度定义为速度的瞬时变化率。通过取Δt → 0的极限，弦长近似弧长以及平均加速度都变得精确。小角度Δθ也趋近于零。我们得到精确表达式：
a = v² / r
This is the centripetal acceleration magnitude. It depends on the square of the speed and inversely on the radius. A faster speed or a smaller circle yields a larger acceleration. The derivation can also be expressed using angular quantities: if the particle’s angular velocity is ω = Δθ/Δt, then |Δv| ≈ v Δθ gives a ≈ v ω. Using the relation v = ωr, we immediately recover a = v²/r.
这就是向心加速度的大小。它依赖于速率的平方，与半径成反比。更快的速率或更小的圆周会产生更大的加速度。推导也可以用角量表示：如果粒子的角速度为ω = Δθ/Δt，那么|Δv| ≈ v Δθ给出a ≈ v ω。利用关系式v = ωr，即可重新得到a = v²/r。
Whichever path you choose – similarity of triangles or angular velocity – you must reach the same conclusion and clearly state the limiting process to secure full marks. Examiners expect to see the step where Δt → 0 and the recognition that average acceleration tends to instantaneous acceleration.
无论选择三角形相似路径还是角速度路径，都必须得出相同结论并明确陈述极限过程以获得满分。考官期望看到Δt → 0的步骤，并认识到平均加速度趋向于瞬时加速度。
6. Direction of Acceleration | 加速度的方向
Equally important is proving that the acceleration points towards the centre of the circle. From the velocity vector triangle, as Δθ becomes infinitesimally small, the vector Δv becomes perpendicular to both v₁ and v₂. Because v₁ and v₂ are tangential to the circle, the direction of Δv lies along the radius, pointing inwards. Therefore, the acceleration is centripetal (centre‑seeking).
同样重要的是证明加速度指向圆心。从速度矢量三角形看，当Δθ变得无限小时，矢量Δv变得同时垂直于v₁和v₂。由于v₁和v₂都与圆相切，Δv的方向沿径向指向圆内。因此，加速度是向心的（指向中心）。
To illustrate this on a diagram, draw the two velocity vectors from a common point and show that the small vector Δv that closes the triangle bisects the angle between them. As Δθ approaches zero, this bisector aligns exactly with the inward radial line at the midpoint of the arc PQ. Stating that the acceleration is directed towards the centre completes the required proof.
在图上，从同一点画出两个速度矢量，并显示闭合三角形的微小矢量Δv平分它们之间的夹角。当Δθ趋近于零时，该平分线恰好与弧PQ中点处的向内径向线对齐。陈述加速度指向圆心即完成了要求的证明。
In the June 2018 mark scheme, a clear arrow indicating the radial direction of acceleration was a common marking point. Do not forget to specify direction whenever a question asks “show that the acceleration is directed towards the centre”. Simply giving the magnitude is insufficient.
在2018年6月的评分方案中，用清晰的箭头指明加速度的径向方向是一个常见评分点。当题目要求“证明加速度指向圆心”时，不要忘记指明方向。仅给出大小是不足够的。
7. Alternative Derivation Using Angular Velocity | 用角速度的替代推导
Once the core formula a = v²/r is derived, exam questions often extend the derivation to other useful forms. Substituting the wheel‑and‑axle relation
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