Derivation of Centripetal Acceleration a = v²/r | 向心加速度 a = v²/r 的推导

📚 Derivation of Centripetal Acceleration a = v²/r | 向心加速度 a = v²/r 的推导

Centripetal acceleration is a cornerstone concept in mechanics, describing the acceleration experienced by an object moving in a circular path at constant speed. Although the speed remains unchanged, the continuous change in direction requires a net inward force and causes an acceleration directed toward the centre of the circle. Understanding why this acceleration equals v²/r is essential for mastering topics from circular motion to gravitational orbits in A‑level Physics. In this article, we will derive the formula step by step using vector geometry and the idea of limiting processes, giving you a clear, exam‑ready explanation.

向心加速度是力学中的一个基础概念，它描述了物体以恒定速率沿圆周运动时所经历的加速度。尽管速率保持不变，但运动方向持续变化，因此需要一个指向圆心的净力，并产生指向圆心的加速度。理解为什么该加速度等于 v²/r 对于掌握从圆周运动到引力轨道等 A‑level 物理主题至关重要。在本文中，我们将利用向量几何和极限过程逐步推导该公式，提供清晰且适用于考试的讲解。

1. Setting the Scene: Uniform Circular Motion | 场景设定：匀速圆周运动

Uniform circular motion refers to the motion of an object travelling in a circle at a steady speed. The magnitude of the velocity vector, the speed v, is constant, but the direction of the velocity changes continuously. This directional change implies that the velocity vector is not constant; therefore, the object is accelerating, even though its speed does not change.

匀速圆周运动是指物体以稳定的速率沿圆周运动。速度矢量的大小（即速率 v）保持不变，但速度的方向不断变化。这种方向的改变意味着速度矢量并非恒定；因此，即使速率不变，物体仍在加速。

Consider a particle moving counter‑clockwise around a circle of radius r. At an instant t, the particle is at point P with velocity vector v₁ tangent to the circle. After a short time interval Δt, the particle reaches point Q with velocity vector v₂, still tangent but pointing in a different direction. Both vectors have the same length v.

考虑一个粒子沿半径为 r 的圆周逆时针运动。在时刻 t，粒子位于 P 点，速度矢量为 v₁，与圆周相切。经过一个短暂的时间间隔 Δt 后，粒子到达 Q 点，速度矢量为 v₂，依然与圆周相切但指向不同方向。两个矢量长度均为 v。

2. Defining the Change in Velocity | 定义速度的变化量

The acceleration a is defined as the rate of change of velocity: a = Δv / Δt in the limit Δt → 0. To find Δv, we subtract the initial velocity vector v₁ from the final velocity vector v₂: Δv = v₂ − v₁. Graphically, this is obtained by placing the tails of v₁ and v₂ together and drawing the vector from the head of v₁ to the head of v₂.

加速度 a 定义为速度的变化率：在 Δt → 0 的极限下 a = Δv / Δt。为了求出 Δv，我们用末速度矢量 v₂ 减去初速度矢量 v₁：Δv = v₂ − v₁。在图形上，可以将 v₁ 和 v₂ 的尾部放在一起，然后从 v₁ 的头部向 v₂ 的头部画一条矢量。

Because the particle moves through an angle Δθ during time Δt, the angle between v₁ and v₂ is also Δθ. The triangle formed by v₁, v₂, and Δv is isosceles, with two sides of length v. For small Δθ, the length of Δv can be approximated from the chord of the circle, which approaches the arc length.

由于粒子在 Δt 时间内转过角度 Δθ，v₁ 和 v₂ 之间的夹角也为 Δθ。由 v₁、v₂ 和 Δv 构成的三角形是等腰三角形，两条边长为 v。当 Δθ 很小时，Δv 的大小可以通过圆的弦长近似得到，而弦长趋近于弧长。

3. Relating Angular Displacement to Arc Length | 将角位移与弧长联系起来

In the circular path, the arc length Δs swept from P to Q is given by Δs = r Δθ, where Δθ is measured in radians. The speed of the particle is v = Δs/Δt, so v = r Δθ/Δt. The quantity Δθ/Δt is the angular speed ω, hence v = r ω.

在圆周路径上，从 P 到 Q 扫过的弧长 Δs 由 Δs = r Δθ 给出，其中 Δθ 以弧度为单位。粒子的速率为 v = Δs/Δt，因此 v = r Δθ/Δt。量 Δθ/Δt 为角速率 ω，因此 v = r ω。

Now look at the velocity vector triangle. The angle between v₁ and v₂ is Δθ, and the two sides adjacent to that angle have magnitude v. The side opposite the angle is |Δv|. For a very small angle, the chord length |Δv| is approximately equal to the arc length of a circle of radius v subtended by angle Δθ, i.e., |Δv| ≈ v Δθ.

现在观察速度矢量三角形。v₁ 和 v₂ 之间的夹角为 Δθ，与该角相邻的两条边大小为 v。夹角对边为 |Δv|。对于非常小的角度，弦长 |Δv| 近似等于半径为 v、圆心角为 Δθ 的圆的弧长，即 |Δv| ≈ v Δθ。

4. The Magnitude of Average Acceleration | 平均加速度的大小

The average acceleration during interval Δt has magnitude a_avg = |Δv|/Δt. Substituting the approximation |Δv| ≈ v Δθ gives a_avg ≈ v Δθ/Δt. Recognising that Δθ/Δt = ω and ω = v/r, we obtain a_avg ≈ v × (v/r) = v²/r.

在时间间隔 Δt 内的平均加速度大小为 a_avg = |Δv|/Δt。代入近似关系 |Δv| ≈ v Δθ，得到 a_avg ≈ v Δθ/Δt。注意到 Δθ/Δt = ω 且 ω = v/r，我们得到 a_avg ≈ v × (v/r) = v²/r。

This expression is independent of Δt and Δθ. As Δt → 0, the approximation becomes exact, and the instantaneous acceleration magnitude is

该表达式与 Δt 和 Δθ 无关。当 Δt → 0 时，该近似变为精确值，瞬时加速度的大小为

a = v² / r

5. Alternative Derivation Using Similar Triangles | 使用相似三角形的另一种推导

A more rigorous geometric method avoids the small‑angle arc approximation by using similar triangles. Draw the position vectors r₁ and r₂ from the centre to P and Q. The triangle formed by r₁, r₂, and the displacement Δs (chord) has sides r and r, included angle Δθ. The velocity triangle has sides v and v, included angle Δθ. These two triangles are similar because they are both isosceles with the same vertex angle.

一种更严格的几何方法通过相似三角形避免小角度弧长近似。画出从圆心到 P 和 Q 的位置矢量 r₁ 和 r₂。由 r₁、r₂ 和弦位移 Δs 构成的三角形边长为 r 和 r，夹角为 Δθ。速度三角形边长为 v 和 v，夹角为 Δθ。这两个三角形相似，因为它们都是等腰三角形且顶角相同。

From similarity, the ratio of the corresponding sides is equal: |Δv|/v = |Δs|/r. Thus, |Δv| = (v/r) × |Δs|. Divide both sides by Δt: |Δv|/Δt = (v/r) × |Δs|/Δt. As Δt → 0, |Δs|/Δt → v (the instantaneous speed), and |Δv|/Δt → a. Substituting yields a = (v/r) × v = v²/r.

由相似性，对应边之比相等：|Δv|/v = |Δs|/r。因此 |Δv| = (v/r) × |Δs|。两边除以 Δt：|Δv|/Δt = (v/r) × |Δs|/Δt。当 Δt → 0 时，|Δs|/Δt → v（瞬时速率），且 |Δv|/Δt → a。代入得到 a = (v/r) × v = v²/r。

This derivation clearly shows that the acceleration magnitude depends only on the speed and the radius, and that it is exact in the limit.

这一推导清楚地表明加速度的大小仅取决于速率和半径，并且在极限情况下是精确的。

6. Direction of the Centripetal Acceleration | 向心加速度的方向

To find the direction of the acceleration vector, examine the limit of Δv as Δt → 0. In the velocity triangle, as Δθ becomes infinitesimally small, the vector Δv becomes perpendicular to both v₁ and v₂. This direction points radially inward toward the centre of the circle. Hence, the acceleration is directed toward the centre, justifying the name ‘centripetal’, meaning ‘centre‑seeking’.

为了确定加速度矢量的方向，考察 Δt → 0 时 Δv 的极限。在速度三角形中，当 Δθ 趋近于无穷小，矢量 Δv 变得与 v₁ 和 v₂ 均垂直。该方向沿径向指向圆心。因此，加速度指向圆心，印证了“向心”（即“指向中心”）的名词。

In vector form, if r̂ is the unit radial vector pointing outward from the centre, the centripetal acceleration can be written as a = − (v²/r) r̂. The negative sign shows it is opposite to the radius vector, i.e., toward the centre.

用矢量形式表示，设 r̂ 为从圆心向外指向的单位径向矢量，则向心加速度可写为 a = − (v²/r) r̂。负号表明其与半径矢量方向相反，即指向中心。

7. Expressing Acceleration in Terms of Angular Velocity | 用角速率表示加速度

Using the relationship v = ω r, we can express the centripetal acceleration in another useful form: a = (ω r)² / r = ω² r. Thus,

利用关系式 v = ω r，我们可以用另一种实用的形式表示向心加速度：a = (ω r)² / r = ω² r。因此，

a = ω² r

This form is particularly handy when the angular speed ω is known directly, for instance from frequency f via ω = 2πf.

当角速率 ω 可直接获知时，例如通过频率 f 由 ω = 2πf 得到，这种形式尤为方便。

Both expressions a = v²/r and a = ω² r are equivalent and extensively used in problems on circular motion, satellites, and banking of curves.

表达式 a = v²/r 和 a = ω² r 两者等价，并广泛用于圆周运动、卫星问题和弯道倾斜角的计算中。

8. Physical Interpretation and the Role of Force | 物理解释与力的作用

According to Newton’s second law, any acceleration requires a net force in the same direction. For circular motion, the net force must point toward the centre: F_net = m a = m v²/r = m ω² r. This centripetal force is not a new kind of force but is provided by real forces such as tension, gravity, or friction.

根据牛顿第二定律，任何加速度都需要一个同方向的净力。对于圆周运动，净力必须指向圆心：F_net = m a = m v²/r = m ω² r。该向心力并非一种新型的力，而是由真实的力（如张力、重力或摩擦力）提供。

It is vital to remember that centripetal acceleration is a kinematic quantity describing the motion; the corresponding force is the cause. Confusing the two leads to common misconceptions, such as the imaginary ‘centrifugal force’ acting on the object in an inertial frame.

必须牢记向心加速度是描述运动的运动学量；相应的力则是原因。混淆两者会导致常见的误解，例如在惯性参考系中想象物体受到虚构的“离心力”作用。

9. Worked Example: Deriving Period from Acceleration | 实例：由加速度推导周期

Consider a satellite in a circular orbit of radius r moving at constant speed v. The centripetal acceleration is provided by gravity: g = v²/r. From this, we can deduce the orbital period T. Since v = 2πr/T, substituting gives g = (2πr/T)² / r = 4π² r / T². Rearranging yields T² = (4π² / g) r. This connects directly to Kepler’s third law for a constant gravitational field approximation.

考虑一颗在半径为 r 的圆形轨道上以恒定速率 v 运动的卫星。向心加速度由重力提供：g = v²/r。由此可推导轨道周期 T。因为 v = 2πr/T，代入得到 g = (2πr/T)² / r = 4π² r / T²。整理得 T² = (4π² / g) r。这直接关联到恒定重力场近似下的开普勒第三定律。

In a realistic gravitational field, the acceleration is GM/r², not g, leading to the more general T² ∝ r³. Nonetheless, the principle remains the same: the centripetal acceleration formula is the bridge between kinematics and dynamics in orbits.

在真实的引力场中，加速度为 GM/r² 而非 g，从而导出更普遍的 T² ∝ r³。尽管如此，原理仍然相同：向心加速度公式是轨道问题中连接运动学与动力学的桥梁。

10. Common Pitfalls and How to Avoid Them | 常见陷阱及规避方法

Students often mistake the speed v for the velocity and assume there is no acceleration if speed is constant. Always differentiate between scalar speed and vector velocity. Acceleration is the rate of change of velocity, not speed.

学生常将速率 v 当作速度，并认为若速率恒定则无加速度。务必区分标量速率和矢量速度。加速度是速度的变化率，而非速率的变化率。

Another pitfall is misapplying the formula a = v²/r when the motion is not genuinely uniform. The formula holds instantaneously only if the particle’s path can be approximated as a circular arc of radius r with instantaneous speed v. In non‑uniform circular motion, a tangential component of acceleration also exists.

另一个陷阱是在运动并非真正匀速的情况下错误应用公式 a = v²/r。该公式仅在粒子的路径可近似为半径为 r、瞬时速率为 v 的圆弧时瞬时成立。在非匀速圆周运动中，还存在切向加速度分量。

Finally, watch out for unit consistency: v in m/s, r in m, ω in rad/s. Using degrees instead of radians for Δθ will give wrong numerical results.

最后，注意单位一致性：v 用 m/s，r 用 m，ω 用 rad/s。若使用角度而非弧度计算 Δθ，将导致错误的数值结果。

11. Step‑by‑Step Summary of the Derivation | 推导的分步总结

Identify: Object moves in a circle of radius r with constant speed v.
识别：物体以恒定速率 v 沿半径为 r 的圆周运动。
Draw vectors: Velocity vectors at two nearby points are v₁ and v₂, equal in magnitude, separated by small angle Δθ.
绘制矢量：相邻两点的速度矢量为 v₁ 和 v₂，大小相等，夹角为小量 Δθ。
Find Δv: Subtract vectors to get Δv = v₂ − v₁, length |Δv|.
求Δv：矢量相减得 Δv = v₂ − v₁，长度 |Δv|。
Similar triangles: Velocity triangle similar to position triangle → |Δv|/v = chord/ r.
相似三角形：速度三角形与位置三角形相似 → |Δv|/v = 弦长/ r。
Limit: As Δt→0, chord → arc length Δs = rΔθ, and |Δv|/Δt → a.
取极限：当 Δt→0，弦长 → 弧长 Δs = rΔθ，且 |Δv|/Δt → a。
Substitute: a = v Δθ/Δt = v ω = v (v/r) = v²/r.
代入：a = v Δθ/Δt = v ω = v (v/r) = v²/r。
Direction: Δv points toward center, so a is centripetal.
方向：Δv 指向中心，故 a 为向心加速度。

12. Applications and Closing Thoughts | 应用与结语

The formula a = v²/r appears throughout physics: in the design of banked roads, roller coasters, centrifuges, and planetary motion. Mastering its derivation gives you not only the ability to recall it confidently but also a deeper understanding of vector analysis and limits. Always remember that a true physical insight comes from following the logic step by step, from definitions to the final elegant result.

公式 a = v²/r 在物理学中随处可见：在弯道设计、过山车、离心机和行星运动中均有应用。掌握其推导不仅能让你自信地记住公式，还能让你更深刻地理解矢量分析和极限思想。请始终记住，真正的物理洞察力源于一步一步遵循逻辑，从定义走向最终的简洁结果。

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Derivation of Centripetal Acceleration a = v²/r | 向心加速度 a = v²/r 的推导