Circular and Periodic Motion Formula Derivation | 圆周与周期运动公式推导

📚 Circular and Periodic Motion Formula Derivation | 圆周与周期运动公式推导

In OxfordAQA International A-Level Physics, circular and periodic motion is a core topic. Understanding how key formulas are derived—rather than just memorising them—gives you a deeper insight and helps you tackle unfamiliar problems with confidence. This article walks you through the derivations step by step, from angular velocity to centripetal acceleration and simple harmonic motion.

在 OxfordAQA 国际 A-Level 物理中，圆周运动与周期运动是一个核心主题。理解关键公式是如何推导出来的，而非死记硬背，能够让你洞察更深，并自信地应对陌生问题。本文带你一步一步完成从角速度到向心加速度以及简谐运动的推导。

1. Angular Displacement and Angular Velocity | 角位移与角速度

For an object moving in a circle, we measure its position by the angle θ swept out from a reference line. Angular displacement Δθ is measured in radians (rad). One full revolution equals 2π rad.

对于沿圆周运动的物体，我们用从参考线扫过的角度 θ 来度量它的位置。角位移 Δθ 以弧度 (rad) 为单位。一整圈等于 2π rad。

The average angular velocity ω_av is defined as the rate of change of angular displacement: ω_av = Δθ / Δt. For uniform circular motion, the instantaneous angular velocity ω is constant and given by:

平均角速度 ω_av 定义为角位移的变化率：ω_av = Δθ / Δt。对于匀速圆周运动，瞬时角速度 ω 是恒定的，由下式给出：

ω = Δθ / Δt

Angular velocity has units of rad s⁻¹. It is a scalar in A-Level specifications but can be treated as a vector directed along the axis of rotation.

角速度的单位是 rad s⁻¹。在 A-Level 的规范中它是一个标量，但可以视为沿转轴方向的矢量。

2. Relationship Between Linear and Angular Velocity | 线速度与角速度的关系

The distance s travelled along the circular arc is related to the angle by the arc length formula: s = rθ, where r is the radius of the circle and θ is in radians.

沿圆弧行进的距离 s 与角度之间由弧长公式联系：s = rθ，其中 r 是圆的半径，θ 的单位是弧度。

The linear speed v (the magnitude of tangential velocity) is the rate at which the object moves along the circumference. Using s = rθ, we get:

线速度 v（切向速度的大小）是物体沿圆周运动的速率。利用 s = rθ，我们得到：

v = Δs / Δt = r (Δθ / Δt) = r ω

Therefore, the instantaneous linear speed and angular velocity are linked by v = rω. This relation only holds when ω is expressed in rad s⁻¹.

因此，瞬时线速度与角速度通过 v = rω 联系起来。这个关系仅在 ω 以 rad s⁻¹ 为单位时才成立。

3. Period and Frequency | 周期与频率

The period T is the time taken to complete one full revolution. In one period the angular displacement is 2π rad, so ω = 2π / T. Rearranging gives T = 2π / ω.

周期 T 是完成一整圈所需的时间。在一个周期内，角位移是 2π rad，所以 ω = 2π / T。整理得 T = 2π / ω。

Frequency f is the number of revolutions per second: f = 1 / T. Consequently, ω = 2πf. The relationship can also be expressed in terms of linear speed: v = 2πr / T.

频率 f 是每秒转动的圈数：f = 1 / T。因此，ω = 2πf。这个关系也可以用线速度表达：v = 2πr / T。

ω = 2π / T f = 1 / T v = 2πr / T

These equations are fundamental for solving any problem involving periodic circular motion.

这些方程是解决任何涉及周期圆周运动问题的基础。

4. Derivation of Centripetal Acceleration: Geometric Approach | 向心加速度的推导：几何方法

Consider an object moving at constant speed v in a circle of radius r. In a short time interval Δt, it moves from point A to point B, subtending a small angle Δθ. The velocity vectors at A and B have the same magnitude v but differ in direction. The change in velocity Δv is obtained by vector subtraction: Δv = v_B − v_A.

考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在很短的 Δt 时间内，它从点 A 运动到点 B，对应的圆心角为小角度 Δθ。在 A 点和 B 点的速度矢量大小同为 v，但方向不同。速度的变化量 Δv 由矢量相减得到：Δv = v_B − v_A。

If we draw the velocity vectors tail-to-tail, the triangle formed by v_A, v_B and Δv is similar to the triangle formed by the two radii and the chord AB. For small Δθ, the chord length is approximately the arc length Δs = r Δθ. Hence, Δv / v ≈ Δs / r, which gives Δv ≈ (v / r) Δs.

若将速度矢量尾尾相接画出，由 v_A、v_B 和 Δv 构成的三角形与两条半径和弦 AB 构成的三角形相似。当 Δθ 很小时，弦长近似等于弧长 Δs = r Δθ。因此，Δv / v ≈ Δs / r，即 Δv ≈ (v / r) Δs。

The magnitude of the average acceleration during Δt is a_av = Δv / Δt ≈ (v / r) (Δs / Δt) = v²/r. In the limit Δt → 0, the instantaneous acceleration points toward the centre of the circle and has magnitude a_c = v²/r.

在 Δt 内的平均加速度大小为 a_av = Δv / Δt ≈ (v / r) (Δs / Δt) = v²/r。在 Δt → 0 的极限下，瞬时加速度指向圆心，大小为 a_c = v²/r。

Using v = rω, we also get a_c = r ω². This acceleration is called the centripetal acceleration.

利用 v = rω，我们还得到 a_c = r ω²。这个加速度就叫做向心加速度。

5. Derivation of Centripetal Acceleration: Calculus Approach | 向心加速度的推导：微积分方法

For completeness, we can derive the same result using position vectors. Define the position of an object in uniform circular motion by r = r cos(ωt) i + r sin(ωt) j, where ω is constant angular velocity. The velocity vector is the first derivative:
v = dr/dt = −rω sin(ωt) i + rω cos(ωt) j.

为了完整性，我们可以用位置矢量得出同样的结果。定义匀速圆周运动中物体的位置为 r = r cos(ωt) i + r sin(ωt) j，其中 ω 为恒定角速度。速度矢量是一阶导数：v = dr/dt = −rω sin(ωt) i + rω cos(ωt) j。

The speed is the magnitude: v = √[(−rω sin(ωt))² + (rω cos(ωt))²] = rω, as expected. Differentiating again gives acceleration:
a = dv/dt = −rω² cos(ωt) i − rω² sin(ωt) j = −ω² r.

速率大小为 v = √[(−rω sin(ωt))² + (rω cos(ωt))²] = rω，与预期一致。再次求导给出加速度： a = dv/dt = −rω² cos(ωt) i − rω² sin(ωt) j = −ω² r。

The acceleration is opposite to the position vector, i.e. directed toward the centre, and its magnitude is a_c = ω²r = v²/r.

加速度与位置矢量方向相反，即指向圆心，其大小为 a_c = ω²r = v²/r。

6. Centripetal Force Formula | 向心力公式

According to Newton’s second law, the net force required to keep an object moving in a circle is directed toward the centre and is called the centripetal force F_c. Using F = ma, we have:

根据牛顿第二定律，维持物体做圆周运动所需的净力指向圆心，称为向心力 F_c。利用 F = ma，我们得到：

F_c = m v² / r = m r ω²

It is crucial to understand that centripetal force is not a new type of force; it is provided by tension, gravity, friction, or other real forces acting toward the centre.

关键是要理解向心力不是一种新的力；它是由指向圆心的拉力、重力、摩擦力或其他真实力提供的。

7. Relating Linear Velocity, Period and Radius | 线速度、周期与半径的关系

Combining v = 2πr / T with the centripetal acceleration formula gives another useful expression:

将 v = 2πr / T 与向心加速度公式结合，得到另一个有用的表达式：

a_c = 4π² r / T²

This form is particularly handy when the period is known. The centripetal force then becomes F_c = 4π² m r / T².

当周期已知时，这种形式特别方便。此时向心力变为 F_c = 4π² m r / T²。

8. Introduction to Simple Harmonic Motion (SHM) | 简谐运动简介

Simple harmonic motion is a type of periodic motion where the acceleration a of the oscillating object is directly proportional to its displacement x from equilibrium and is always directed toward that equilibrium position:

简谐运动是一种周期运动，其中振动物体的加速度 a 与它偏离平衡位置的位移 x 成正比，并且总是指向平衡位置：

a ∝ −x or a = −ω² x

Here ω is called the angular frequency of the oscillation, and it is constant for a given system. The negative sign indicates that acceleration and displacement are in opposite directions.

这里的 ω 称为振动的角频率，对于给定系统是一个常数。负号表示加速度与位移方向相反。

The connection to circular motion is elegant: if a particle moves uniformly in a circle, its projection onto any diameter performs SHM. This lets us transfer the maths of circular motion directly to oscillations.

与圆周运动的联系十分精妙：如果一个质点做匀速圆周运动，它在任意直径上的投影做简谐运动。这使得我们可以将圆周运动的数学直接迁移到振动上。

9. Deriving Displacement, Velocity and Acceleration in SHM | 简谐运动位移、速度、加速度的推导

Take the projection of circular motion. If a particle rotates with angular velocity ω starting from an initial phase φ, its x-coordinate is x = A cos(ωt + φ), where A is the amplitude (maximum displacement, equal to the radius of the reference circle). Often we choose φ = 0 for simplicity, giving x = A cos(ωt).

取圆周运动的投影。若一个质点以角速度 ω 从初相 φ 开始旋转，其 x 坐标为 x = A cos(ωt + φ)，其中 A 是振幅（最大位移，等于参考圆的半径）。为简便，通常取 φ = 0，即 x = A cos(ωt)。

The velocity v is the time derivative of displacement:

速度 v 是位移对时间的导数：

v = dx/dt = −Aω sin(ωt)

The maximum speed occurs when sin(ωt) = ±1, giving v_max = ωA. The acceleration is the derivative of velocity:

当 sin(ωt) = ±1 时速率最大，得到 v_max = ωA。加速度是速度的导数：

a = dv/dt = −Aω² cos(ωt) = −ω² x

This confirms the defining equation of SHM. The maximum acceleration is a_max = ω²A, occurring at the extreme positions.

这证实了简谐运动的定义方程。最大加速度为 a_max = ω²A，出现在极端位置。

10. Deriving Period Formulas for Mass-Spring and Pendulum | 弹簧振子与单摆的周期公式推导

For a mass-spring system obeying Hooke’s law, the restoring force is F = −kx, where k is the spring constant. Newton’s second law gives ma = −kx, so a = −(k/m)x. Comparing with a = −ω²x yields ω² = k/m, hence:

对于遵从胡克定律的弹簧振子，回复力为 F = −kx，其中 k 是劲度系数。由牛顿第二定律 ma = −kx，得 a = −(k/m)x。与 a = −ω²x 对比，得 ω² = k/m，因此：

T = 2π / ω = 2π √(m/k)

For a simple pendulum of length L, the restoring force for small angles (θ small, sinθ ≈ θ) is −mg sinθ ≈ −mg (x/L). Thus a = −(g/L)x. Comparing gives ω² = g/L, so:

对于长度为 L 的单摆，小角度 (θ 小, sinθ ≈ θ) 下回复力为 −mg sinθ ≈ −mg (x/L)。因此 a = −(g/L)x。对比得 ω² = g/L，所以：

T = 2π √(L/g)

These derivations are standard and show how the angular frequency ω folds into the period for any SHM system.

这些是标准推导，展示了角频率 ω 如何融入任何简谐运动系统的周期之中。

11. Worked Examples | 典型例题

Example 1: A car goes round a roundabout of radius 15 m at 10 m s⁻¹. Calculate its centripetal acceleration and the coefficient of friction needed if the car’s mass is 1200 kg and g = 9.81 m s⁻².

例题 1： 一辆汽车以 10 m s⁻¹ 的速度绕半径为 15 m 的环岛行驶。计算向心加速度，并求当汽车质量为 1200 kg、g = 9.81 m s⁻² 时所需的摩擦系数。

a_c = v²/r = (10)² / 15 = 6.67 m s⁻². The centripetal force is provided by friction: f = m a_c = 1200 × 6.67 = 8000 N. Since f = μ m g, we get μ = 8000 / (1200 × 9.81) ≈ 0.68.

a_c = v²/r = (10)² / 15 = 6.67 m s⁻²。向心力由摩擦力提供：f = m a_c = 1200 × 6.67 = 8000 N。因为 f = μ m g，得 μ = 8000 / (1200 × 9.81) ≈ 0.68。

Example 2: A mass of 0.5 kg on a spring oscillates with amplitude 3 cm and period 0.6 s. Determine the spring constant k and the maximum acceleration.

例题 2： 一个 0.5 kg 的物体在弹簧上振动，振幅为 3 cm，周期为 0.6 s。求劲度系数 k 和最大加速度。

From T = 2π √(m/k), k = 4π² m / T² = 4π² × 0.5 / (0.6)² ≈ 54.8 N m⁻¹. a_max = ω²A = (2π/T)² × 0.03 = (2π/0.6)² × 0.03 ≈ 3.29 m s⁻².

由 T = 2π √(m/k) 得 k = 4π² m / T² = 4π² × 0.5 / (0.6)² ≈ 54.8 N m⁻¹。a_max = ω²A = (2π/T)² × 0.03 = (2π/0.6)² × 0.03 ≈ 3.29 m s⁻²。

12. Summary of Key Formulas | 主要公式总结

The following table collects the essential equations you need for circular and periodic motion in the OxfordAQA examination.

下表汇总了你在 OxfordAQA 考试中所需的圆周运动与周期运动的基本方程。

Quantity / 物理量	Formula / 公式
Angular velocity	ω = Δθ / Δt = 2π / T = 2πf
Linear speed	v = r ω = 2πr / T
Centripetal acceleration	a_c = v² / r = r ω² = 4π² r / T²
Centripetal force	F_c = m v² / r = m r ω²
SHM defining equation	a = −ω² x
SHM displacement	x = A cos(ωt + φ)
SHM maximum speed	v_max = ωA
Mass-spring period	T = 2π √(m/k)
Simple pendulum period	T = 2π √(L/g)

Remember that for all derived equations, angles must be in radians and the small-angle approximation holds for the pendulum.

请记住，所有导出方程中角度必须以弧度为单位，且单摆适用小角度近似。

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Circular and Periodic Motion Formula Derivation | 圆周与周期运动公式推导