Circular Motion for A-Level Physics | 圆周运动（A-Level 物理）

📚 Circular Motion for A-Level Physics | 圆周运动（A-Level 物理）

This article covers the essential concepts of circular motion that appear in the A-Level physics syllabus, specifically pages 303-337 of the Unit 6 textbook. We will explore angular quantities, centripetal acceleration and force, vertical circles, and real-world applications.

本文涵盖 A-Level 物理教学大纲中圆周运动的核心概念，对应 Unit 6 教材第 303-337 页。我们将深入探讨角量、向心加速度与向心力、竖直圆周运动以及实际应用。

1. Angular Displacement and Radians | 角位移与弧度

When an object moves along a circular path, its position can be described by the angle swept out from a reference line. This angle is called the angular displacement, θ, measured in radians.

当物体沿圆周路径运动时，其位置可以通过从参考线扫过的角度来描述。这个角度称为角位移 θ，单位为弧度。

One radian is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius. The full revolution around a circle is 2π radians, since the circumference is 2πr.

一弧度定义为：在圆心处所对的弧长等于半径时对应的角度。绕圆周一整圈为 2π 弧度，因为周长是 2πr。

θ = s / r

where s is the arc length and r is the radius. This relationship is fundamental to converting between linear and angular measurements.

其中 s 为弧长，r 为半径。这一关系是从线性量到角量转换的基础。

2. Angular Speed and Angular Velocity | 角速率与角速度

Angular speed, ω, is the rate of change of angular displacement with respect to time. For uniform circular motion, the angular speed is constant.

角速率 ω 是角位移随时间的变化率。对于匀速圆周运动，角速度恒定不变。

ω = Δθ / Δt = 2π / T = 2πf

where T is the period (time for one complete revolution) and f is the frequency. Angular speed is measured in radians per second (rad s⁻¹).

其中 T 为周期（完成一整圈所需时间），f 为频率。角速度的单位是弧度每秒（rad s⁻¹）。

The linear speed v at a distance r from the centre is related to the angular speed by:

距圆心 r 处的线速度 v 与角速度的关系为：

v = ωr

This equation shows that points farther from the rotation axis move faster for the same angular speed. For example, a point on the rim of a spinning wheel moves faster than a point near the axle.

该式表明，在相同角速度下，离转轴更远的点运动更快。例如，正在旋转的车轮轮缘上的点比靠近车轴的点运动得更快。

3. Centripetal Acceleration | 向心加速度

An object moving in a circle at constant speed is not moving at constant velocity, because velocity includes direction. The direction of motion changes continuously, so the object is accelerating.

以恒定速率做圆周运动的物体并非在做匀速运动，因为速度包括方向。运动方向持续改变，因此物体具有加速度。

This acceleration is directed towards the centre of the circle and is called centripetal acceleration. Its magnitude is given by:

该加速度指向圆心，称为向心加速度。其大小为：

a = v² / r = ω²r

Notice that centripetal acceleration increases with the square of the linear speed and decreases with the radius. Alternatively, it increases linearly with the radius when using angular speed.

注意，向心加速度随线速度的平方增大而增大，随半径增大而减小。若用角速度表示，则随半径线性增大。

The direction of centripetal acceleration at any instant is perpendicular to the velocity vector, pointing inward along the radius. It changes direction continuously but always points to the centre.

任意时刻向心加速度的方向垂直于速度矢量，沿半径指向圆心。其方向不断变化，却始终指向圆心。

4. Centripetal Force | 向心力

According to Newton’s second law, any acceleration requires a resultant force. The resultant force that produces centripetal acceleration is called the centripetal force.

根据牛顿第二定律，任何加速度都需要合外力。产生向心加速度的合外力称为向心力。

F = ma = mv² / r = mω²r

Despite the name, “centripetal” is not a separate type of force. It is provided by real forces such as tension, gravity, friction, or the normal reaction, depending on the situation.

尽管被称为“向心力”，但它并非一种独立的力。它由真实力提供，如拉力、重力、摩擦力或法向反作用力，具体取决于情境。

For a ball tied to a string and whirled horizontally, the tension provides the centripetal force.

对于系在绳上并水平旋转的小球，绳的张力提供向心力。
For a car turning on a flat road, static friction between the tyres and the road provides the necessary centripetal force.

对于在平路上转弯的汽车，轮胎与路面之间的静摩擦力提供所需的向心力。
For a satellite orbiting the Earth, the gravitational attraction provides the centripetal force.

对于绕地球运行的卫星，万有引力提供向心力。

5. Derivation of a = v² / r | 推导 a = v² / r

The derivation of centripetal acceleration uses vector subtraction of velocities at two nearby points on the circular path. Consider a particle moving with constant speed v around a circle of radius r.

向心加速度的推导利用圆周路径上两个邻近点的速度矢量相减。考虑一个以恒定速率 v 绕半径为 r 的圆运动的质点。

At time t, the velocity is v₁, and after a short interval Δt, the velocity is v₂. Both have the same magnitude v, but different directions. The angle between them is Δθ, which equals the angular displacement during Δt.

在时刻 t，速度为 v₁，经过短暂时间间隔 Δt 后，速度为 v₂。两者大小均为 v，但方向不同。它们之间的夹角为 Δθ，等于 Δt 内的角位移。

The change in velocity Δv = v₂ – v₁ forms an isosceles triangle with the two velocity vectors. The magnitude of Δv is approximately vΔθ for small angles.

速度变化量 Δv = v₂ – v₁ 与两个速度矢量构成等腰三角形。对于小角度，Δv 的大小近似为 vΔθ。

Dividing by Δt and using Δθ/Δt = ω:

除以 Δt，并利用 Δθ/Δt = ω：

a = Δv / Δt = vΔθ / Δt = vω = v² / r

Since v = ωr, both forms a = v² / r and a = ω²r are equivalent. The direction of Δv (and hence a) points radially inward in the limit as Δt approaches zero.

由于 v = ωr，a = v² / r 与 a = ω²r 两种形式等价。在 Δt 趋近于零的极限下，Δv（从而 a）的方向指向径向内。

6. Motion in a Horizontal Circle: Conical Pendulum | 水平圆周运动：圆锥摆

The conical pendulum consists of a mass on a string that describes a horizontal circle while the string traces out a cone. This is a classic example of horizontal circular motion.

圆锥摆由一根系着质量的绳子组成，质量在水平面内画圆，而绳子则描绘出一个圆锥。这是水平圆周运动的经典例子。

Let θ be the angle between the string and the vertical, and l be the string length. The radius of the circular path is r = l sin θ.

设 θ 为绳子与竖直方向的夹角，l 为绳长。圆周运动的半径为 r = l sin θ。

Resolving tensions: vertically, T cos θ = mg; horizontally, T sin θ = mω²r = mω² l sin θ.

对张力进行分解：竖直方向，T cos θ = mg；水平方向，T sin θ = mω²r = mω² l sin θ。

Dividing the two equations eliminates T:

两式相除消去 T：

tan θ = ω²r / g = ω² l sin θ / g → cos θ = g / (ω²l)

Therefore, for a given angular speed ω and a fixed string length l, the angle θ is uniquely determined. The faster the bob rotates, the larger the angle θ becomes.

因此，对于给定的角速度 ω 和固定的绳长 l，角度 θ 是唯一确定的。摆球旋转越快，角度 θ 就越大。

7. Motion in a Vertical Circle: Key Forces | 竖直圆周运动：关键力

Motion in a vertical circle is more complex because the gravitational force has a component tangential to the path. The speed is not constant; it is maximum at the bottom and minimum at the top, assuming energy is conserved.

竖直圆周运动更为复杂，因为重力的一个分量沿路径切向方向。若能量守恒，则速度并非恒定：在最低点最大，在最高点最小。

At the top of a vertical circle of radius r, with the mass m moving at speed v_top, the forces on the mass are its weight mg downward and the tension (or normal reaction) T_top also downward. So the resultant downward force provides the centripetal acceleration.

在半径为 r 的竖直圆周的最高点，质量为 m 的物体速度为 v_top，其所受的力包括竖直向下的重力 mg，以及同样向下的张力（或法向反作用力）T_top。因此，向下的合力提供向心加速度。

T_top + mg = mv_top² / r

At the bottom, the weight is downward but the tension is upward. The upward resultant force provides the centripetal acceleration:

在最低点，重力向下，但张力向上。向上的合力提供向心加速度：

T_bottom – mg = mv_bottom² / r

Thus T_bottom is always greater than T_top for the same speed, and the speed is already greater at the bottom in practice.

因此，在相同速率下 T_bottom 总是大于 T_top，而且实际上最低点的速率本身就更大。

8. Minimum Speed at the Top of a Vertical Circle | 竖直圆周最高点的最小速率

For the mass to just complete a full loop, the tension at the top can reduce to zero. In this critical case, the centripetal force is provided entirely by the weight.

为使物体恰好完成整圈运动，最高点处的张力可以减至零。在这种临界情况下，向心力完全由重力提供。

mg = mv_min² / r

v_min = √(gr)

If the speed at the top is less than √(gr), the object will lose contact with the track or the string will go slack before reaching the top. This critical speed does not depend on the mass.

若最高点速率小于 √(gr)，物体将在到达最高点之前脱离轨道，或绳子会松弛。该临界速率与质量无关。

Using conservation of energy, the minimum speed required at the bottom can be found. The height difference between top and bottom is 2r. Thus:

利用能量守恒，可以求出在最低点所需的最小速率。最高点与最低点的高度差为 2r。因此：

½mv_bottom² = ½mv_top² + mg(2r)

v_bottom_min = √(5gr)

This is a well-known result for the “loop-the-loop” problem: to complete a vertical circle, the initial speed at the bottom must be at least √(5gr).

这是著名的“过山车回环”问题结论：要完成竖直圆周运动，底部初始速率至少为 √(5gr)。

9. Banked Curves | 倾斜弯道

When a road or track is banked at an angle θ, the normal reaction from the surface has a horizontal component that can provide the centripetal force. This reduces or eliminates the reliance on friction.

当道路或轨道以角度 θ 倾斜时，来自路面的法向反作用力具有水平分量，可以提供向心力。这减少或消除了对摩擦力的依赖。

For ideal banking (where no friction is needed), the horizontal component of the normal reaction N sin θ provides the centripetal force, and N cos θ balances the weight:

对于理想倾斜（无需摩擦力时），法向反作用力的水平分量 N sin θ 提供向心力，而 N cos θ 平衡重力：

N sin θ = mv² / r

N cos θ = mg

Dividing gives the ideal speed:

相除得理想速率：

tan θ = v² / (rg)

For a given radius r and banking angle θ, there is exactly one speed v for which no friction is required. Above or below this speed, friction acts to provide or remove extra centripetal force.

对于给定的半径 r 和倾斜角 θ，恰好存在一个无需摩擦力的速率 v。高于或低于该速率时，摩擦力将起提供或抵消额外向心力的作用。

10. Applications: Satellites and Orbits | 应用：卫星与轨道

Circular motion applies directly to satellite orbits. A satellite in a circular orbit around the Earth experiences gravitational attraction as the centripetal force.

圆周运动直接适用于卫星轨道。绕地球做圆周轨道的卫星，其受到的万有引力充当向心力。

Equating gravitational force to centripetal force:

将引力与向心力相等：

GMm / r² = mv² / r

where M is the mass of the Earth, m is the mass of the satellite, r is the orbital radius from the Earth’s centre, and G is the gravitational constant.

其中 M 为地球质量，m 为卫星质量，r 为从地心算起的轨道半径，G 为引力常数。

Simplifying, the orbital speed is independent of the satellite mass:

化简后，轨道速率与卫星质量无关：

v = √(GM / r)

For geostationary satellites, the period T equals 24 hours, so the satellite remains above a fixed point on the equator. A higher orbit (larger r) corresponds to a slower orbital speed and a longer period.

对于地球静止卫星，周期 T 等于 24 小时，因此卫星始终位于赤道上方同一点。更高的轨道（更大的 r）对应更慢的轨道速率和更长的周期。

11. Non-Uniform Circular Motion | 非匀速圆周运动

In many real situations, the speed along a circular path changes. A roller coaster loop is an example. The total acceleration has two components: centripetal acceleration (changing direction) and tangential acceleration (changing speed).

在许多实际情境中，沿圆周路径的速率会变化。过山车回环就是一个例子。总加速度有两个分量：向心加速度（改变方向）和切向加速度（改变速率大小）。

The centripetal component is a_c = v² / r, directed toward the centre. The tangential component a_t is equal to the rate of change of speed, dv/dt, directed along the tangent to the path.

向心分量为 a_c = v² / r，指向圆心。切向分量 a_t 等于速率变化率 dv/dt，沿路径的切线方向。

The resultant acceleration magnitude is the vector sum:

合加速度大小为二者的矢量和：

a = √(a_c² + a_t²)

In uniform circular motion, a_t = 0 and only centripetal acceleration exists. In vertical circular motion, gravity provides both tangential and centripetal components depending on the position.

在匀速圆周运动中，a_t = 0，仅存在向心加速度。在竖直圆周运动中，重力根据位置同时提供切向和向心分量。

12. Common Examination Points | 常见考点

A-level exam questions on circular motion typically test the following skills:

A-level 考试中圆周运动的题目通常考查以下技能：

Converting between degrees and radians, and using θ = s / r.

度与弧度之间的换算，以及使用 θ = s / r。
Applying v = ωr and relating period, frequency, and angular speed.

应用 v = ωr，并联系周期、频率与角速度。
Identifying the real source of the centripetal force in a given physical situation (tension, gravity, friction, normal reaction).

识别给定物理情境中向心力的真实来源（张力、重力、摩擦力、法向反作用力）。
Solving problems involving conical pendulums and banked curves, including the derivation of tan θ = v² / (rg).

解决涉及圆锥摆和倾斜弯道的问题，包括推导 tan θ = v² / (rg)。
Calculating the critical speed for a vertical circle, v = √(gr) at the top and v = √(5gr) at the bottom.

计算竖直圆周的临界速率：顶部 v = √(gr)，底部 v = √(5gr)。
Using energy conservation in conjunction with circular motion equations.

结合能量守恒与圆周运动方程进行求解。

Always start by drawing a free-body force diagram, choose an appropriate radial direction, and apply Newton’s second law along the radius. Check units carefully: rad s⁻¹ for ω, m s⁻¹ for v, m s⁻² for a, and N for F.

解题时务必先画受力分析图，选择适当的径向方向，并沿半径方向应用牛顿第二定律。仔细检查单位：ω 用 rad s⁻¹，v 用 m s⁻¹，a 用 m s⁻²，F 用 N。

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Circular Motion for A-Level Physics | 圆周运动（A-Level 物理）