Circular Motion in CIE A-Level Physics | CIE A-Level 物理中的圆周运动

📚 Circular Motion in CIE A-Level Physics | CIE A-Level 物理中的圆周运动

Circular motion appears whenever an object moves along a circular path at constant speed. Although the speed may be constant, velocity is not, because direction changes continuously. This topic links kinematics, dynamics and energy, and is central to CIE A-Level Physics Paper 2 and Paper 4 questions.

当物体以恒定速率沿圆周路径运动时，就会出现圆周运动。尽管速率可能不变，但由于方向不断改变，速度并非恒定。本专题将运动学、动力学和能量联系起来，是 CIE A-Level 物理 Paper 2 和 Paper 4 的核心考点。

1. Describing Circular Motion | 描述圆周运动

In circular motion, an object travels along the circumference of a circle. Uniform circular motion means the speed is constant, but the direction of motion changes at every instant, so velocity is changing and the object is accelerating.

在圆周运动中，物体沿圆的周长运动。匀速圆周运动指速率恒定，但运动方向时刻改变，因此速度在变化，物体具有加速度。

Even if the word “uniform” is used, never assume acceleration is zero. CIE exam questions often ask why an object moving in a circle at constant speed is still accelerating.

即使用到“匀速”一词，也不能认为加速度为零。CIE 考试常问：物体以恒定速率做圆周运动，为什么仍有加速度？

---

2. Angular Displacement and Radian Measure | 角位移与弧度制

Angular displacement θ is the angle swept out by the radius. In A-Level physics, θ must be measured in radians, where one radian is the angle subtended at the centre of a circle by an arc length equal to the radius.

角位移 θ 是半径扫过的角度。A-Level 物理中 θ 必须用弧度表示：1 rad 是弧长等于半径时在圆心所张的角度。

The definition can be written as:

其定义可写为：

θ = s / r

where s is arc length and r is radius. This gives a dimensionless ratio, but we label it radians for clarity.

其中 s 为弧长，r 为半径。这是一个无量纲比值，但为了清晰，我们标记为弧度。

• Full circle: s = 2πr, so θ = 2π rad.
• 完整圆周：s = 2πr，因此 θ = 2π rad。

---

3. Angular Velocity and Period | 角速度与周期

Angular velocity ω is the rate of change of angular displacement. For uniform circular motion it is constant and given by:

角速度 ω 是角位移的变化率。在匀速圆周运动中，它恒定不变，公式为：

ω = θ / t = 2π / T

where T is the period, the time for one complete revolution. Frequency f is the number of revolutions per second, so f = 1/T and ω = 2πf.

其中 T 是周期，即完成一整圈所需的时间。频率 f 是每秒转过的圈数，因此 f = 1/T，且 ω = 2πf。

The SI unit of angular velocity is rad s⁻¹. Since radian is dimensionless, it can also be written as s⁻¹, but CIE expects rad s⁻¹ in calculations involving v = rω.

角速度的国际单位是 rad·s⁻¹。由于弧度无量纲，也可写作 s⁻¹，但在涉及 v = rω 的计算中 CIE 考试期望使用 rad s⁻¹。

---

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

Linear speed v along the tangent is related to angular velocity and radius by:

沿切线方向的线速度 v 与角速度和半径的关系为：

v = rω

This follows directly from s = rθ by dividing both sides by time. All points on a rigid rotating body have the same ω, but the linear speed increases with distance from the centre.

该关系可由 s = rθ 两边同时除以时间直接得出。同一刚体上各点的 ω 相同，但距圆心越远，线速度越大。

Substituting ω = 2π/T gives:

代入 ω = 2π/T 可得：

v = 2πr / T

This is often used for satellites, wheels and rotating discs.

这常用于卫星、车轮和转盘问题。

---

5. Centripetal Acceleration | 向心加速度

Because velocity direction changes, an object in uniform circular motion has an acceleration directed towards the centre. Its magnitude is:

由于速度方向不断变化，做匀速圆周运动的物体具有指向圆心的加速度。其大小为：

a = v² / r

Using v = rω, this can also be written as:

利用 v = rω，也可写作：

a = rω²

The acceleration vector always points towards the centre of the circle, which is why it is called centripetal acceleration. It changes velocity direction without changing speed.

加速度矢量始终指向圆心，因此称为向心加速度。它只改变速度方向，不改变速率。

---

6. Centripetal Force and Newton’s Second Law | 向心力与牛顿第二定律

From Newton’s second law, a resultant force must act towards the centre to produce centripetal acceleration. This force is:

根据牛顿第二定律，必须有指向圆心的合力来产生向心加速度。该力为：

F = mv² / r = mrω²

Centripetal force is not a new type of force. It is the name given to the resultant of existing forces such as tension, gravity, friction or the normal contact force, directed towards the centre.

向心力不是新类型的力，它是现有力（如张力、重力、摩擦力或法向接触力）指向圆心的合力名称。

A common error is to draw “centripetal force” as an extra force on a free-body diagram. CIE mark schemes penalise adding a separate centripetal force arrow. Identify the real force causing the circular path.

常见错误是在受力图中额外画出“向心力”箭头。CIE 评分标准会对单独添加向心力的画法扣分。应找出真正提供圆周运动的力。

---

7. Horizontal Circular Motion Examples | 水平圆周运动实例

For a car turning on a flat road, the horizontal friction between tyres and road provides centripetal force. The maximum safe speed is found by setting friction equal to mv²/r.

汽车在水平路面上转弯时，轮胎与路面之间的水平摩擦力提供向心力。最大安全速度可通过令摩擦力等于 mv²/r 求得。

For a stone whirled in a horizontal circle on a string, tension in the string provides centripetal force. If the string breaks, the stone flies off along the tangent, not radially outwards, because there is no longer a force to change its velocity direction.

用绳子在水平面甩动石块时，绳的张力提供向心力。若绳子断裂，石块将沿切线方向飞出，而不是沿半径向外，因为不再有改变速度方向的力。

For a coin placed on a rotating turntable, static friction provides centripetal force. If the turntable rotates too fast, the required mv²/r exceeds the maximum static friction, so the coin slips outwards along the tangent.

对于放在旋转圆盘上的硬币，静摩擦力提供向心力。若圆盘旋转太快，所需的 mv²/r 超过最大静摩擦力，硬币就会沿切线向外滑出。

Typical CIE questions ask you to calculate tension, maximum speed before skidding, or explain why an object slips when friction is insufficient.

典型的 CIE 题目要求计算张力、打滑前的最大速度，或解释为何摩擦力不足时物体会滑出。

---

8. Conical Pendulum | 圆锥摆

A conical pendulum consists of a mass moving in a horizontal circle at the end of a string, with the string tracing out a cone. The vertical component of tension balances weight, while the horizontal component provides centripetal force.

圆锥摆是由系在绳端的质量在水平面内做圆周运动组成，绳子扫出一个圆锥面。张力的竖直分量平衡重力，水平分量提供向心力。

The governing equations are:

其控制方程为：

T cos θ = mg

T sin θ = mv² / r

Dividing the two equations gives tan θ = v²/(rg), which can be used to find the angle, speed or radius.

两式相除得 tan θ = v²/(rg)，可用来求角度、速度或半径。

The period of a conical pendulum depends only on the vertical height h of the cone, not directly on the mass: T = 2π√(h/g). CIE may ask you to derive this relation.

圆锥摆的周期只取决于圆锥的竖直高度 h，与质量无直接关系：T = 2π√(h/g)。CIE 可能要求推导这一关系。

---

9. Vertical Circular Motion and Energy Conservation | 竖直圆周运动与能量守恒

In vertical circular motion, speed is not constant because gravitational potential energy changes with height. The object is fastest at the lowest point and slowest at the highest point.

在竖直圆周运动中，速率并不恒定，因为重力势能随高度变化。物体在最低点最快，在最高点最慢。

For a mass on a string or a roller coaster loop, the minimum speed at the top is found by requiring the string to remain taut or the track to remain in contact. At the top, weight alone can supply centripetal force at the minimum condition:

对于绳子系着的物体或过山车圆环，通过要求绳子保持张紧或轨道保持接触可求得最高点的最小速度。在最高点，最小条件下重力刚好提供向心力：

mg = mv² / r → v = √(gr)

Using v₁ for the speed at the bottom and v₂ for the speed at the top, conservation of mechanical energy gives:

用 v₁ 表示最低点速度，v₂ 表示最高点速度，机械能守恒给出：

½ m v₁² = ½ m v₂² + 2mgr

Substituting v₂ = √(gr) gives v₁ = √(5gr).

代入 v₂ = √(gr) 可得 v₁ = √(5gr)。

At the bottom of a vertical circle, tension minus weight provides centripetal force: T − mg = mv₁²/r. At the top, tension plus weight provides centripetal force: T + mg = mv₂²/r.

在竖直圆周最低点，张力减重力提供向心力：T − mg = mv₁²/r。在最高点，张力加重力提供向心力：T + mg = mv₂²/r。

For a rigid rod, the minimum speed at the top is zero because the rod can support compression, unlike a string.

对于刚性杆，最高点的最小速度为零，因为杆可以承受压缩，而绳子不能。

---

10. Banked Curves and Common Misconceptions | 倾斜弯道与常见误区

A banked road or track is inclined towards the centre of a bend. The horizontal component of the normal contact force can help provide centripetal force, reducing reliance on friction.

倾斜道路或轨道向弯道圆心方向倾斜。法向接触力的水平分量有助于提供向心力，从而减少对摩擦的依赖。

For ideal banking with no friction, the design speed satisfies:

对于无摩擦的理想倾斜情况，设计速度满足：

tan θ = v² / rg

where θ is the banking angle. This is the same geometric relationship as in a conical pendulum.

其中 θ 为倾斜角。这与圆锥摆中的几何关系相同。

Published by TutorHao | A-Level Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）