Circular Motion | 圆周运动考点精讲

📚 Circular Motion | 圆周运动考点精讲

In A-Level Edexcel Physics, circular motion is a fundamental topic that describes the movement of an object along a circular path at constant speed. Understanding angular displacement, angular velocity, centripetal acceleration, and centripetal force is essential for tackling real-world applications such as banked tracks, conical pendulums, and vertical loops.

在A-Level Edexcel 物理中，圆周运动是描述物体沿圆形路径以恒定速率运动的基本主题。理解角位移、角速度、向心加速度和向心力，对于解决弯道倾斜、圆锥摆和竖直环形轨道等实际应用至关重要。

1. Radian Measure | 弧度制

Angular displacement in circular motion is measured in radians (rad). One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

圆周运动中的角位移以弧度 (rad) 为单位。1弧度的定义为：当圆弧的长度等于半径时，该圆弧所对的圆心角的大小。

The relationship between arc length s, radius r, and angular displacement θ is given by:

弧长 s、半径 r 和角位移 θ 之间的关系为：

θ (rad) = s / r

Since the circumference of a circle is 2πr, one complete revolution corresponds to an angular displacement of 2π radians. Therefore, 360° = 2π rad, which allows conversion between degrees and radians using θ(rad) = (π/180) × θ(°).

由于圆的周长为2πr，一整圈对应的角位移为2π弧度。因此，360° = 2π rad，可以借助 θ(rad) = (π/180) × θ(°) 实现度与弧度的转换。

Using radians simplifies the analysis of circular motion and is the standard unit in all equations that link angular and linear quantities.

使用弧度可以简化圆周运动的分析，在所有联系角量与线量的方程中，弧度都是标准单位。

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

Angular velocity ω is the rate of change of angular displacement. For an object moving with uniform circular motion, the magnitude of ω is constant and is measured in rad s⁻¹.

角速度 ω 是角位移的变化率。对于做匀速圆周运动的物体，ω 的大小恒定，单位为 rad s⁻¹。

ω = Δθ / Δt

The period T is the time taken for one complete revolution, and the frequency f is the number of revolutions per unit time (f = 1/T). Angular velocity can be expressed in terms of T or f:

周期 T 是物体完成一整圈所需的时间，频率 f 是单位时间内的转数 (f = 1/T)。角速度可以用 T 或 f 表示：

ω = 2π / T = 2πf

These expressions are derived from the fact that one revolution corresponds to an angular displacement of 2π radians in a time T.

这些表达式源自于：一整圈在时间 T 内对应 2π 弧度的角位移。

It is important to ensure that your calculator is in radian mode when performing calculations involving ω and transformations between linear and angular variables.

进行涉及 ω 及线量与角量转换的计算时，务必确保计算器处于弧度模式。

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

For an object moving along a circular path of radius r with constant angular velocity ω, its instantaneous linear speed v is constant and directed tangentially to the circle.

对于沿半径为 r 的圆形路径以恒定角速度 ω 运动的物体，其瞬时线速度 v 大小不变，方向沿圆的切线。

The link between linear speed v and angular velocity ω is:

线速度 v 与角速度 ω 之间的联系为：

v = rω

This relationship holds for all particles on a rigid rotating body and is fundamental when comparing points at different distances from the centre of rotation.

这一关系适用于刚体上所有旋转的点，并且在比较距旋转中心不同距离的点时十分关键。

The following table summarises the key variables:

下表总结了关键变量：

Quantity	Symbol	Unit
Angular velocity	ω	rad s⁻¹
Linear speed	v	m s⁻¹
Radius	r	m
Period	T	s

Linear speed can also be expressed as v = 2πr / T, which is consistent with v = rω because ω = 2π / T.

线速度也可表示为 v = 2πr / T，这与 v = rω 一致，因为 ω = 2π / T。

4. Centripetal Acceleration | 向心加速度

Although the speed of a body in uniform circular motion is constant, its velocity is continuously changing because the direction of motion changes. This change in velocity implies an acceleration directed towards the centre of the circle, called centripetal acceleration.

尽管物体在匀速圆周运动中的速率恒定，但由于运动方向不断变化，其速度矢量一直在改变。这种速度变化指向圆心的加速度，称为向心加速度。

The magnitude of centripetal acceleration a is given by two equivalent formulas:

向心加速度 a 的大小可用两个等价的公式表示：

a = v² / r

a = rω²

These expressions are derived from the geometry of circular motion and from the fact that the velocity vector sweeps out a circle at a constant angular rate.

这些表达式源于圆周运动的几何关系，以及速度矢量以恒定角速率扫出一个圆的事实。

Centripetal acceleration is always perpendicular to the velocity, so it changes the direction of motion but never the speed in uniform circular motion.

向心加速度始终与速度垂直，因此在匀速圆周运动中，它只改变运动方向，不改变速率大小。

The direction of a is towards the centre, so the vector form is a = -ω²r (directed radially inward).

加速度 a 的方向指向圆心，其矢量形式为 a = -ω²r（径向向内）。

5. Centripetal Force | 向心力

According to Newton’s second law, an object that undergoes centripetal acceleration must experience a net force towards the centre of the circle. This force is called the centripetal force and its magnitude is:

根据牛顿第二定律，任何产生向心加速度的物体必然受到一个指向圆心的合力。该力称为向心力，其大小为：

F = mv² / r = mrω²

It is crucial to understand that ‘centripetal force’ is not a new type of force; it is simply the name given to the resultant force that causes circular motion. Always identify the real forces (tension, friction, gravity, normal reaction) that provide this resultant.

必须理解，“向心力”并不是一种新的力，它只是为了产生圆周运动而作用于物体上的合力的名称。始终要找出提供该合力的真实力（如张力、摩擦力、重力、支持力）。

For example, in a car turning on a flat road, the centripetal force is provided by friction between the tyres and the road. In a conical pendulum, it arises from the horizontal component of the string tension.

例如，汽车在水平路面上转弯时，向心力由轮胎与路面之间的摩擦力提供；在圆锥摆中，向心力来源于绳子张力的水平分量。

A common misconception is the concept of ‘centrifugal force’, which is a fictitious force experienced only in a rotating reference frame. In an inertial frame, there is no outward force pushing the object away from the centre.

一个常见的误区是“离心力”，它是一种只在旋转参考系中感受到的虚拟力。在惯性系中，并没有将物体推离圆心的向外作用力。

6. Horizontal Circular Motion: The Conical Pendulum | 水平面内的圆周运动：圆锥摆

A conical pendulum consists of a mass suspended from a string that moves in a horizontal circle with the string tracing out a cone. The forces acting on the mass are tension T and weight mg.

圆锥摆由一个悬挂在绳子上的摆球组成，摆球在水平面内做圆周运动，绳子划出一个锥面。作用在摆球上的力有拉力 T 和重力 mg。

The tension can be resolved into a vertical component T cosθ = mg, and a horizontal component T sinθ that provides the centripetal force:

拉力可以分解为竖直分量 T cosθ = mg，以及提供向心力的水平分量 T sinθ：

T sinθ = mrω²

Dividing the two equations gives tanθ = rω² / g, and using r = L sinθ (where L is the string length), the period of the pendulum is T_period = 2π √(L cosθ / g). The exam may ask for derivations or analysis of how the angle θ depends on angular speed.

将两式相除得到 tanθ = rω² / g，再利用 r = L sinθ（L 为绳长），摆的周期为 T_period = 2π √(L cosθ / g)。考试可能要求推导或分析角度 θ 如何依赖于角速度。

When the angular speed increases, the bob rises (θ increases), and the radius of the circle increases. These relationships can be tested through multiple-choice or structured questions.

当角速度增大时，摆球上升（θ增大），圆周半径增大。这些关系可通过选择题或结构化问题来考查。

7. Banked Tracks | 弯道倾斜

For vehicles to negotiate a curved path safely at high speed, roads or tracks are often banked at an angle. The normal reaction and friction (if present) together provide the necessary centripetal force.

为使车辆高速安全地通过弯道，路面或赛道通常设计成倾斜的。支持力以及（可能存在的）摩擦力共同提供所需的向心力。

In the idealised case of no friction, the horizontal component of the normal reaction R alone supplies the centripetal force, while the vertical component balances the weight:

在无摩擦的理想情况下，支持力 R 的水平分量独自提供向心力，竖直分量与重力平衡：

R sinθ = mv² / r

R cosθ = mg

Dividing yields the banking condition for a frictionless turn:

两式相除得出无摩擦转弯的倾斜条件：

tanθ = v² / (rg)

This equation shows that the required banking angle increases with speed and decreases with the radius of curvature.

该方程表明，所需的倾斜角随速度增大而增大，随曲率半径增大而减小。

When friction is included, the analysis becomes more complex. The frictional force can act up or down the slope depending on whether the car is moving faster or slower than the ideal speed.

当考虑摩擦力时，分析变得更为复杂。摩擦力的方向可能沿斜面向上或向下，这取决于车速高于还是低于理想速度。

8. Vertical Circular Motion | 竖直面内的圆周运动

Vertical circular motion is more complicated because the speed is not normally constant due to gravitational work. Common examples include a bucket of water swung in a vertical circle or a roller coaster loop.

竖直面内的圆周运动更为复杂，因为重力做功通常导致速率不再恒定。常见的例子包括在竖直面内甩动水桶或过山车环形轨道。

At the top and bottom of a circular path, the net force towards the centre is still mv²/r, but the contributing forces change. For an object attached to a string of length r, moving at speed v at the bottom, the tension T must overcome weight and supply centripetal force:

在圆周路径的最高点和最低点，指向圆心的合力仍为 mv²/r，但参与力有所不同。对于系在绳长 r 上的物体，以速度 v 通过最低点时，拉力 T 必须克服重力并提供向心力：

T − mg = mv² / r (bottom)

At the top, both tension and weight act towards the centre:

在最高点，拉力和重力均指向圆心：

T + mg = mv² / r (top)

For the water not to fall out of the bucket, the reaction (or tension) at the top must be greater than or equal to zero. This leads to the minimum speed condition: v_min = √(rg). Below this speed, the bucket would need a downward reaction that the string cannot supply.

要使水不从桶中洒出，最高点的支持力（或拉力）必须大于等于零。这给出了临界最小速率条件：v_min = √(rg)。低于此速率，绳子无法提供向下的拉力，水就会洒出。

In a roller coaster loop, the normal reaction at the top must satisfy N ≥ 0 for the cart to stay on the track, giving the same minimum speed v_min = √(rg) for a circular loop.

在过山车环形轨道中，最高点的支持力 N 必须满足 N ≥ 0，车体才能紧贴轨道，同样给出圆形回路的最小速率 v_min = √(rg)。

Energy conservation between top and bottom is often used to find speeds at different points, assuming no loss: (1/2)mv_top² + mg(2r) = (1/2)mv_bottom².

经常利用最高点与最低点之间的能量守恒（假设无能量损失）来求各点的速率：(1/2)mv_top² + mg(2r) = (1/2)mv_bottom²。

9. Energy Considerations in Uniform and Non-Uniform Circular Motion | 圆周运动中的能量

In uniform circular motion, the speed remains constant, so the kinetic energy is unchanged. The centripetal force does no work because it acts perpendicular to the displacement at every instant.

在匀速圆周运动中，速率恒定，因此动能不变。向心力不做功，因为它在每一时刻都与位移垂直。

However, in vertical circular motion or any non-uniform circular motion, the tangential component of the net force changes the speed, and work is done by forces such as gravity or an externally applied torque.

然而，在竖直圆周运动或任何非匀速圆周运动中，合力的切向分量会改变速率，重力或外部力矩等力做功。

Understanding energy conversion helps solve problems involving minimum speeds, tension variations, and power in rotating systems. For instance, a pendulum bob reaches its maximum speed at the lowest point, where gravitational potential energy is fully converted into kinetic energy.

理解能量转化有助于解决涉及最小速率、拉力变化及旋转系统功率的问题。例如，摆锤在最低点速率最大，此时重力势能完全转化为动能。

Many A-Level questions ask you to link centripetal force conditions with energy conservation to find critical speeds or forces at specific positions.

许多A-Level题目要求学生将向心力条件与能量守恒联系起来，求出关键速率或特定位置的力。

10. Common Pitfalls and Exam Tips | 常见误区与应试技巧

Pitfall 1: Believing that a centrifugal force pushes the object outward. Always analyse forces from an inertial frame; the only force pulling towards the centre is the resultant of real forces.

误区1： 认为存在离心力将物体向外推。务必从惯性参考系分析力；指向圆心的力只是真实力的合力。

Pitfall 2: Confusing v = rω with angular velocity ω. Remember that ω must be in rad s⁻¹. If given rev min⁻¹, convert to rad s⁻¹ first.

误区2： 混淆 v = rω 中的 ω 与角速度。记住 ω 必须以 rad s⁻¹ 为单位。如果题目给的转速单位是 rev min⁻¹，需要先转换成 rad s⁻¹。

Pitfall 3: Assuming speed is constant in all circular motion problems. In vertical circles, speed varies; use energy methods to find the speed at different points.

误区3： 认为所有圆周运动问题中速率恒定。在竖直面内的圆周运动中速率是变化的；要使用能量方法求出不同位置的速率。

Pitfall 4: Forgetting that the centripetal force is the net force, not an individual force. Always resolve forces and then set the resultant towards the centre equal to mv²/r or mrω².

误区4： 忘记向心力是合力而不是单独的一个力。始终需要分解力，然后令指向圆心的合力等于 mv²/r 或 mrω²。

Exam tip: When drawing free-body diagrams, show all real forces, then indicate the direction of the net force towards the centre. This makes it easier to write the correct equation.

应试技巧： 画受力图时，标出所有真实力，然后标明指向圆心的合力方向，这有助于写出正确的方程。

Exam tip: For banked track problems, decide whether friction acts up or down the slope by comparing the car’s speed to the ideal frictionless speed v_ideal = √(rg tanθ).

应试技巧： 对于倾斜轨道问题，通过比较汽车的实际速度与理想无摩擦速度 v_ideal = √(rg tanθ) 来判定摩擦力的方向。

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Circular Motion | 圆周运动考点精讲