Centripetal Force: Analysing the Resultant Force for Circular Motion | 向心力：圆周运动所需的合力分析

📚 Centripetal Force: Analysing the Resultant Force for Circular Motion | 向心力：圆周运动所需的合力分析

When an object moves along a circular path, its direction changes continuously. This change in direction means the object is constantly accelerating, even if its speed remains constant. According to Newton’s first law, an external resultant force must act on the object to produce this acceleration. That resultant force is called the centripetal force, and it always points toward the centre of the circle.

当物体沿圆周路径运动时，其方向持续改变。方向的改变意味着物体时刻处于加速状态，即使其速率保持不变。根据牛顿第一定律，必须有一个外部的合力作用在物体上才能产生这种加速度。这个合力称为向心力，它始终指向圆心。

1. Defining Circular Motion Terminology | 圆周运动基本术语定义

In A-Level CIE Physics, understanding the precise definitions of angular quantities is essential before analysing forces. The angular displacement θ is measured in radians, while the angular velocity ω represents the rate of change of angular displacement. The linear speed v at any point on the circle is related to the angular velocity by the equation v = ωr, where r is the radius of the circular path.

在CIE A-Level物理中，准确理解角量定义是分析力的前提。角位移θ以弧度为单位测量，而角速度ω表示角位移的变化率。圆周上任意一点的线速率v与角速度的关系为v = ωr，其中r为圆周路径的半径。

The period T is the time taken for one complete revolution, and the frequency f is the number of revolutions per second. These are linked by T = 1/f and ω = 2π/T = 2πf. For example, a particle rotating at 5 revolutions per second has an angular velocity of 31.4 rad s⁻¹.

周期T是完成一整圈转动所需的时间，频率f是每秒转动的圈数。它们的关系为T = 1/f，且ω = 2π/T = 2πf。例如，一个以每秒5圈旋转的质点，其角速度为31.4 rad s⁻¹。

v = ωr, ω = 2π/T = 2πf

2. Why Does Circular Motion Require a Force? | 为什么圆周运动需要力？

Newton’s first law of motion states that a body remains at rest or in uniform motion in a straight line unless acted upon by a resultant force. In circular motion, although the speed may be constant, the velocity is not constant because its direction changes at every instant. Therefore, a resultant force must act toward the centre to continuously ‘pull’ the velocity vector around the circle.

牛顿第一定律指出，若没有外力作用，物体将保持静止或沿直线做匀速运动。在圆周运动中，尽管速率可能恒定，但由于方向不断改变，速度并不恒定。因此，必须有一个指向圆心的合力持续“牵引”速度矢量绕圆周转动。

This inward acceleration is called centripetal acceleration. The word ‘centripetal’ means ‘centre-seeking’. It is important to note that this acceleration is not caused by a separate ‘centripetal force’ that appears only in circular motion; rather, it is provided by existing forces such as tension, gravity, friction, or a normal reaction, which together produce the required resultant force.

这种向内的加速度称为向心加速度。“向心”（centripetal）意为“指向圆心”。需要特别指出的是，这种加速度并非由某种仅在圆周运动中出现的特殊“向心力”引起；它是由已有的力——如张力、重力、摩擦力或法向反力——共同提供的合力的结果。

If the resultant force disappears, the object will immediately fly off along the tangent to the circle, obeying Newton’s first law. For instance, when a string breaks during the whirling of an object, the object moves in a straight line tangent to the circle at the breaking point, not radially outward.

如果合力消失，物体将立即沿圆周的切线方向飞出，这服从牛顿第一定律。例如，在旋转物体时绳子断裂，物体会沿断裂点的切线方向做直线运动，而不会沿径向向外飞出。

3. Deriving Centripetal Acceleration and Force | 推导向心加速度与向心力公式

Consider an object moving with constant speed v along a circle of radius r. At two very close instants, its velocities v₁ and v₂ have the same magnitude but different directions. The change in velocity Δv points toward the centre of the circle. Using the relationship between linear and angular speed, and applying calculus or geometry, the magnitude of the centripetal acceleration is found to be a = v²/r.

考虑一个以恒定速率v沿半径为r的圆运动的物体。在两个非常接近的时刻，其速度v₁和v₂大小相同但方向不同。速度变化量Δv指向圆心。利用线速度与角速度的关系，并通过几何或微积分方法，可得向心加速度的大小为a = v²/r。

Since angular velocity ω = v/r, we can also write a = ω²r. Combining this with Newton’s second law F = ma, the centripetal force is given by:

由于角速度ω = v/r，我们也可以写成a = ω²r。结合牛顿第二定律F = ma，向心力公式为：

F = mv²/r = mω²r

A common derivation involves drawing a velocity vector diagram: for a small angular displacement Δθ, the change in velocity Δv ≈ v·Δθ. Dividing by Δt and letting Δt approach zero gives a = v·(dθ/dt) = vω = v²/r = ω²r.

一个常见的推导方法涉及绘制速度矢量图：对于很小的角位移Δθ，速度变化量Δv ≈ v·Δθ。将两边除以Δt并令Δt趋于零，得到a = v·(dθ/dt) = vω = v²/r = ω²r。

4. Identifying the Source of Centripetal Force | 识别向心力的来源

Centripetal force is not an independent force; it is the name given to the resultant force that happens to point toward the centre. In different physical situations, this resultant force comes from different interactions. The table below summarises common scenarios.

向心力不是一种独立的力；它是恰好指向圆心的合力所获得的名称。在不同的物理情境中，这个合力来自不同的相互作用。下表总结了常见的场景。

Scenario 场景	Source of centripetal force 向心力来源
Object on a string revolving horizontally 水平圆周运动	Tension in the string 绳的张力
Car turning on a flat road 汽车在平路上转弯	Friction between tyres and road 轮胎与路面间的摩擦力
Car turning on a banked road 汽车在倾斜路面转弯	Horizontal component of the normal reaction (and friction) 法向反力的水平分量（以及摩擦力）
Satellite orbiting Earth 卫星绕地球运行	Gravitational force 万有引力
Electron orbiting a nucleus (Bohr model) 电子绕核运动（玻尔模型）	Electrostatic (Coulomb) force 静电力（库仑力）
Object rotating on a turntable 物体在转台上旋转	Static friction 静摩擦力

To solve problems, first identify all real forces acting on the object, then resolve them into radial and tangential components. The radial resultant must equal the centripetal force. The tangential component, if present, changes the speed of the object, not its direction.

解题时，首先要识别物体所受的所有真实力，然后将它们分解为径向和切向分量。径向合力必须等于向心力。若有切向分量，它改变的是物体的速率，而不是方向。

5. Horizontal Circular Motion with Tension | 水平圆周运动中的张力分析

A classic example is a mass m attached to a string of length r, revolving at constant speed v in a horizontal circle. If the string remains horizontal, the tension T provides the entire centripetal force:

一个经典例子是质量为m的物体系在长度为r的绳端，以恒定速度v在水平面内做圆周运动。若绳子保持水平，则张力T完全提供向心力：

T = mv²/r = mω²r

If the speed doubles, the tension increases by a factor of four, because T is proportional to v². This explains why a fast-spinning object is much harder to control and why the string may snap when the speed exceeds a critical value.

如果速率加倍，张力将增大为原来的四倍，因为T与v²成正比。这解释了为什么旋转很快的物体更难控制，也解释了当速度超过临界值时绳子可能拉断的原因。

In practice, a string whirled horizontally is never perfectly horizontal because gravity pulls the mass downward. The string makes an angle θ with the horizontal, and the vertical component of tension balances the weight: T·sinθ = mg. The horizontal component provides the centripetal force: T·cosθ = mv²/r. This arrangement is known as a conical pendulum.

实际上，水平旋转的绳子不可能是完全水平的，因为重力会向下拉物体。绳子与水平方向成一定角度θ，张力的竖直分量与重力平衡：T·sinθ = mg。水平分量提供向心力：T·cosθ = mv²/r。这种装置称为锥摆。

6. Vertical Circular Motion | 竖直圆周运动分析

When a mass is whirled in a vertical circle, gravity constantly changes the speed of the object. At the top of the circle, weight and tension both point downward, so the centripetal force is the sum mg + T = mv²/r. At the bottom, tension acts upward and weight acts downward, so T − mg = mv²/r.

当物体在竖直平面内做圆周运动时，重力会不断改变物体的速率。在圆周顶端，重力与张力均向下，因此向心力为二者之和mg + T = mv²/r。在圆周底端，张力向上而重力向下，因此T − mg = mv²/r。

At the top of the circle, the minimum possible speed occurs when the tension is zero, i.e., when the weight alone provides the centripetal force:

在圆周顶端，最小可能速率出现在张力为零时，即仅由重力提供向心力：

mg = mv²_min/r → v_min = √(gr)

For example, a bucket of water whirled in a vertical circle of radius 0.5 m must have a minimum top speed of √(9.81 × 0.5) ≈ 2.21 m s⁻¹ for the water not to fall out. If the speed is lower, the bucket will not complete the circle, and the water spills.

例如，一个水桶在半径为0.5 m的竖直圆周中旋转，其顶端最小速率为√(9.81 × 0.5) ≈ 2.21 m s⁻¹，水才不会倒出。如果速率更低，水桶将无法完成整圈，水就会溢出。

7. Motion on Banked Tracks | 倾斜弯道上的运动分析

A banked track is designed so that a car (or cyclist or train) can turn without relying on friction. The track surface is inclined at an angle θ to the horizontal. The normal reaction N is perpendicular to the surface. Its horizontal component N·sinθ provides the centripetal force, while its vertical component N·cosθ balances the weight.

倾斜弯道（banked track）的设计目的是让汽车（或自行车、火车）在没有摩擦力的情况下也能转弯。跑道表面与水平方向成θ角。法向反力N垂直于表面。其水平分量N·sinθ提供向心力，竖直分量N·cosθ与重力平衡。

For a design speed v, the ideal banking angle satisfies:

对于设计速度v，理想倾斜角满足：

tan θ = v²/(rg)

At this speed, no friction is needed. If the car travels faster than the design speed, friction acts down the slope to provide additional centripetal force; if slower, friction acts up the slope to prevent sliding downward. An exam question may ask you to derive this equation using components of the normal reaction.

在此速率下，无需摩擦力。如果汽车比设计速度快，摩擦力沿斜面向上提供额外向心力；如果比设计速度慢，摩擦力沿斜面向上防止汽车向下滑。考试题可能会要求你利用法向反力的分量推导此方程。

8. Satellites and Gravitational Orbital Motion | 卫星与万有引力轨道运动

A satellite in a stable circular orbit of radius r around Earth experiences a gravitational force given by Newton’s law of gravitation:

在距地心半径为r的稳定圆轨道上运行的卫星，所受万有引力由牛顿万有引力定律给出：

F = GMm/r²

This gravitational force provides the centripetal force, so GMm/r² = mv²/r. Cancelling m gives the orbital speed:

该万有引力提供向心力，即GMm/r² = mv²/r。消去m可得轨道速度：

v = √(GM/r)

Thus, the orbital speed is independent of the satellite’s mass and depends only on the central body’s mass M and the orbital radius r. The period can be obtained from v = 2πr/T, leading to T² = (4π²/GM)r³, which is Kepler’s third law.

因此，轨道速度与卫星质量无关，仅取决于中心天体质量M和轨道半径r。周期可由v = 2πr/T得到，从而推出T² = (4π²/GM)r³，即开普勒第三定律。

For geostationary satellites, the orbital period equals 24 hours, so they appear stationary above a fixed point on the equator. Using the equation above with T = 86 400 s, students can calculate the geostationary orbital radius to be approximately 42 000 km from Earth’s centre.

对于地球同步卫星，其轨道周期为24小时，因此看起来静止在赤道上方某个固定点。利用上述方程，将T = 86 400 s代入，可以计算出地球同步轨道半径约为距地心42 000 km。

9. Worked Example 1: Horizontal Circle | 例题一：水平圆周运动

A 0.50 kg object is whirled in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m s⁻¹ on a frictionless table. Calculate (a) the centripetal acceleration, (b) the tension in the string.

一个0.50 kg的物体在无摩擦桌面上，以4.0 m s⁻¹的恒定速率在半径为0.80 m的水平圆中旋转。求：(a) 向心加速度；(b) 绳中张力。

(a) a = v²/r = (4.0)²/0.80 = 20 m s⁻².
(b) T = mv²/r = 0.50 × 20 = 10 N.

(a) a = v²/r = (4.0)²/0.80 = 20 m s⁻²。
(b) T = mv²/r = 0.50 × 20 = 10 N。

Note that the centripetal force is entirely provided by the tension because the normal reaction and weight cancel vertically. On a horizontal table, no friction is present, so the analysis is straightforward.

注意，由于法向反力与重力在竖直方向相抵消，向心力完全由张力提供。在水平桌面上没有摩擦力，因此分析较为直接。

10. Worked Example 2: Vertical Circle | 例题二：竖直圆周运动

A 0.20 kg ball on a string of length 0.60 m is swung in a vertical circle. At the top of the circle, the speed is 3.0 m s⁻¹. Calculate the tension in the string at that instant. Given g = 9.81 m s⁻².

一个0.20 kg的小球系在0.60 m长的绳上，在竖直平面内摆动。在圆周顶端，速率为3.0 m s⁻¹。求此刻绳中的张力。已知g = 9.81 m s⁻²。

At the top, both weight and tension act downward:

在顶端，重力与张力均向下：

mg + T = mv²/r

T = mv²/r − mg = 0.20 × (3.0)²/0.60 − 0.20 × 9.81 = 3.0 − 1.96 ≈ 1.0 N.

T = mv²/r − mg = 0.20 × (3.0)²/0.60 − 0.20 × 9.81 = 3.0 − 1.96 ≈ 1.0 N。

If the speed at the top were only 2.42 m s⁻¹, we would find T = 0, which represents the critical condition where the string just goes slack. Computing √(gr) = √(0.60 × 9.81) ≈ 2.42 m s⁻¹ confirms this boundary.

如果顶端速率仅为2.42 m s⁻¹，则T = 0，这是绳子刚好松弛的临界条件。计算√(gr) = √(0.60 × 9.81) ≈ 2.42 m s⁻¹即可验证此边界。

11. Common Misconceptions and Exam Pitfalls | 常见误解与考试陷阱

‘Centrifugal force’ is a fictitious force. In an inertial frame, no outward force acts on the object. The object tends to move in a straight line; the centripetal force is required to bend its path.

“离心力”是虚拟力。在惯性系中，没有向外的力作用在物体上。物体倾向于沿直线运动；向心力是弯曲其路径所必需的力。
Confusing speed with velocity. A constant speed does not imply zero acceleration; in circular motion, the velocity vector changes direction continuously.

混淆速率与速度。恒定速率并不意味着加速度为零；在圆周运动中，速度矢量方向不断改变。
Forgetting to resolve forces. In vertical or banked cases, all forces must be resolved along the radial direction (toward the centre) before applying F = mv²/r.

忘记对力进行分解。在竖直或倾斜情况下，必须先沿径向（指向圆心）分解所有力，才能应用F = mv²/r。
Using the wrong radius. The radius r in the centripetal force equation is the radius of the circular path, not necessarily the length of the string or the height of the cone.

用错半径。向心力公式中的r是圆周路径的半径，不一定是绳长或锥体的高度。
Dropping the direction of centripetal force. Centripetal force always points toward the centre; at the top of a vertical circle it points downward, while at the bottom it points upward.

忽略向心力的方向。向心力始终指向圆心；在竖直圆周顶端指向下，而在底端指向上。

12. Key Equations Summary and Revision Tips | 关键公式总结与复习建议

The following equations are essential for any circular motion question in CIE A-Level Physics. Practise deriving them and identifying when each form is most convenient.

以下是CIE A-Level物理中所有圆周运动问题必备的公式。请练习推导它们，并识别在什么情况下使用哪种形式最方便。

v = ωr, a = v²/r = ω²r, F = mv²/r = mω²r

For vertical circles, remember to include weight in the radial resultant. For satellites, use gravitational force as the centripetal force. For banked tracks, resolve the normal reaction into horizontal and vertical components, and recall that tan θ = v²/(rg) for the ideal banking angle.

对于竖直圆周，记住在径向合力中包含重力。对于卫星，使用万有引力作为向心力。对于倾斜弯道，将法向反力分解为水平和竖直分量，并记住理想倾斜角满足tan θ = v²/(rg)。

Finally, always start a problem by drawing a free-body diagram showing all real forces. Then choose the radial direction (positive toward the centre) and write down the resultant force equation. This systematic approach prevents most errors and is exactly what examiners expect in CIE mark schemes.

最后，解任何题目时都要先画受力图，标出所有真实力。然后选择径向方向（以指向圆心为正），写出合力方程。这种系统化方法能避免大多数错误，也正是CIE评分标准所期望的答题方式。

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Centripetal Force: Analysing the Resultant Force for Circular Motion | 向心力：圆周运动所需的合力分析