Sources of Centripetal Force: Worked Examples | 向心力的来源：实例分析

📚 Sources of Centripetal Force: Worked Examples | 向心力的来源：实例分析

In A-Level Physics, circular motion is a classic topic that tests your ability to identify the real force providing the centripetal force. This article walks through common physical situations, from a car turning on a flat road to a satellite orbiting Earth, and shows how to apply Newton’s second law toward the centre.

在 A-Level 物理中，圆周运动是一个经典考点，考查你能否找出真正提供向心力的那个力。本文将通过汽车转弯、卫星绕地球等常见情境，讲解如何对圆心方向应用牛顿第二定律。

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1. What Is Centripetal Force? | 什么是向心力？

Centripetal force is not a new, independent force. It is any force or combination of forces that directs an object toward the centre of its circular path. The name simply describes the role of an existing force.

向心力并不是一种新的、独立的力。它是任何把物体拉向圆周运动轨迹中心的力或合力。“向心力”仅仅是对某个已有力所起作用的描述。

For an object of mass m moving at speed v in a circle of radius r, the required centripetal force is:

F_c = m × (v² / r) = m × ω² × r

Here ω is the angular speed in rad s⁻¹. This equation gives the required force; the actual force must come from a physical interaction such as tension, friction, gravity, or a normal reaction.

其中 ω 是角速度，单位为 rad s⁻¹。这个公式给出的是“所需”的力；而真正的力必须来自某种物理相互作用，如张力、摩擦力、重力或支持力。

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2. A Simple Framework: Resolve Toward the Centre | 简单框架：向圆心方向分解力

The most reliable method for circular motion problems is:

解圆周运动问题最可靠的方法是：

• Draw a free-body diagram showing all real forces.
• Draw a free-body diagram showing all real forces.

• Choose the radial direction pointing toward the centre.
• 选取指向圆心的径向方向。

• Apply F_net(radial) = m × v²/r.
• 应用 F_net(径向) = m × v²/r。

If the required centripetal force exceeds the maximum real force available, the object cannot stay on the circular path — it will slip, skid, or fly off.

如果所需的向心力超过实际能提供的最大力，物体就不能保持圆周运动——它会滑动、侧滑或飞出去。

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3. Example 1: Car Turning on a Flat Road | 实例1：汽车在平直路面上转弯

When a car turns on a level road, the friction between the tyres and the road provides the centripetal force. The normal reaction balances the weight vertically, so it plays no role in turning.

汽车在水平路面上转弯时，轮胎与路面之间的摩擦力提供向心力。支持力在竖直方向平衡重力，因此与转弯无关。

For a car of mass m and tyres with coefficient of friction μ, the maximum friction is:

f_max = μ × N = μ × m × g

Setting this equal to the required centripetal force gives the maximum safe speed:

μ × m × g = m × v² / r → v_max = √(μ × g × r)

Notice that m cancels: a heavier car does not have a higher safe speed if the friction coefficient is the same.

注意 m 会被消去：若摩擦系数相同，更重的车并不会获得更高的安全速度。

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4. Example 2: Banking of Roads | 实例2：倾斜路面（弯道外侧高）

On a banked curve, the road surface is tilted. The normal reaction now has a horizontal component that points toward the centre of the turn, so less friction is needed.

在倾斜弯道上，路面是斜的。支持力现在有一个指向弯道圆心的水平分量，因此所需的摩擦力更小。

For a perfectly banked curve with angle θ and no reliance on friction, the horizontal component of the normal reaction provides the centripetal force:

N × sin θ = m × v² / r, N × cos θ = m × g

Dividing the two equations gives:

tan θ = v² / (r × g)

This is an important result: for a given radius and speed, the optimum banking angle is determined solely by v, r, and g. If the car goes faster than this design speed, friction acts down the slope; if slower, friction acts up the slope.

这是一个重要的结论：对给定的半径和速度，最佳倾斜角仅由 v、r 和 g 决定。如果车速大于设计速度，摩擦力沿斜面向下；如果车速小于设计速度，摩擦力沿斜面向上。

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5. Example 3: Conical Pendulum | 实例3：圆锥摆

A conical pendulum consists of a bob of mass m attached to a string that traces out a horizontal circle. The string makes a constant angle θ with the vertical.

圆锥摆由一个质量为 m 的小球组成，小球系在绳上，在水平面内画圆。绳与竖直方向保持恒定角度 θ。

The two forces on the bob are tension T and weight mg. Vertically, the bob has no acceleration:

T × cos θ = m × g

Horizontally, the component of tension provides the centripetal force:

T × sin θ = m × v² / r

Dividing these equations yields:

tan θ = v² / (r × g)

Notice this has the same mathematical form as the ideal banking angle. The radius r is the horizontal distance from the bob to the vertical axis, not the length of the string.

注意这个式子和理想倾斜角的数学形式相同。半径 r 是小球到竖直轴的水平距离，而不是绳长。

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6. Example 4: Satellite in Circular Orbit | 实例4：圆轨道卫星

For a satellite of mass m orbiting Earth at radius r from Earth’s centre, the only force is gravity. It both provides the centripetal force and acts as the weight of the satellite.

对于绕地球做圆周运动、质量为 m 的卫星来说，它受到的唯一力是万有引力。这个力既提供向心力，也充当卫星的“重力”。

G × (M × m) / r² = m × v² / r

Simplifying gives the orbital speed:

v = √(G × M / r)

where G is the gravitational constant and M is the mass of Earth. The satellite’s own mass cancels out, so all satellites at the same orbital radius have the same speed.

其中 G 是引力常量，M 是地球质量。卫星自身的质量被消去，因此在同一轨道半径上的所有卫星速度相同。

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7. Example 5: Charged Particle in a Magnetic Field | 实例5：带电粒子在磁场中的运动

A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force given by F = q × v × B. This force is always perpendicular to the velocity, so it acts as a centripetal force without doing work.

带电粒子垂直进入匀强磁场时，会受到洛伦兹力 F = q × v × B。该力始终垂直于速度，因此提供向心力，且不对粒子做功。

q × v × B = m × v² / r → r = (m × v) / (q × B)

This result explains why a faster particle or a heavier particle moves in a larger circle, while a stronger magnetic field gives a tighter circle. The period of the motion is T = 2π × m / (q × B), which is independent of speed.

这个结果解释了为什么速度更大或质量更大的粒子运动半径更大，而磁场越强则半径越小。运动周期为 T = 2π × m / (q × B)，与速度无关。

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8. Example 6: Vertical Circular Motion | 实例6：竖直圆周运动

The classic vertical circle problem involves a ball attached to a string or a bucket of water swung in a vertical plane. Here both tension and gravity contribute to the centripetal force, and the required force changes with position.

典型的竖直圆周运动问题涉及用绳系着小球或水桶在竖直平面内摆动。此时张力和重力都参与提供向心力，且所需的向心力随位置变化。

At the bottom of the circle, tension must overcome the weight and provide the centripetal force:

T_bottom − m × g = m × v² / r

At the top, both tension and weight point downward toward the centre, so:

T_top + m × g = m × v² / r

For the object to just complete a full circle with a string, the tension cannot be negative. The critical minimum speed at the top occurs when T_top = 0:

m × g = m × v_min² / r → v_min = √(g × r)

If the speed at the top is lower than this, the string goes slack and the object falls. This is why a bucket of water can stay in the bucket at the top only if the speed is high enough.

如果顶部速度低于这个值，绳子会松弛，物体将掉落。这就是为什么水桶在最高点要不洒水，速度必须足够大。

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9. Common Mistakes and Exam Tips | 常见错误与考试技巧

• Confusing real forces with centripetal force. Centripetal force is the result, not a separate force such as “centrifugal force”. Do not add it to a free-body diagram.
• 将实际力与向心力混淆。 向心力是合力效果，不是“离心力”那样的独立力；不要在受力分析图中把它们加上去。

• Forgetting that direction matters. At the top of a vertical circle, gravity and tension act in the same direction; at the bottom, they oppose each other.
• 忘记方向的重要性。 在竖直圆周最高点，重力与张力同向；在最低点，它们是反向的。

• Using the wrong radius. In a conical pendulum, r is the horizontal radius of the circular path, not the string length.
• 半径选错。 在圆锥摆中，r 是圆周运动的水平半径，而不是绳长。

• Skipping the unit check. Always convert revolutions per minute to rad s⁻¹ by multiplying by 2π/60.
• 忽视单位换算。 把每分钟转数换算为 rad s⁻¹ 时，要乘以 2π/60。

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10. Summary Table of Centripetal Force Sources | 向心力来源汇总表

The table below summarises the physical source of centripetal force in different situations.

下表总结了不同情境下向心力的实际来源。

Situation | 情境	Source of centripetal force | 向心力的来源	Key equation | 关键方程
Car on flat curve | 汽车在水平弯道	Static friction | 静摩擦力	f = m × v² / r
Banked road (ideal) | 理想倾斜路面	Horizontal component of normal reaction | 支持力的水平分量	tan θ = v² / (r × g)
Conical pendulum | 圆锥摆	Horizontal component of tension | 张力的水平分量	T × sin θ = m × v² / r
Satellite orbit | 卫星轨道	Gravitational force | 万有引力	G × M × m / r² = m × v² / r
Charged particle in B-field | 带电粒子在磁场中	Magnetic force | 洛伦兹力	q × v × B = m × v² / r

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11. Worked Exam-Style Question | 典型考题解析

Problem: A 1200 kg car travels around a circular track of radius 50 m. The coefficient of static friction between the tyres and the road is 0.60. Calculate the maximum speed at which the car can turn without sliding.

题目：一辆 1200 kg 的汽车在半径 50 m 的圆形跑道上行驶。轮胎与路面之间的静摩擦系数为 0.60。求汽车不侧滑的最大速度。

Step 1: Write the condition for maximum friction.

第1步：写出最大摩擦力的条件。

f_max = μ × m × g = 0.60 × 1200 × 9.81 ≈ 7063 N

Step 2: Equate to the required centripetal force.

第2步：令其等于所需的向心力。

7063 = m × v² / r = 1200 × v² / 50

Step 3: Solve for v.

第3步：解出 v。

v² = (7063 × 50) / 1200 ≈ 294.3 → v ≈ 17.2 m s⁻¹

The maximum safe speed is about 17 m s⁻¹ (about 62 km h⁻¹). In an exam, always state the direction of the centripetal force toward the centre and identify that static friction, not kinetic friction, is providing it.

最大安全速度约为 17 m s⁻¹（约 62 km h⁻¹）。考试中要明确指出向心力方向指向圆心，并且是静摩擦力而不是动摩擦力提供向心力。

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12. Final Advice for A-Level Physics | 给 A-Level 物理的最后建议

When solving any circular motion problem, first ask: “What physical force acts toward the centre?” Then write the radial component equation. Do not invent a new force; instead, resolve weight, tension, normal reaction, or friction along the radial direction.

解任何圆周运动问题时，先问自己：“什么真实力指向圆心？”然后写出径向分量方程。不要凭空捏造新的力；而应把重力、张力、支持力或摩擦力沿径向分解。

Finally, check limiting cases. If the required centripetal force is greater than the maximum available force, the circular path cannot be maintained. This idea appears in many A-Level multiple-choice and data-response questions.

最后，检查极限情况。如果所需向心力大于实际能提供的最大力，圆周运动就无法维持。这个思路经常出现在 A-Level 选择题和数据处理题中。

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