📚 Centripetal Forces | 向心力
Objects moving in a circular path, even at constant speed, are continuously changing direction. This change in velocity demands an acceleration directed towards the centre of the circle, and consequently a net inward force. Understanding centripetal force is essential for analysing everything from a car rounding a bend to the orbit of a satellite. In CIE A-Level Physics, you are expected to derive expressions, apply them to various scenarios, and clearly distinguish between centripetal force and the fictitious centrifugal effect.
沿圆周运动的物体，即使速率恒定，其运动方向也在不断改变。速度方向的改变需要一个指向圆心的加速度，因而也需要一个指向圆心的净力。理解向心力是分析汽车过弯、卫星轨道等众多现象的基础。在 CIE A-Level 物理中，你需要推导相关公式，将其应用于各种场景，并清晰地区分向心力与虚设的离心效应。
1. Uniform Circular Motion | 匀速圆周运动
y. Since velocity is a vector, any change in direction implies an acceleration. The path is a circle of radius r, and the object completes one revolution in a period T. The speed v is related to the circumference by v = 2π r / T. The angular displacement per unit time is the angular speed ω, defined as ω = Δθ / Δt, measured in rad s⁻¹. For one complete revolution, ω = 2π / T, leading to the useful relation v = ω r.
在匀速圆周运动中，速度的大小（速率）保持不变，但速度矢量的方向持续变化。由于速度是矢量，方向的任何改变都意味着存在加速度。运动轨迹是半径为 r 的圆，物体每转动一周所用的时间称为周期 T。速率 v 与圆周长的关系为 v = 2π r / T。单位时间内的角位移称为角速度 ω，即 ω = Δθ / Δt，单位为 rad s⁻¹。对于完整的一周，ω = 2π / T，由此得到常用的关系式 v = ω r。
Even though the speed is constant, the object is accelerating because the direction of motion is changing. The acceleration vector points towards the centre of the circle, perpendicular to the velocity vector at every instant. This is called centripetal acceleration. Without this acceleration, the object would move off along a straight-line tangent, as predicted by Newton's first law.
尽管速率恒定，但物体因运动方向不断变化而具有加速度。加速度矢量始终指向圆心，在任何时刻都与速度矢量垂直。这就是向心加速度。如果没有这个加速度，根据牛顿第一定律，物体将沿切线方向做匀速直线运动。
2. Centripetal Acceleration | 向心加速度
The magnitude of the centripetal acceleration a꜀ al Δt. The direction of a is radially inward, towards the centre of the circle. In CIE examinations, you are expected to be able to derive the formula a = v² / r using vector diagrams and the small-angle approximation, or by considering radial components of motion.
向心加速度 a 的大小有两个等价的表达式：a = v² / r 和 a = ω² r。这两个公式都可以从短时间 Δt 内速度矢量三角形导出。加速度 a 的方向沿径向指向圆心。在 CIE 考试中，你需要能够利用矢量图和小角度近似、或通过运动的径向分量推导出 a = v² / r。
ve a = (2π f)² r. This shows that doubling the frequency quadruples the required inward acceleration. Remember that although the object' celeration is non-zero; it is responsible for continuously deflecting the velocity vector.
对于以恒定角速度运动的物体，向心加速度也可用频率 f 表示：由 ω = 2π f 可得 a = (2π f)² r。这表明频率加倍将使所需的向心加速度增至原来的四倍。务必牢记，尽管速率不变，加速度并不为零；它持续地偏转着速度矢量的方向。
3. Centripetal Force | 向心力
From Newton's second law, a net force must act to produce the centripetal acceleration. This net force is called the centripetal force: F = m a = m v² / r = m ω² r The direction of the centripetal force is the same as the acceleration — towards the centre of the circular path. If the resultant force towards the centre ceases, the object will travel tangentially in a straight line at constant speed.
根据牛顿第二定律，必须有一个净力来产生向心加速度。这个净力即为向心力：F = m a = m v² / r = m ω² r。向心力的方向与加速度相同——指向圆形轨迹的圆心。一旦指向圆心的合力消失，物体就将沿切线方向以恒定速率做直线运动。
It is vital to understand that 'centripetal force' is not a new type of force; it is simply the name given to the resultant force that acts towards the centre of the circle. This resultant may be provided by tension, friction, gravitational attraction, the normal reaction, or a combination of several forces. In problem-solving, always identify the physical force(s) supplying the centripetal requirement.
必须清晰地认识到，“向心力”并不是一种新型的力；它不过是指向圆心的合力。这个合力可以由张力、摩擦力、万有引力、法向反作用力或多种力的组合来提供。在解题时，务必先识别出哪些真实的力充当了向心力。
4. Sources of Centripetal Force | 向心力的来源
Different situations involve different physical forces acting as the centripetal force. For a ball whirled on a string, the tension in the string provides the inward pull. For a car turning on a flat road, the static friction between tyres and road points towards the centre of the curve. For a planet orbiting the Sun, the gravitational attraction supplies the centripetal force. In a vertical loop-the-loop, the normal reaction and the weight combine to give the resultant inward force at each point.
不同情景中，由不同的真实力充当向心力。系在绳子一端旋转的小球，绳子张力提供向内的拉力；在水平路面上转弯的汽车，轮胎与路面间的静摩擦指向弯道中心；绕太阳运行的行星，万有引力充当向心力；在竖直的环形轨道上，法向反作用力与重力的合力在每个位置提供所需的向内合力。
When several forces act, you must resolve them into radial and tangential components. Only the net radial component contributes to the centripetal force. For instance, on a banked track, both the normal reaction and friction have radial components. A free-body diagram is essential: draw all forces, choose a coordinate system with one axis pointing to the centre, and apply Fnet(radial) = m v²/r.
当多个力作用时，必须将它们分解为径向分量和切向分量。只有沿径向的净力才构成向心力。例如，在倾斜轨道上，法向反作用力和摩擦力都有径向分量。受力图至关重要：画出所有力，选取一个坐标轴指向圆心，然后应用径向净力等于 m v²/r。
5. Centripetal vs. Centrifugal Force | 向心力与离心力的区分
Many learners confuse centripetal force with the so-called centrifugal force. Centripetal force is a real force acting towards the centre, observed in an inertial (non-accelerating) frame of reference. Centrifugal force, on the other hand, is a fictitious force that appears only when the motion is described from a rotating reference frame. In the rotating frame, objects seem to be pushed outward, but this is a consequence of inertia rather than a genuine interaction.
许多学生将向心力与所谓的离心力混淆。向心力是真实存在的力，方向指向圆心，可在惯性（非加速）参考系中观测到。而离心力是一种虚设力，仅在从转动参考系描述运动时才出现。在转动参考系中，物体看起来像是被向外推，但这其实是惯性的表现，并非真正的相互作用。
In your A-Level answers, you should never state that there is an outward centrifugal force acting on a body in circular motion when using an inertial frame. Instead, explain that the body would move off tangentially if the centripetal force were removed. For example, when a car turns sharply, a passenger feels pushed against the door because the passenger's body tends to continue in a straight line, while the car turns inward. The door provides the centripetal force, not a centrifugal push.
在 A-Level 的作答中，采用惯性系时切勿说物体受到向外的离心力。相反，应解释如果向心力消失，物体将沿切线飞出。例如，汽车急转弯时，乘客感觉被压向车门，这是因为乘客身体倾向于沿直线运动，而汽车向内转动；车门提供的是向心力，而不是向外的离心推力。
6. Vehicles on a Level Curve | 水平弯道上的车辆
m possible friction is fₘₐₓ = μₛ N, where μₛ is the coefficient of static friction and N is the normal reaction. On a level road, N = mg, so fₘₐₓ = μₛ mg. Equating this to the required centripetal force gives μₛ mg = m v²/r. Hence the maximum safe speed to avoid skidding is vₘₐₓ = √μₛ g r. If the speed exceeds this value, the friction is insufficient to keep the car on the circular path, and it slides outward.
车辆在水平弯道上行驶时，向心力由轮胎与路面的静摩擦力提供。最大静摩擦为 fₘₐₓ = μₛ N，其中 μₛ 为静摩擦系数，N 为法向反作用力。在水平路面上 N = mg，因此 fₘₐₓ = μₛ mg。令其等于所需的向心力，得 μₛ mg = m v²/r。故最大安全速度（避免侧滑）为 vₘₐₓ = √μₛ g r。若车速超过此值，摩擦力不足以维持圆周运动，汽车将向外侧滑出。
Note that the maximum safe speed does not depend on the vehicle's mass; it depends only on the coefficient of friction, the radius of the bend, and g. This explains why all vehicles, regardless of weight, must slow down on tight corners. If the road is wet or icy, μₛ is reduced, lowering the maximum safe speed dramatically.
值得注意的是，最大安全车速与车辆质量无关，仅取决于摩擦系数、弯道半径和重力加速度 g。这说明了为什么无论车重如何，所有车辆在急弯处都必须减速。路面湿滑或结冰时 μₛ 减小，最大安全车速将大幅降低。
7. Banked Curves | 倾斜弯道
component of the normal reaction can provide the necessary centripetal force, even in the absence of friction. For a given speed v and radius r, the ideal banking angle is given by tan θ = v² / (g r). This is derived by resolving the normal reaction into horizontal and vertical components and setting N cos θ = mg and N sin θ = m v²/r.
为减少对摩擦的依赖，公路和铁路弯道常以角度 θ 向内倾斜。在平滑的倾斜弯道上，即使没有摩擦力，法向反作用力的水平分量也能提供必要的向心力。对于给定的速度 v 和半径 r，理想倾斜角由 tan θ = v² / (g r) 决定。这是通过将法向反作用力分解为水平和竖直分量，并令 N cos θ = mg 和 N sin θ = m v²/r 推导得出的。
At the ideal banking angle, a vehicle can negotiate the curve without any lateral frictional force, making the turn safer and reducing tyre wear. If a vehicle travels slower than the design speed, friction acts up the slope to prevent sliding inward; if faster, friction acts down the slope. The banking formula also explains why high-speed racetracks have steeply banked curves.
在理想倾斜角时，车辆过弯无需任何侧向摩擦力，这既安全又能减少轮胎磨损。若车速低于设计速度，摩擦力沿斜坡向上以防止向内滑动；若高于设计速度，摩擦力则沿斜坡向下。倾斜公式也解释了为何高速赛道的弯道具有很大的倾角。
8. Vertical Circular Motion | 竖直平面内的圆周运动
A mass whirled in a vertical circle represents a classic problem where both gravity and tension vary in their contribution. At the top of the circle, both weight mg and tension T act downwards towards the centre, so the net centripetal force is Ttop the bottom, tension acts upwards towards the centre while weight acts downwards, giving Tbottom ich requires T ≥ 0 at all points. The critical case occurs at the top: the minimum speed vmin at the top is found by setting Ttop = 0, yielding vmin = √g r.
系在绳端的小球在竖直平面内做圆周运动是一个经典问题，其中重力和拉力的贡献随位置变化。在圆的最高点，重力 mg 和拉力 T 均向下指向圆心，故净向心力为 Ttop + mg = m v²/r。在最低点，拉力向上指向圆心而重力向下，因此 Tbottom − mg = m v²/r。为了维持圆周运动，绳子必须始终绷紧，这意味着所有位置都要求 T ≥ 0。临界情况出现在最高点：令 Ttop = 0 得最小速度 vmin = √g r。
If the object has a speed at the top greater than √g r, the excess speed increases the tension in the string. Using energy conservation between top and bottom, one finds vbottom² = vtop² + 4 g r, and the tension at the bottom is Tbottom = m (vbottom² / r + g). These relations are often tested in conjunction with energy considerations, highlighting the interplay of kinematics, dynamics, and energy.
若物体在最高点的速度大于 √g r，多余的速度会增加绳子拉力。利用最高点与最低点之间的能量守恒可得 vbottom² = vtop² + 4 g r，最低点拉力为 Tbottom = m (vbottom² / r + g)。这些关系常与能量问题结合考查，突出了运动学、动力学和能量之间的相互作用。
9. The Conical Pendulum | 圆锥摆
t angle θ with the vertical. The forces acting are tension T and weight mg. Resolving vertically: T cos θ = mg. Radially, the horizontal component of tension provides the centripetal force: T sin θ = m v²/r, where the radius of the circle is r = L sin θ, L being the string length. Combining these yields tan θ = v²/(g r) and also v = √g r tan θ ingly, the period of rotation Tₚ (to distinguish from tension) can be expressed as Tₚ = 2π √L cos θ / g, independent of the mass.
圆锥摆由一根悬挂重物的绳子构成，重物以恒定角速度在水平面内画圆。绳子与竖直线成恒定夹角 θ。作用力为拉力 T 和重力 mg。竖直方向分解：T cos θ = mg。径向水平分量提供向心力：T sin θ = m v²/r，其中圆周半径 r = L sin θ，L 为绳长。联立两式可得 tan θ = v²/(g r) 以及 v = √g r tan θ。有趣的是，旋转周期 Tₚ = 2π √L cos θ / g 与质量无关。
period shortens. If the angle becomes very small, cos θ ≈ 1, and the period approaches 2π √L/g, resembling a simple pendulum. The conical pendulum provides a neat demonstration of how centripetal force can be supplied entirely by a component of tension.
对于给定绳长 L，周期仅取决于 cos θ 和 g。θ 增大时 cos θ 减小，周期变短。若摆角非常小，cos θ ≈ 1，周期趋近 2π √L/g，与单摆相似。圆锥摆清晰展示了向心力可以完全由拉力的一个分量来提供。
10. Gravitation as Centripetal Force | 万有引力作为向心力
ing the Sun, or satellites orbiting the Earth, the centripetal force is provided by Newton' l to m v²/r gives v² = G M / r. The orbital speed thus decreases with increasing radius: v ∝ 1/√r. Using v = 2π r / T, we obtain T² = (4π² / G M) r³, which is Kepler' eriod squared is proportional to the cube of the orbital radius, independent of the planet's mass.
对于绕太阳运行的行星或绕地球运行的卫星，向心力由牛顿万有引力定律提供：F = G M m / r²，其中 M 为中心天体质量，r 为轨道半径。令其等于 m v²/r 可得 v² = G M / r。轨道速率因此随半径增大而减小：v ∝ 1/√r。利用 v = 2π r / T，可得 T² = (4π² / G M) r³，这正是开普勒第三定律。这表明轨道周期的平方与轨道半径的立方成正比，且与行星质量无关。
have a period T = 24 hours and orbit in the equatorial plane. By substituting T and Earth' to the expression for r, one finds a fixed orbital radius of about 4.23 × 10⁷ m from Earth' also lets us calculate the mass of the Sun or other celestial bodies by measuring the orbital parameters of their satellites.
地球同步卫星是一个特例：其周期 T = 24 小时，且轨道位于赤道平面内。将 T 和地球质量 Mₑ 代入半径表达式，可求得一个固定的轨道半径，约为 4.23 × 10⁷ m（从地心算起）。此类卫星相对于地面静止，广泛用于通信。利用引力与向心力的联系，我们还可以通过测量卫星的轨道参数来计算太阳或其他天体的质量。
In all these applications, the key principle remains the same: identify the real force directed towards the centre, equate it to m v²/r (or m ω² r), and solve for the unknown quantity. Practising a wide range of problems — from conical pendulums to banked curves and satellite motion — will build the fluency required for the CIE A-Level examination. Always draw a clear free-body diagram and label the radial direction before writing equations.
在所有这些应用中，基本原理始终不变：找出真实存在的指向圆心的力，令其等于 m v²/r（或 m ω² r），然后解出未知量。通过大量练习——从圆锥摆到倾斜弯道再到卫星运动——你将在 CIE A-Level 考试中游刃有余。在列方程之前，务必画出清晰的受力图并标出径向方向。
Published by TutorHao | Physics Revision Series | ale
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