📚 GCSE OCR Physics: Circular Motion – Essential Revision | GCSE OCR 物理：圆周运动考点精讲
In your GCSE OCR Physics course, circular motion appears in topics covering forces, motion and even space physics. Understanding why an object moving in a circle at constant speed still accelerates, what provides the centripetal force, and how to use key equations such as v = rω, a = v²/r, and F = mv²/r is essential for top marks. This revision guide breaks down every concept clearly, from velocity vectors to satellite orbits, helping you gain confidence for your exams.
在 GCSE OCR 物理课程中，圆周运动出现在力、运动乃至空间物理等主题中。理解为何物体以恒定速率沿圆周运动依然有加速度、什么提供向心力，以及如何使用核心公式如 v = rω、a = v²/r 和 F = mv²/r，对取得高分至关重要。这篇考点精讲将一步步剖析每个概念，从速度矢量到卫星轨道，帮助你自信应对考试。
1. What is Circular Motion? | 什么是圆周运动？
Circular motion describes an object travelling along a circular path. In uniform circular motion, the object moves with constant speed. However, because the direction of the object’s velocity is continuously changing, its velocity is not constant. Velocity is a vector quantity, so both magnitude (speed) and direction matter. Therefore, any object in uniform circular motion is accelerating, even if its speed does not change.
圆周运动描述的是物体沿圆周路径运动。在匀速圆周运动中，物体的速率保持不变。但是，由于速度方向不断变化，速度并不恒定。速度是矢量，既有大小（速率）也有方向。因此，任何做匀速圆周运动的物体都在加速，即使它的速率没有变化。
The acceleration is directed towards the centre of the circle. This is called centripetal acceleration. The word ‘centripetal’ means ‘centre-seeking’. The object’s inertia tends to keep it moving in a straight line, so a resultant force (centripetal force) must pull it inward to maintain the circular path.
加速度的方向指向圆心，这称为向心加速度。’向心’一词意为’指向中心’。物体的惯性使其趋向直线运动，因此必须有一个指向圆心的合力（向心力）不断将其向内拉，才能维持圆周路径。
2. Velocity Vector and Direction Change | 速度矢量与方向变化
At any instant, the velocity vector of an object in circular motion is tangent to the circle. As the object moves from one point to another, the direction of this tangent changes. Even if the arrow length (speed) stays the same, the velocity changes because its direction changes. This is why we say there is an acceleration.
在任意时刻，圆周运动中物体的速度矢量与圆周相切。当物体从一点移到另一点时，这条切线的方向发生了改变。即使箭头的长度（速率）不变，由于方向变化，速度也发生了变化。这就是为何我们说存在加速度。
Compare this with straight-line motion at constant speed: there the velocity direction is fixed, so no acceleration. For circular motion, a net force must act perpendicular to the velocity to keep changing its direction without altering its magnitude.
将此与匀速直线运动比较：后者速度方向固定，因此没有加速度。而在圆周运动中，必须有一个垂直于速度的合力不断改变速度方向，同时不改变其大小。
3. Centripetal Force – The Inward Pull | 向心力——指向圆心的拉力
According to Newton’s second law, a resultant force is needed to cause any acceleration (F = ma). In circular motion, the centripetal force is this resultant force, always directed towards the centre of the circle. The centripetal force is not a new type of force; it is simply the name we give to whichever force is pulling or pushing the object inward.
根据牛顿第二定律，任何加速度都需要合力（F = ma）。在圆周运动中，向心力就是这个合力，始终指向圆心。向心力不是一种新的力，它只是我们对任何将物体向圆心拉或推的力的称呼。
A common misconception is that there is an outward ‘centrifugal’ force. In reality, if the centripetal force is removed (e.g., a string breaks), the object flies off along a tangent due to its inertia — not because an outward force pushed it. The sensation of being pushed outward in a turning car is your body’s inertia trying to continue in a straight line while the car turns inward.
一个常见的误解是存在向外的’离心力’。实际上，如果向心力消失（例如绳子断开），物体会由于惯性沿切线方向飞出——而不是因为有一个向外的力推它。乘车转弯时感觉自己被向外推，其实是身体惯性试图保持直线运动，而车向内转弯的结果。
4. Examples of Centripetal Force | 向心力的实例
Different situations provide centripetal force from different sources. Identifying the correct force is essential for exam questions.
不同情境下，向心力由不同的力提供。正确识别这些力是考试解题的关键。
- Ball on a string: The tension in the string acts as the centripetal force. If the string breaks, the ball moves off in a straight line tangent to the circle at the point of release.
- 绳子上的小球： 绳的拉力充当向心力。如果绳子断裂，小球会从断开点沿圆的切线方向直线飞出。
- Car turning a corner: Friction between the car’s tyres and the road surface provides the centripetal force. On an icy road, friction is reduced, so the car may skid outward.
- 汽车转弯： 轮胎与路面之间的摩擦力提供向心力。在结冰路面上，摩擦力减小，汽车可能向外侧滑。
- Planet orbiting the Sun: The gravitational pull of the Sun on the planet acts as the centripetal force. The planet’s tangential velocity keeps it from falling directly into the Sun.
- 行星绕太阳公转： 太阳对行星的万有引力充当向心力。行星的切线速度使其不至于直接落入太阳。
- Fairground rides (e.g., rotor or ‘spinning drum’): The normal force from the wall of the drum acts as the centripetal force, pushing the rider towards the centre.
- 游乐场旋转项目（如转筒）： 转筒内壁提供的支持力指向圆心，充当向心力。
5. Period and Frequency | 周期与频率
The period, T, is the time taken to complete one full revolution. It is measured in seconds (s). Frequency, f, is the number of complete revolutions per second, measured in hertz (Hz). They are inversely related.
周期 T 是完成一圈所需的时间，单位是秒（s）。频率 f 是每秒完成的圈数，单位是赫兹（Hz）。两者互为倒数。
f = 1 / T and T = 1 / f
For an object rotating with a period of 0.5 s, the frequency is 2 Hz. Understanding this relationship is vital because angular velocity depends directly on frequency. Many exam questions ask you to convert between period, frequency and angular speed.
若一个物体的旋转周期为 0.5 秒，频率就是 2 赫兹。理解这层关系至关重要，因为角速度直接取决于频率。许多考题要求你在周期、频率和角速度之间进行转换。
6. Angular Velocity (ω) | 角速度（ω）
Angular velocity, symbol ω (omega), measures how quickly an object rotates. It is defined as the rate of change of angular displacement (angle swept out per unit time). The SI unit is radian per second (rad/s).
角速度符号 ω（omega），衡量物体转动的快慢。它定义为角位移的变化率（单位时间内扫过的角度）。SI 单位是弧度每秒（rad/s）。
One full circle corresponds to an angle of 2π radians. If an object completes a full circle in time T, its angular velocity is:
一个完整的圆对应 2π 弧度。如果一个物体在时间 T 内完成一整圈，其角速度为：
ω = 2π / T = 2π f
So if an object rotates twice per second (f = 2 Hz), then ω = 2π × 2 = 4π rad/s ≈ 12.6 rad/s. This quantity is the same for all points on a rigid rotating disc, regardless of how far they are from the centre.
因此，如果一个物体每秒转两圈（f = 2 Hz），那么 ω = 2π × 2 = 4π rad/s ≈ 12.6 rad/s。对于刚性转盘上的所有点，无论离中心多远，这个角速度都是相同的。
7. Relationship Between Linear and Angular Velocity | 线速度与角速度的关系
Linear velocity (v) and angular velocity (ω) are connected by the radius (r) of the circular path. The linear speed of a point on a rotating object equals the product of its angular speed and the distance from the centre.
线速度（v）和角速度（ω）通过圆周运动的半径（r）关联起来。旋转物体上某点的线速度等于其角速度与到中心的距离的乘积。
v = r ω
This makes sense: if you double the radius while keeping the same angular speed, the point has to travel a larger circumference in the same time, so its linear speed doubles. For example, a point on a spinning CD near the edge moves faster than a point near the centre, even though the disc’s angular speed is the same everywhere.
这很合理：如果你在保持相同角速度的同时将半径加倍，该点需要在相同时间内走过更大的圆周，因此线速度也加倍。例如，旋转 CD 上靠近边缘的点比靠近中心的点移动得更快，尽管盘的角速度处处相同。
8. Centripetal Acceleration | 向心加速度
An object in uniform circular motion accelerates towards the centre. The magnitude of this centripetal acceleration (a) depends on the linear speed and the radius. It can also be expressed in terms of angular velocity.
匀速圆周运动的物体具有指向圆心的加速度。这个向心加速度（a）的大小取决于线速度和半径。也可以用角速度表示。
a = v² / r and a = ω² r
These two expressions are equivalent because substituting v = rω gives a = (rω)²/r = ω²r. Notice that if you keep angular velocity constant, acceleration increases with radius. If you keep linear speed constant, acceleration is larger for a tighter (smaller radius) circle.
这两个表达式是等价的，因为代入 v = rω 可得 a = (rω)²/r = ω²r。注意，若保持角速度恒定，加速度随半径增大而增大；若保持线速度恒定，则轨道半径越小（越急的弯），加速度越大。
Worked example: A model car moves at 3.0 m/s around a circular track of radius 1.5 m. Its centripetal acceleration is a = (3.0)² / 1.5 = 9.0 / 1.5 = 6.0 m/s². Even though the speed is constant, the car experiences an acceleration of 6.0 m/s² towards the centre.
计算示例：一辆模型汽车以 3.0 m/s 的速度沿半径 1.5 m 的圆形轨道行驶。其向心加速度 a = (3.0)² / 1.5 = 9.0 / 1.5 = 6.0 m/s²。即使速率不变，汽车仍然承受着指向中心的 6.0 m/s² 的加速度。
9. Centripetal Force Equation | 向心力公式
From Newton’s second law (F = ma), we obtain the centripetal force required to keep an object of mass m moving in a circular path of radius r at speed v. Since a = v²/r, we have:
根据牛顿第二定律（F = ma），我们可以得出使质量为 m 的物体以速度 v 在半径为 r 的圆周运动所需的向心力。由于 a = v²/r，有：
F = m × a = m v² / r = m ω² r
The force required is directly proportional to the mass and to the square of the speed, and inversely proportional to the radius. This explains why taking a sharp bend at high speed requires much greater force. If friction or tension cannot provide this force, the object will fail to follow the circular path.
所需的向心力与质量以及速度的平方成正比，与半径成反比。这就解释了为何高速过急弯需要大得多的力。如果摩擦力或拉力不能提供这个力，物体将无法维持圆周路径。
Exam tip: Always check that you are using the correct mass and radius. The radius is the distance from the centre of the circle to the object’s centre of mass. Convert all quantities to SI units (kg, m, s). A common exam question provides the frequency or period and expects you to find v or ω first, then compute the force.
解题技巧：务必检查使用的是否为正确的质量和半径。半径是从圆心到物体质心的距离。将所有量转换为 SI 单位（kg、m、s）。常见考题会给出频率或周期，要求你先求出 v 或 ω，再计算力。
10. Circular Orbits of Satellites | 卫星的圆轨道
For a satellite in a circular orbit around Earth, the centripetal force is provided by the gravitational attraction between the Earth and the satellite. We can equate gravitational force to the required centripetal force:
对于绕地球做圆周运动的卫星，向心力由地球与卫星之间的万有引力提供。我们可以将引力与所需的向心力建立等量关系：
G M m / r² = m v² / r
Here G is the gravitational constant, M is the Earth’s mass, m is the satellite mass, and r is the orbital radius (distance from the centre of the Earth). Cancelling m and simplifying gives the orbital speed:
其中 G 是引力常量，M 是地球质量，m 是卫星质量，r 是轨道半径（距地心的距离）。消去 m 并化简，得到轨道速度：
v = √(G M / r)
This shows that for a given planet, the orbital speed depends only on the orbital radius. Satellites in lower orbits move faster. Geostationary satellites have an orbital period of 24 hours, matching Earth’s rotation, and their orbital radius must be about 42 000 km from Earth’s centre.
由此可见，对于给定行星，轨道速度只取决于轨道半径。低轨道卫星移动得更快。地球同步卫星的轨道周期为 24 小时，与地球自转同步，其轨道半径必须距地心约 42 000 公里。
11. Common Misconceptions and Exam Tips | 常见误解与解题技巧
Many students lose marks by misunderstanding centripetal force or velocity. Below is a
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