IB Physics: Circular Motion Force Analysis and Critical Problems | IB物理：圆周运动受力分析与临界问题

📚 IB Physics: Circular Motion Force Analysis and Critical Problems | IB物理：圆周运动受力分析与临界问题

In IB Physics, circular motion is a rich application of Newton’s laws. Students must identify the real forces acting on an object, resolve them along the radial direction, and recognise when the required centripetal force exceeds what the forces can supply. This article gives a complete framework for force analysis in uniform circular motion and solves the critical speed problems that often appear in IB Paper 2 and internal assessments.

在IB物理中，圆周运动是牛顿定律的综合应用。学生需要找出物体受到的真实力，沿径向进行分解，并判断当所需向心力超过系统能提供的最大值时会出现什么情况。本文为匀速圆周运动的受力分析提供完整框架，并系统讲解IB Paper 2和内部评估中常见的临界速度问题。

1. Key Concepts: Angular Velocity and Centripetal Acceleration | 关键概念：角速度与向心加速度

For an object moving in a circle of radius r, the linear speed v and angular velocity ω are related by v = ωr. The angular velocity is also related to the period T and frequency f by ω = 2π/T = 2πf.

对于半径为r的圆周运动，线速度v与角速度ω满足v = ωr。角速度还与周期T和频率f有关：ω = 2π/T = 2πf。

Even when the speed is constant, the direction of the velocity changes continuously. This change in direction corresponds to a centripetal acceleration directed toward the centre of the circle.

即使速度大小不变，速度方向也在连续改变。这种方向变化对应着指向圆心的向心加速度。

a_c = v² / r = ω² r

The centripetal acceleration is never zero in circular motion, because the velocity vector is never parallel to the acceleration vector except in the limiting radial sense of the change in direction.

在圆周运动中向心加速度永远不会为零，因为速度方向始终在改变，而向心加速度正是描述这种方向变化快慢的物理量。

2. Centripetal Force Is a Net Force | 向心力是合力

Centripetal force is not a new physical force. It is the name given to the net force component that points toward the centre of the circle. Newton’s second law along the radial direction gives:

向心力并不是一种新的相互作用力，而是指指向圆心的合力分量。沿径向应用牛顿第二定律得到：

F_c = m a_c = m v² / r = m ω² r

This equation is always true for uniform circular motion, but it does not tell us which real force provides the centripetal force. In every question, you must first draw a free-body diagram and then determine whether the real forces can combine to produce the required F_c.

这个方程对匀速圆周运动总是成立，但它并不告诉我们哪一个真实力提供向心力。做每道题时，必须先画受力分析图，再判断真实力能否合成为所需的F_c。

A very common error is to add “centrifugal force” as an extra force on the diagram. Centrifugal force is an inertial pseudoforce that appears only in a rotating reference frame; in an inertial frame it does not exist.

一个非常常见的错误是在受力图中额外加上“离心力”。离心力只在转动参考系中才作为惯性力出现；在惯性参考系中它并不存在。

3. Sources of Centripetal Force | 向心力的常见来源

Different physical situations use different real forces as the centripetal force. A single force may act alone, or several forces may combine to point toward the centre.

不同物理情境中，充当向心力的真实力各不相同。可以是某个力单独起作用，也可以是几个力合成后指向圆心。

Situation	Centripetal force source
Car turning on a flat road	Static friction between tyres and road
Satellite orbiting a planet	Gravitational attraction
Ball on a horizontal string	Tension in the string
Cyclist on a banked track	Horizontal component of normal force and friction

When identifying the source, always ask: which real force or combination of real forces has a component toward the centre? The answer is the centripetal force.

在判断来源时，要问：哪一个真实力或哪些真实力的合力具有指向圆心的分量？这个答案就是向心力。

4. Horizontal Circular Motion: Tension and Friction | 水平圆周运动：拉力与摩擦力

Consider a mass attached to a string moving on a horizontal frictionless table in a circle of radius r. Vertically, the normal force N balances the weight mg, so N = mg. Horizontally, the only force is the tension T, which points toward the centre.

考虑一个系在绳子上的物体在水平光滑桌面上做半径为r的圆周运动。竖直方向支持力N与重力mg平衡，因此N = mg。水平方向唯一受到的力是拉力T，方向指向圆心。

T = m v² / r

If the same object is placed on a rotating turntable, the static friction f_s provides the centripetal force. Static friction can increase up to a maximum value f_s,max = μ_s N = μ_s mg.

如果同一个物体放在旋转转盘上，则由静摩擦力f_s提供向心力。静摩擦力最大只能达到f_s,max = μ_s N = μ_s mg。

m v² / r ≤ μ_s m g

This inequality immediately gives the maximum speed before slipping. You will see the same structure in many critical speed problems.

这个不等式直接给出了物体开始滑动前的最大速度。很多临界速度问题都具有相同结构。

5. Vertical Circular Motion: Top and Bottom Forces | 竖直圆周运动：最高点与最低点受力

A mass attached to a string moving in a vertical circle has a changing speed: it slows down going up and speeds up going down. The forces are still radial, but gravity has a different direction relative to the radius at each position.

用绳子系住物体在竖直平面内做圆周运动时，速度大小不断变化：上升时减速，下降时加速。受力仍然沿径向，但重力在每一位置与半径方向的关系不同。

At the highest point, tension T and gravitational force mg both point downward, so Newton’s second law gives:

在最高点，拉力T和重力mg都向下，因此牛顿第二定律给出：

T_top + m g = m v_top² / r

At the lowest point, tension points upward while gravity points downward, so:

在最低点，拉力向上，重力向下，因此：

T_bottom − m g = m v_bottom² / r

Notice that the tension at the bottom is always greater than at the top for the same mass and radius, because it must do more work against gravity and provide the centripetal force.

注意到对于同一质量和半径，最低点的拉力总是大于最高点，因为它既要克服重力，又要提供向心力。

6. Critical Speed at the Highest Point | 最高点的临界速度

For a ball attached to a string, tension cannot push; it can only pull. If the required centripetal force at the top is larger than mg, the string must pull downward. If it is smaller, gravity alone is too large, and the string would go slack before the ball reaches the top.

对于绳子系住的小球，拉力只能拉，不能推。如果最高点所需的向心力大于mg，绳子需要向下拉；如果所需向心力小于mg，重力本身过大，小球还没到达最高点绳子就会松弛。

The critical condition is that tension becomes exactly zero at the highest point. Then gravity alone provides the centripetal force:

临界条件是在最高点拉力恰好为零。此时重力单独提供向心力：

m g = m v_c² / r ⇒ v_c = √(g r)

This is the minimum speed at the top for a string or for any object that can only be pulled inward while in contact with the track, such as a ball rolling inside an inverted U-shaped track without a retaining rail.

这是绳子模型或只能被向内拉而不能被向外推的轨道模型在最高点的最小速度，例如没有内侧导轨、小球沿外轨内侧滚动的情况。

If the mass is attached to a rigid rod, the rod can push outward at the top. In that case, the minimum speed at the top can be zero, because the rod can support the ball against gravity.

如果物体连接在一根刚性杆上，杆在最高点可以向外推。此时最高点最小速度可以为零，因为杆可以支撑物体对抗重力。

7. Energy and the Minimum Speed at the Bottom | 能量关系与最低点所需最小速度

To complete a full vertical circle with a string, the speed at the bottom must be large enough to still leave enough speed at the top. Using conservation of mechanical energy between the lowest point and the highest point, with height difference 2r:

要用绳子完成整个竖直圆周运动，最低点的速度必须足够大，以保证到达最高点时仍有足够速度。利用最低点和最高点之间的机械能守恒，高度差为2r：

½ m v_bottom² = ½ m v_top² + m g (2r)

At the critical limit v_top = √(gr), so:

在临界极限下v_top = √(gr)，因此：

v_bottom = √(v_top² + 4 g r) = √(g r + 4 g r) = √(5 g r)

This famous result, v_bottom = √(5gr), is often tested in loop-the-loop questions. It shows that the bottom speed must be about 2.24 times the minimum top speed for a taut-string vertical circle.

这个著名结论v_bottom = √(5gr)经常出现在“竖直圆环”问题中。它说明在绳子绷紧的竖直圆周运动中，最低点速度大约是最高点最小速度的2.24倍。

8. Conical Pendulum | 圆锥摆

A conical pendulum is a mass moving in a horizontal circle while the string traces out a cone. The string makes an angle θ with the vertical. The mass moves with constant speed but its acceleration is horizontal, directed toward the centre of the horizontal circle.

圆锥摆是指小球在水平面内做圆周运动，同时绳子扫过一个圆锥面。绳子与竖直方向夹角为θ。小球速度大小不变，但加速度水平指向水平圆的圆心。

The vertical component of tension balances gravity, while the horizontal component provides the centripetal force:

拉力的竖直分量与重力平衡，水平分量提供向心力：

T cos θ = m g
T sin θ = m v² / r

Dividing the second equation by the first gives a direct expression for the angle:

将第二个方程除以第一个方程，得到夹角的正切表达式：

tan θ = v² / (r g)

Using r = L sin θ, where L is the string length, the angular velocity satisfies ω² = g/(L cos θ). As θ increases, the pendulum must spin faster.

利用r = L sin θ（L为绳长），可得角速度满足ω² = g/(L cos θ)。θ越大，圆锥摆转动越快。

9. Banked Curves | 倾斜弯道

On an ideally banked curve, no friction is needed if the road is inclined at the correct angle for a particular speed. The normal force N is inclined, so it has both vertical and horizontal components.

在理想倾斜弯道上，如果路面倾斜角与速度相匹配，就不需要摩擦力。支持力N是倾斜的，因此同时具有竖直和水平分量。

N cos θ = m g
N sin θ = m v² / r

Dividing the radial equation by the vertical equation gives:

用径向方程除以竖直方程得到：

tan θ = v² / (r g)

This equation is identical in form to the conical pendulum result. It gives the ideal banking angle for a given speed and radius. If the actual speed is higher, additional static friction is required; if lower, friction acts up the slope to prevent sliding down.

这个方程在形式上与圆锥摆相同。它给出了给定速度和半径下的理想倾斜角。如果实际速度更大，则需要额外的静摩擦力；如果实际速度更小，摩擦力会沿斜面向上，防止物体向下滑动。

10. Flat Curve with Friction | 平路弯道与摩擦力

On a flat curve, friction is the only horizontal force available to turn the car. The maximum centripetal force is limited by the maximum static friction:

在平路弯道上，水平方向只有摩擦力能让汽车转弯。最大向心力受最大静摩擦力限制：

f_s,max = μ_s m g

For a car of mass m moving at speed v on a turn of radius r, the requirement is:

对于质量m、速度v、转弯半径r的汽车，要求为：

m v² / r ≤ μ_s m g

Thus the maximum safe speed is:

因此最大安全速度为：

v_max = √(μ_s g r)

If the car exceeds this speed, it will skid outward because static friction is no longer large enough. On ice, μ_s is much smaller, so v_max decreases dramatically.

如果汽车超过这个速度，就会因为静摩擦力不足而向外侧滑。在冰面上μ_s大大减小，因此v_max也会急剧降低。

11. Problem-Solving Strategy | 解题策略

Follow these steps for any circular motion problem in IB Physics:

在IB物理中解决圆周运动问题时，请按以下步骤操作：

Identify the object and draw a full free-body diagram showing all real forces.

确定研究对象并画出完整受力图，标出所有真实力。
Choose the radial direction as pointing toward the centre of the circle, and the perpendicular direction vertically or tangentially as appropriate.

选择指向圆心的方向为径向，必要时选择竖直或切向为垂直方向。
Resolve forces along the radial direction and set the net radial force equal to mv²/r.

沿径向分解力，并令径向合力等于mv²/r。
Identify the physical limit: tension cannot be negative, normal force cannot be negative, static friction has a maximum value.

找出物理极限：拉力不能为负，支持力不能为负，静摩擦力存在最大值。
If v at different heights is involved, use conservation of mechanical energy to connect the speeds.

如果涉及不同高度的速度，使用机械能守恒来联系各位置的速度。

This procedure changes a seemingly complicated problem into a series of simple algebraic steps.

这个方法可以把看似复杂的问题简化成一系列代数步骤。

12. Worked Example and Common Mistakes | 例题精讲与常见错误

Worked example: A small block slides from rest along a frictionless track and enters a vertical loop of radius R. Find the minimum release height h required for the block to just complete the loop without leaving the track.

例题：一个小滑块从静止开始沿光滑轨道滑下，进入半径为R的竖直圆环。求滑块刚好能完成整个圆环而不脱离轨道时的最小释放高度h。

At the top of the loop, the critical condition is that the normal force N from the track is zero. Gravity alone provides the centripetal force:

在圆环最高点，临界条件是轨道对滑块的支持力N为零。这时重力单独提供向心力：

m g = m v_top² / R ⇒ v_top² = g R

Using energy conservation from the release point to the top of the loop, where the block has risen a vertical distance 2R:

从释放点到最高点应用能量守恒，释放点到最高点的竖直高度为2R：

m g h = m g (2R) + ½ m v_top²

m g h = 2 m g R +

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IB Physics: Circular Motion Force Analysis and Critical Problems | IB物理：圆周运动受力分析与临界问题