Would an Increase in Mass of a Sphere-Shaped Plastic Object Affect Its Terminal Velocity? Formula Derivation | 增加球形塑料物体的质量会影响其终端速度吗？公式推导

📚 Would an Increase in Mass of a Sphere-Shaped Plastic Object Affect Its Terminal Velocity? | 增加球形塑料物体的质量会影响其终端速度吗？

Terminal velocity is a classic topic in IB Physics, combining forces, motion, and fluid resistance. A common inquiry is how changing the mass of a spherical object, such as a plastic ball, influences the speed it ultimately reaches when falling through a fluid. In this article, we derive the relevant equations and analyse different scenarios where mass is increased.

终端速度是IB物理中的一个经典话题，结合了力、运动和流体阻力。一个常见的问题是，改变球形物体（如塑料球）的质量，会如何影响它在流体中下落时最终达到的速度。本文将推导相关公式，并分析增加质量的不同情形。

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1. Introduction to Terminal Velocity | 终端速度简介

When an object falls through a fluid, it experiences a drag force that grows with speed. Initially, the net force accelerates the object downward. As velocity rises, the resistive drag force increases until it balances the net driving force. At this equilibrium, acceleration ceases, and the object moves at a constant speed known as the terminal velocity.

当物体在流体中下落时，会受到随速度增大而增强的阻力。一开始，净力使物体向下加速。随着速度增加，阻力增大，直到与净驱动力平衡。达到平衡时加速度为零，物体以恒定速度运动，即终端速度。

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2. Forces Acting on a Falling Sphere | 作用在下落球体上的力

Consider a homogeneous sphere of radius r, density ρ_s, and mass m falling through a fluid of density ρ_f. Three vertical forces act on it: (i) Weight W = mg acting downwards; (ii) Buoyant force F_b = ρ_f V g acting upwards, where V is the volume of the sphere; (iii) Drag force F_d opposing the motion, which acts upwards when the sphere falls downwards.

考虑一个半径为 r、密度为 ρ_s、质量为 m 的均质球体，在密度为 ρ_f 的流体中下落。它受到三个竖直方向的力：（i）向下的重力 W = mg；（ii）向上的浮力 F_b = ρ_f V g，其中 V 为球体体积；（iii）阻碍运动的拖拽力 F_d，球体向下运动时该力向上。

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3. Drag Force and Drag Coefficient | 拖拽力与拖拽系数

For a sphere moving at moderate to high speeds, the drag force is well described by the quadratic drag equation: F_d = ½ C_d ρ_f A v², where C_d is the drag coefficient (about 0.47 for a smooth sphere), A = π r² is the cross-sectional area, and v is the instantaneous speed. At very low Reynolds numbers, Stokes’ law (F_d = 6π η r v) applies, but this article focuses on the quadratic regime typical of plastic balls falling in air.

对于以中高速度运动的球体，拖拽力可由二次阻力方程描述：F_d = ½ C_d ρ_f A v²，式中 C_d 为拖拽系数（光滑球体约为0.47），A = π r² 为截面积，v 为瞬时速度。在极低雷诺数下，应使用斯托克斯定律（F_d = 6π η r v），但本文聚焦塑料球在空气中下落的典型二次阻力区。

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4. Net Force and Equation of Motion | 净力与运动方程

Applying Newton’s second law in the vertical direction gives: m a = mg – F_b – F_d. Substituting F_b = ρ_f V g and F_d = ½ C_d ρ_f A v², we obtain m a = mg – ρ_f V g – ½ C_d ρ_f A v². At terminal velocity v_t, the acceleration a is zero, so the net force vanishes:

mg – ρ_f V g – ½ C_d ρ_f A v_t² = 0

沿竖直方向应用牛顿第二定律：m a = mg – F_b – F_d。代入浮力和拖拽力的表达式，得 m a = mg – ρ_f V g – ½ C_d ρ_f A v²。在终端速度 v_t 时，加速度 a 为零，净力为零：

mg – ρ_f V g – ½ C_d ρ_f A v_t² = 0

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5. Deriving Terminal Velocity Expression | 终端速度表达式推导

Rearrange the equilibrium condition to isolate v_t: ½ C_d ρ_f A v_t² = mg – ρ_f V g. The right-hand side is the net gravitational force (weight minus buoyancy). Substitute the sphere’s volume V = (4/3) π r³ and cross-sectional area A = π r²:

v_t² = (2 (mg – ρ_f (4/3)π r³ g)) / (ρ_f C_d π r²)

Simplify by cancelling π r² and expressing mass as m = ρ_s (4/3) π r³. After algebra, an elegant form emerges:

v_t = √[ (8/3) · (ρ_s – ρ_f)/ρ_f · (g r / C_d) ]

Equivalently, using the net weight W_net = mg – F_b:

v_t = √( 2 W_net / (ρ_f C_d A) )

整理平衡条件以解出 v_t：½ C_d ρ_f A v_t² = mg – ρ_f V g。右侧为净重力（重力减浮力）。代入球体体积 V = (4/3) π r³ 及截面积 A = π r²，可消去 π r²，并利用 m = ρ_s (4/3) π r³。经代数运算，得到简洁形式：

v_t = √[ (8/3) · (ρ_s – ρ_f)/ρ_f · (g r / C_d) ]

等价地，用净重 W_net = mg – F_b 表示为：

v_t = √( 2 W_net / (ρ_f C_d A) )

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6. Role of Mass in the Derived Formula | 质量在推导公式中的作用

Mass appears explicitly in the net weight term: a larger mass directly increases W_net, provided the buoyant force does not increase proportionally. However, mass is also linked to the sphere’s radius and density. The relationship m = ρ_s (4/3) π r³ shows that an increase in mass can result from either a larger radius (at constant density) or a greater density (at constant radius). Each case affects terminal velocity differently.

质量直接出现在净重项中：更大的质量会使 W_net 增大，前提是浮力没有成比例增加。但质量也与球体的半径和密度相关联。关系式 m = ρ_s (4/3) π r³ 表明，质量增加可能源于半径增大（密度不变）或密度增大（半径不变）。两种情形对终端速度的影响不尽相同。

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7. Case 1: Constant Volume, Increased Density | 情况1：体积不变，密度增加

Imagine taking the same plastic sphere but using a denser plastic formulation, so ρ_s increases while r remains fixed. Then V and A are unchanged, and the buoyant force F_b stays constant. The net weight W_net = (ρ_s – ρ_f) V g grows linearly with density. Since v_t ∝ √(ρ_s – ρ_f), the terminal velocity increases with the square root of the density excess. An increase in mass here unambiguously raises v_t.

设想使用相同的塑料球但采用更致密的塑料配方，使得 ρ_s 增大而 r 不变。此时 V 与 A 保持不变，浮力 F_b 恒定。净重 W_net = (ρ_s – ρ_f) V g 随密度线性增长。因 v_t ∝ √(ρ_s – ρ_f)，终端速度随密度差值的平方根增大。这种情况下质量的增加无疑会提高终端速度。

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8. Case 2: Constant Density, Increased Radius (Mass Increases) | 情况2：密度不变，半径增大（质量增加）

Now consider keeping the material the same (ρ_s constant) but using a larger ball. As radius r grows, mass increases as r³, volume increases, and the cross-sectional area A increases as r². From the derived formula v_t = √[ (8/3)(ρ_s – ρ_f) g r / (ρ_f C_d) ], we see that v_t ∝ √r. Expressing r in terms of mass: r ∝ m^(1/3), thus v_t ∝ m^(1/6). Therefore, a larger mass (due to a bigger ball) still increases terminal velocity, but the growth is slower—only the sixth root of mass. Doubling the radius makes the ball eight times heavier, yet terminal velocity rises only by a factor of about √2 ≈ 1.41.

现在保持材料不变（ρ_s 恒定），但使用更大的球。随着半径 r 增大，质量以 r³ 增加，体积增大，截面积 A 以 r² 增大。由推导公式 v_t = √[ (8/3)(ρ_s – ρ_f) g r / (ρ_f C_d) ] 可见，v_t ∝ √r。用质量表示半径：r ∝ m^(1/3)，于是 v_t ∝ m^(1/6)。因此，因球体变大而引起的质量增加仍会提高终端速度，但增长较缓——仅与质量的六次方根成正比。半径翻倍使球体重八倍，而终端速度仅增至约√2≈1.41倍。

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9. Additional Considerations: Buoyancy Negligence in Air | 附加考量：空气中的浮力可忽略

For a plastic sphere (density ~1000–1500 kg m⁻³) falling in air (density ~1.2 kg m⁻³), the buoyant force is less than 0.2% of the weight. Therefore, W_net ≈ mg, and the terminal velocity expression simplifies to v_t = √(2mg / (ρ_air C_d A)). In this approximation, mass appears directly inside the square root. If volume is fixed, v_t ∝ √m. If density is fixed, substituting A ∝ r² and m ∝ r³ yields v_t ∝ √(r³/r²) = √r ∝ m^(1/6) again, consistent with the full derivation.

对于在空气（密度约1.2 kg m⁻³）中下落的塑料球（密度约1000–1500 kg m⁻³），浮力不足重力的0.2%。因此 W_net ≈ mg，终端速度表达式简化为 v_{Published by TutorHao | IB Physics Revision Series | aleveler.com}

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