Terminal Velocity's Relationship to Mass

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SUMMARY

Terminal velocity increases linearly with mass, as demonstrated through the relationship between weight and drag force. According to Newton's second law, the forces acting on an object at terminal velocity balance out, leading to the equation mg = F_{D}. The drag force, defined by F_{D} = (1/2)ρv²C_{D}A, indicates that a heavier object experiences a greater drag force, resulting in a higher terminal velocity. This principle applies universally, regardless of the object's material composition, as long as the mass remains constant.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with drag force equations
  • Knowledge of mass density and its role in fluid dynamics
  • Basic grasp of terminal velocity concepts
NEXT STEPS
  • Study the implications of varying drag coefficients (C_{D}) on terminal velocity
  • Explore the effects of different fluid densities (ρ) on falling objects
  • Investigate the relationship between cross-sectional area (A) and terminal velocity
  • Examine real-world applications of terminal velocity in various materials
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Physics students, educators, and anyone interested in the dynamics of falling objects and fluid mechanics.

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Homework Statement


As shown in the plot above, terminal velocity is shown to increase linearly with the number of coffee filters dropped in a turbulent (air) medium. Therefore, terminal velocity depends on mass. Give an explanation for this starting from Newton's laws.

Homework Equations


\vec{F} = m\vec{a}
\vec{F}_{ab}=-\vec{F}_{ba}
Drag force F_{D}=\frac{1}{2}\rho v^{2}C_{D}A where \rho is the mass density of the fluid, v is the velocity of the object relative to the fluid, A is the reference (cross-sectional in our case) area of the object, and C_{D} is the drag coefficient of the object in the fluid.

The Attempt at a Solution


By Newton's second law: \sum F=\sum ma=mg-F_{D}=mg-\frac{1}{2}\rho v_{T}^{2}C_{D}A
By definition of terminal velocity: \sum F=0 \Rightarrow 0=mg-\frac{1}{2}\rho v_{T}^{2}C_{D}A

This means that the sum of the drag force on the object and the weight of the object must be 0 for the object to have reached terminal velocity. Therefore an object with greater mass has a greater weight, and so the drag force on that object must be equally greater to bring the sum of the forces to 0.

Since the drag force is a function of the square of the terminal velocity on the object, the terminal velocity is greater for an object with greater drag force, and therefore is greater for an object with greater mass.


 
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bistan said:
Since the drag force is a function of the square of the terminal velocity on the object, the terminal velocity is greater for an object with greater drag force, and therefore is greater for an object with greater mass.

What does that statement say about the difference between the terminal velocity of a large styrofoam ball v.s. a small metal ball, assuming they have the same mass?
 
phinds said:
What does that statement say about the difference between the terminal velocity of a large styrofoam ball v.s. a small metal ball, assuming they have the same mass?

If I understand correctly, I must also state that \rho, C_{D}, A must also remain constant?
 

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