Terminal Velocity's Relationship to Mass

  • Thread starter Thread starter bistan
  • Start date Start date
  • Tags Tags
    Mass Relationship
AI Thread Summary
Terminal velocity increases with mass due to the relationship between weight and drag force, as described by Newton's laws. When an object reaches terminal velocity, the forces acting on it, including weight and drag, balance out to zero. Heavier objects experience greater drag forces, which results in higher terminal velocities. The drag force is proportional to the square of the terminal velocity, reinforcing that greater mass leads to greater terminal velocity. The discussion also highlights the need for constants like fluid density, drag coefficient, and cross-sectional area to remain unchanged for accurate comparisons.
bistan
Messages
11
Reaction score
0

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.


 
Last edited:
Physics news on Phys.org
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?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top