How do I find the terminal speed of a body in water?

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Homework Help Overview

The discussion revolves around determining the terminal speed of a baseball in water, given its terminal speed in air. Participants are exploring the implications of different densities and the role of buoyancy in the calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to relate the terminal velocities in air and water, questioning the assumptions made in the problem. Some suggest working symbolically rather than numerically, while others express confusion about the reasoning behind the ratios being discussed.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem analytically. There is a recognition of the need to clarify assumptions and the importance of understanding the relationships between the variables involved.

Contextual Notes

Some participants note that the problem may lack sufficient information, such as the density of the baseball, which could affect the calculations. There is also mention of potential misconceptions in the problem's construction.

Ushitha Dissanayake
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Homework Statement


A baseball has a terminal speed of 42 m/s in air (ρ = 1.2 kg/m^3). What would be its terminal speed in water (ρ = 1.0 x 10^3 kg/m^3)?
A) 0.05 m/s
B) 1.5 m/s
C) 18 m/s
D) 42 m/s
E) 1200 m/s

Homework Equations


D=½Cρ*Av^2
where the magnitude of the drag force (D), relative speed = v, drag co-efficient = C, ρ is the air density (mass per volume) and A is the effective cross-sectional area of the body (the area of a cross section taken perpendicular to the
velocity )

v_t=√2*F_g/Cρ*A
where: v_t is the terminal speed, F_g is force gravity, and C, ρ and A is the same as above

The Attempt at a Solution


Im not sure how to solve this at all, the example that my book gave me involved finding terminal speed by giving me the drag co-efficient as well as the R to find the volume and area for a sphere
 
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I believe this is a badly constructed problem because the obvious way to "solve" it (i.e., comparing the force of gravity on the ball with the drag force of water) is wrong and the proper way of solving it (also taking the buoyancy of the ball into account) requires more information than what is given in the problem (i.e., the density of the ball itself).

Either way, let us solve it the wrong way, which I suspect is what the problem constructor intended. Work symbolically, i.e., do not look for numerical values of middle steps until you reach a final expression. What would be the relation between the terminal velocities in air and water, respectively?
 
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I'm not following
 
Ushitha Dissanayake said:
I'm not following
This is not helpful. Exactly what are you not following?
 
I don't really understand what you said
 
Still not helpful. You need to be specific in what it was you did not understand and why.
 
"What would be the relation between the terminal velocities in air and water, respectively?" I don't understand what you're implying here
 
What is the terminal velocities in water and air, respectively? What is the ratio between them? You have already given an expression for the terminal velocity so this should not be hard.
 
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i tried finding a ratio by dividing (ρ = 1.2 kg/m^3) by (ρ = 1.0 x 10^3 kg/m^3) which gave 1.2*10^-3, then multiplying that with the terminal velocity of it in the air which gave 0.0504/ but the answer is 1.5
 
  • #10
You are not listening. You cannot just go around randomly constructing ratios and expect other ratios to behave the same way. I asked you to find a very specific ratio by dividing the two expressions you have for the terminal velocity in air and water, respectively.
 
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  • #11
can you explain why we have to do it this way i still don't understand how this ratio thing works
 
  • #12
A ratio is one quantity divided by another. Please use the analytical expressions you have, do not insert numbers. What do you get? It will be much clearer to you if you do it.
 
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  • #13
thanks for ur explanation
 
  • #14
This still does not show that you have gotten the problem correctly. Please write out your solution. Also, as already mentioned, the ”correct” solution relies on false assumptions and we have not yet begun to discuss that. You will not learn properly if you are happy just getting the ”right” answer in these cases. You need to examine the problem assumptions.
 
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