How to determine air-resitance for a dropped coin

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Discussion Overview

The discussion revolves around determining air resistance for a coin dropped from a height of 2 meters onto a concrete floor. Participants explore the implications of air resistance in this scenario, considering both theoretical and practical aspects.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant suggests that the terminal velocity equation could be used to find the drag coefficient, referencing a Mythbusters episode that measured terminal velocity.
  • Another participant notes that when dropped from 2 meters, air resistance will have an insignificant effect, as the coin reaches the ground quickly at approximately 6.5 m/s, while terminal velocity is much higher, around 40 m/s.
  • A different viewpoint highlights the difficulty in measuring air resistance due to the coin's flipping motion during the drop, which affects the orientation and thus the resistance experienced.

Areas of Agreement / Disagreement

Participants express differing views on the significance of air resistance in this scenario. While some argue it is negligible, others emphasize the complexities introduced by the coin's orientation during the drop.

Contextual Notes

Participants do not resolve the implications of the coin's flipping motion on air resistance, nor do they clarify the assumptions behind their calculations or references.

16physicist
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Can anyone tell me how I can determine air-resitance for a coin being dropped from 2 meters onto a concrete floor?

Thanks a lot.

-Arash Ay
 
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Depends on what, exactly, you are looking do figure out. Mythbusters did an episode on this, iirc, where they actually measured the terminal velocity, so you could take that and plug it into the terminal velocity equation to find the drag coefficient.

Realistically, though, when dropped from 2 meters, drag will play an insignificant role in what happens. The coin hits the ground in about 2/3 of a second, traveling at 6.5 m/s. The terminal velocity, on the other hand, is probably somwhere around 40 m/s.
 
This would be very difficult because when the coin is dropped it will flip over and over so the same surface will not always be facing the ground. When the flat side is down the resistance will be more than when the thin edge is facing down. But like Russ said it would be so small and would not make much difference.
 
Okay, I apperciate that, thank you.
 

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