Calculating Drag (having circular reference problems)

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Homework Help Overview

The discussion revolves around calculating drag in a physics experiment involving a square open-sided cube dropped through a vertical distance. The original poster is attempting to understand the relationship between drag, velocity, and displacement, particularly in the context of reaching terminal velocity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for instantaneous velocity to calculate drag and the circular reference issue that arises when trying to find displacement without knowing velocity. There are attempts to apply kinematic equations to find velocity and questions about the implications of negative values in calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the equations involved. Some have provided calculations and insights into the relationship between forces acting on the cube, while others are questioning the assumptions made in the calculations.

Contextual Notes

Participants note the specific parameters of the experiment, including the dimensions of the cube, its mass, and the conditions under which the drag is being calculated. There is an acknowledgment of the potential for reaching terminal velocity and the implications of the calculations leading to negative values.

josiahseto
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Homework Statement



I don't know if it's possible but I am doing a physics EEI on drag.
I know to find drag you need instantaneous velocity which can be found through (change in x/ change in t) but i cannot find displacement unless i find velocity :S..

Homework Equations


d = 0.5 x p x (V^2) x Cd x A
im taking p = 1.2
Cd = 1.05 (for square shape)
A = 0.0225 sq meters

The Attempt at a Solution


I have dropped a square open sided cube 15cm x 15cm through a vertical distance of 2.5m (u = 0m/s) and it took an average of 2.33s
 
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also, its mass is 0.0029 kg
so w = 0.02842N

therefore at terminal velocity point drag = 0.02842N (thats if it does reach TV)

therefore at that point acceleration = 0

so.. we have

u = 0m/s

v = 1.41m/s (calculated through Fd = Fg)

and a = 0m/s at TV point
 
the velocity can be found using v2=u2-2gy
 
v^2 = 0^2 - 2 x 9.8 x 2.5
V^2 = -49
v = sqrt -49 = impossible?
 
josiahseto said:
v^2 = 0^2 - 2 x 9.8 x 2.5
V^2 = -49
v = sqrt -49 = impossible?

Sorry, down is positive, so it would be v^2=0^2+2*9.81*2.5.
 

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