What is the relationship between velocity and air resistance?

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Homework Help Overview

The discussion revolves around modeling air resistance for an object in free fall, specifically focusing on the relationship between velocity and air resistance, which is stated to be proportional to the square of the velocity. Participants are exploring the implications of this relationship in the context of energy conservation and forces acting on the object.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive equations that incorporate air resistance into the motion of a falling object. There are discussions on using conservation of energy versus applying Newton's second law to analyze the forces involved. Some participants question the validity of the original poster's approach and suggest solving a differential equation instead.

Discussion Status

The discussion is active, with participants providing corrections and alternative perspectives on the modeling of air resistance. There is a recognition that air resistance alters the dynamics of the falling object, and some guidance has been offered regarding the appropriate equations to use. Multiple interpretations of the problem are being explored without a clear consensus.

Contextual Notes

Participants note that air resistance is not a conservative force, which complicates the conservation of energy approach. There are also distinctions made regarding the behavior of air resistance for different weights of objects, suggesting that assumptions about the model may need to be reconsidered.

finchie_88
My maths teacher before we broke up for christmas asked the class to work out a simple model for air resistance for an object that is dropped given that the resistive force is proportional to the square of the velocity. I've spent about 10 minutes thinking about it, and this is what I came up with. (I hav'ent read anything on the matter, so any suggestions, corrections, improvements etc would be kind)

Since without air resistance, 0.5mv^2 = mg(h0 - h1)
(h0 > h1)
therefore with air resistance that is proportional to the square of the velocity, that means that 0.5mv^2 + mg(h1 - h0) + k(h0 - h1)v^2 = 0
Therefore, v^2 = (mg(h0 - h1)) / (0.5m + k(h0 - h1))

I tried this when m = 5 kg, h0 - h1 = 1 metre, k = 1.5, so if no air resistance, v = sqrt(19.6), but if air resistance is taken into account, then v = 3.5 m/s.
This makes sense, but ideas would be helpful.
 
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Air resistance is proportional to the square of the velocity, or R = kv². That all. There is no h0 or h1 involved.

The air resistance is a force which reduces the accelerating force on a moving body.

If a body is falling through the air, then the forces acting on it are 1: its weight = mg, acting downwards, and 2: the air resistiance R = kv², acting up, opposite to the direction of motion.

So, the effective accelerating force is: F = mg - R.
From Newton's 2nd law,

F = ma
mg - kv² = ma
or,
mg - kv² = mdv/dt
==============

Edit: corrected the typo. Thanks gamma.
 
Last edited:
therefore with air resistance that is proportional to the square of the velocity, that means that 0.5mv^2 + mg(h1 - h0) + k(h0 - h1)v^2 = 0


I see why you wrote the above. You tried to use the conservation of energy. However, the problem is the air resistance is propotional to the velocity and velocity is not constant. Rather it increases with ((h1 -h0). So your equation is incorrect. Right way is to do what Fermat has suggested. You have to solve the differential equation.

There is a typo in Fermat's posting. DE should be

mg - kv² = m dv/dt


regards,

Gamma.
=============
 
To phrase what Gamma just said in a different way, air resistance is not a "conservative force" and so total kinetic and potential energy is not conserved. Better to set mass times acceleration equal to total force- the equation Gamma gave. By the way, for small, relatively light, objects, air resistance is proportional to the square of speed. For heavier objects it is proportional to speed itself.
 

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