Retarding Force proportional to the velocity falling particl

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Discussion Overview

The discussion revolves around a classical mechanics problem involving the motion of a particle in a medium with a retarding force proportional to its velocity. Participants explore the mathematical derivation of displacement and velocity, addressing the interpretation of forces and the manipulation of equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a mathematical derivation involving the equation F = m(dv/dt) = -mg - kmv and expresses confusion about the transformation of e^(-kt + kc) into ((k(Vo) + g)/k).
  • Another participant points out a potential sign error in the interpretation of forces, clarifying that mg is the gravitational force acting downward while kmv is the retarding force acting upward.
  • A different participant discusses the interpretation of the retarding force -kmv as a positive upward force when considering the direction of motion and velocity, suggesting that it should be treated as opposing gravity.
  • One participant suggests a clearer phrasing for the description of vertical motion and recommends a specific approach to manipulate the equation to clarify the derivation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of forces and the mathematical steps involved in the derivation. There is no consensus on the correct approach or interpretation of the equations presented.

Contextual Notes

Participants highlight ambiguities in the problem statement and the mathematical derivation, particularly regarding the treatment of forces and the integration constant.

Decypher
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Example problem from a Classical Mechanics book.
Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.

F = m(dv/dt) = -mg-kmv

dv/(kv + g) = -dt

(1/k)ln(kv + g) = -t + c

kv + g = e^(-kt+kc)

v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)

I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but I am not sure about where the other half of the exponent goes.
 
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Decypher said:
F = m(dv/dt) = -mg-kmv
Hi Decypher:

I think y ou have a sigh error. mg is the force of gravity pulling the mass towards rthe Earth, and kmv is a retarding force, slowing down the pull. The direction of these forces are opposite.

Hope this is helpful.

Regards,
Buzz
 
I spent a great amount of time on this part you are mentioning. The book does not provide a great description on this statement.
"Where -kmv represents a positive upward force since we take z and v = z(dot) to be positive upward, and the motion is downward-that is, v<0, so that -kmv > 0."
I think they are saying that because we are treating a positive velocity as one that is going higher on the z axis, whe have to treat -kmv as kmv because when a negative velocity is plugged in, kmv becomes an opposing force once again. It makes more sense to me to have it be g-kv just like a force opposing gravity should be.
 
Decypher said:
Find the displacement and velocity of a particle undergoing vertical motion
Decypher said:
we are treating a positive velocity as one that is going higher on the z axis
Hi Decypher:

OK. I think the first quoted sentence would have been clearer if the word "upward" had preceded "vertical".

Decypher said:
I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
The above quote is not what you want to do. I suggest you take the equation
Decypher said:
kv + g = e^(-kt+kc)
and fill in the missing step: subtract g from both sides, and divide both sides by k. Then make the substitution v = dz/dt.

Then you should be able to make a substitution for c (which is an arbitrary constant of integration) which will give you what you want.

Hope this helps.

Regards,
Buzz
 

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