Mechanics - Velocity over Time

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Homework Help Overview

The problem involves a particle moving through a medium with a frictional force proportional to the square of its velocity. The original poster seeks to express the velocity of the particle over time, given an initial velocity and no gravitational forces.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve a differential equation related to the motion of the particle but questions the necessity of the mass in the solution. Some participants suggest keeping the mass as a parameter, while others propose alternative forms for the velocity equation.

Discussion Status

Participants are exploring different formulations of the velocity equation, with some providing corrections to earlier attempts. There is an ongoing examination of the assumptions regarding the frictional force and its implications for the solution.

Contextual Notes

There is a noted confusion regarding the type of friction (linear vs. quadratic) and its effect on the velocity equation. The mass of the particle is treated as an unknown parameter throughout the discussion.

golanor
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Homework Statement


A particle is moving, through a medium in which the friction is: f = bv^2 (v is the velocity).
In t=0 the velocity is [itex]v{0}[/itex]. There is no gravity.
Write the velocity of the particle over time.

Homework Equations


I used: F = ma.


The Attempt at a Solution


-f=-m*a
Solving the differential equation - v(t)=m/(b*t-m/v0)
The thing is, the mass is not known in the question.
Is there a different way of solving this?
 
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Just keep m as parameter, similar to b. There is no way to get rid of it.
 
Thanks!
 
Indeed, you don't have to get rid of the mass. But the solution is not correct; it should be:

m[itex]\dot{v}[/itex] = - b v [itex]\Rightarrow[/itex] v(t) = v0 exp(-γ t )

where γ = b/m
 
cosmic dust said:
Indeed, you don't have to get rid of the mass. But the solution is not correct; it should be:

m[itex]\dot{v}[/itex] = - b v [itex]\Rightarrow[/itex] v(t) = v0 exp(-γ t )

where γ = b/m
Don't forget the square in the force.
 
Ooops... I used linear friction! Sorry, the velocity will be:

v = m v0 / (b v0 t + m)
 
cosmic dust said:
Ooops... I used linear friction! Sorry, the velocity will be:

v = m v0 / (b v0 t + m)

Which is what I wrote :)
 

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