• Support PF! Buy your school textbooks, materials and every day products Here!

Find position function of freefall with air resistence ODE

  • Thread starter 1s1
  • Start date
  • #1
1s1
20
0

Homework Statement


An object of mass ##5##kg is released from rest ##1000##m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant ##b=50##N-sec/m, determine the equation of motion (for the position ##x(t)##) of the object.


Homework Equations


$$m\frac{dv}{dt} = mg-kv$$


The Attempt at a Solution


The given differential equation describes the velocity with respect to time. How do I turn it into a differential equation that describes the position with respect to time?
 

Answers and Replies

  • #2
Zondrina
Homework Helper
2,065
136

Homework Statement


An object of mass ##5##kg is released from rest ##1000##m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant ##b=50##N-sec/m, determine the equation of motion (for the position ##x(t)##) of the object.


Homework Equations


$$m\frac{dv}{dt} = mg-kv$$


The Attempt at a Solution


The given differential equation describes the velocity with respect to time. How do I turn it into a differential equation that describes the position with respect to time?
Your relevant equation is a bit off, shouldn't it read ##m\frac{dv}{dt} = (-1)(mg+kv)##?

If that's the case then solving it with the constraint that ##v(0) = 0## should simplify things a bit.
 
  • #3
1s1
20
0
You may be right about the relevent equation being negative on the RHS. My book has it written as I did, but my prof mentioned that physicists sometimes make sign changes to the constant.

But I am still confused about how to make the function into a function of position? I assume you use the relationship ##d=vt##? Or maybe there is an entirely different formula derived for the position rather than the velocity?
 
  • #4
Zondrina
Homework Helper
2,065
136
You may be right about the relevent equation being negative on the RHS. My book has it written as I did, but my prof mentioned that physicists sometimes make sign changes to the constant.

But I am still confused about how to make the function into a function of position? I assume you use the relationship ##d=vt##? Or maybe there is an entirely different formula derived for the position rather than the velocity?
Notice this was part of your question:

An object of mass 5kg is released from rest
This means at ##t=0##, the velocity of the object is zero. That's why I mentioned solving the equation with the constraint ##v(0) = 0##. Right now you have an acceleration equation since it involves derivatives of velocity, but if you integrate acceleration, you get velocity;if you integrate vel...
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,377
5,971
the sign is given by the direction of the [itex]x[/itex] axis. If you define it to point "down", i.e., in direction of the constant gravitation force, the positive sign [itex]+mg[/itex] is correct. The sign in [itex]-k v[/itex] (assuming [itex]k>0[/itex]) is the same for both directions of the [itex]x[/itex] axis, because the friction force must always be such that it tends to decrease [itex]v[/itex].

Further note, that you have a nice inhomogeneous linear differential equation of first order with constant coefficients, which is pretty easy to solve by separation of variables. The gives you [itex]v(t)[/itex].

Then you have to solve another very simple equation
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}.[/tex]
Given [itex]v[/itex], it's very clear, how to get [itex]x(t)[/itex]!
 
  • #6
612
23
Your relevant equation is a bit off, shouldn't it read ##m\frac{dv}{dt} = (-1)(mg+kv)##?
Why? Air RESISTANCE implies that it will go against the direction of acceleration. This is a physics problem, primarily.

We're supposed to solve the equation and make a position function, right? I will denote ##\frac{dv}{dt}## as ##v'##, for clarity and ease of notation (id est, laziness. It's a friday. Give me a break :tongue:).

I'm assuming we have ##k=b##, right? So, we have $$mv'=mg-bv \\ v' +\frac{b}{m}v=g.$$ In this form, we can introduce an integrating factor. Can you take it from there?

Edit: vanhees71 got there first by noting the inhomogeneous equation.

Given [itex]v[/itex], it's very clear, how to get [itex]x(t)[/itex]!
Indeed. Finding ##\Gamma(x(t)+1)## is easy once we have ##v##. :biggrin:
 
  • #7
1s1
20
0
This is very helpful, thanks!
 
  • #8
Zondrina
Homework Helper
2,065
136
the sign is given by the direction of the [itex]x[/itex] axis. If you define it to point "down", i.e., in direction of the constant gravitation force, the positive sign [itex]+mg[/itex] is correct. The sign in [itex]-k v[/itex] (assuming [itex]k>0[/itex]) is the same for both directions of the [itex]x[/itex] axis, because the friction force must always be such that it tends to decrease [itex]v[/itex].

Further note, that you have a nice inhomogeneous linear differential equation of first order with constant coefficients, which is pretty easy to solve by separation of variables. The gives you [itex]v(t)[/itex].

Then you have to solve another very simple equation
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}.[/tex]
Given [itex]v[/itex], it's very clear, how to get [itex]x(t)[/itex]!
I suppose it depends on how you decide to define up and down.

For me that's dependent on the day of the week :P.
 

Related Threads on Find position function of freefall with air resistence ODE

Replies
1
Views
1K
Replies
14
Views
3K
  • Last Post
Replies
0
Views
690
Replies
8
Views
24K
Replies
2
Views
7K
Replies
14
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
0
Views
2K
Replies
2
Views
2K
Replies
14
Views
2K
Top