Non constant acceleration due to one constant force, and one non constant force

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 4K views
ex81
Messages
73
Reaction score
0

Homework Statement


This is actually related to another https://www.physicsforums.com/showthread.php?t=531596"
basically, I have a Constance force of gravity, and a non constant force of a rocket going the opposite way.

the net force would be something like F(booster) -Force(gravity). Assumed (-) direction, and Force(booster) is smaller. I know when acceleration isn't constant we have to integrate the equation.



Homework Equations


a = dx/dt


The Attempt at a Solution



a=( F(booster)- F(gravity) )/m

a=( F(booster)- mg) )/m

but I'm not sure how i should go about integrating this mess.
 
Last edited by a moderator:
Physics news on Phys.org
Hi there...
This is a tad more complex than you put it.
A rocket, in your case has to have a constant rate of ejection of its gases, let's call it, u;
Now, recall the proper definition of a force; You're right to state that F=ma, when the mass is constant, but in our case, our rocket propels upwards, and in doing that it loses its fuel, therefore, F has to be taken as: [itex]\large \vec{F}=\frac{d\vec{p}}{dt}=-mg, p = m(t)\cdot\vec{v}[/itex]
Loss of mass means: m = m_0-u*t;
Differentiating:
[itex]\large \vec{F}= \frac{d}{dt}m(t)\vec{v}(t) = \frac{dm}{dt}v(t)+m(t)\cdot\frac{d\vec{v}}{dt} = -uv+ma=mg[/itex]
Assuming motion only on the vertical axis, which we'll call y, we get a very pleasant differential equation:
-uy'+(m_0+ut)*y'' = (m0-ut)g;
That you can solve, either by exponential substitution, or, simply through this:
http://www.wolframalpha.com/input/?i=-u*y'[t]+++(m0+-+u*t)*y''[t]+==+(m0-u*t)*g"
But you probably know better(feel free to ask about the technique of resolution to be employed here, if you care to do this manually).
Don't forget to specify your initial conditions,
Daniel
 
Last edited by a moderator:
Also, note that this solution does not account for what happens after all the fuel is exhausted. So its only valid till t=m_0/u; afterwards, standard projectile motion with an initial velocity laws apply...
Daniel
 
Thanks this helps a lot more than the lack of info in my text!

And yes, my book doesn't have any of that in it. I'm "Supposed to figure that out"
 
Glad to be of service,
Do you know how to slove such differential equations? It might be worth your while to learn, if you're in the mood?
Daniel