Force to average acceleration

  • #1

Main Question or Discussion Point

I know that there is an equation for an accelerating object that is already a notable fraction of the speed of light (force = gamma^3 * mass * acceleration), but gamma is changing while the object accelerates
so what i want to know is how to calculate the average acceleration of an object when you already know the force acting on it, the mass of the object, and the starting velocity of the object
 

Answers and Replies

  • #2
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The easiest case to analyse is the case where the proper acceleration (as measured by an accelerometer attached to the accelerating object) is constant. Is that what you mean?
 
  • #3
not quite I'm afraid
I want a calculation for how much it has accelerated with a changing acceleration, because the acceleration is changing as it accelerates
in an example where force and mass are constant (lets say 100,000 newtons and 100 kilograms) at any given point it can be calculated as
100,000 = 100 * acceleration * gamma^3
100,000 / 100 = acceleration * gamma^3
1,000 = acceleration * gamma^3
which is useful when gamma is almost exactly the same the entire time, but when gamma changes a notable amount the calculation requires an average for gamma
which is what I failed to find on the internet
 
  • #4
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In general, you would have to perform an integration [tex]v = v_0 + \int_0^{\Delta t} a \, dt = v_0 + \int (\frac{f}{\gamma^3m})dt[/tex] then you can calculate the average acceleration by [itex]a=\frac{v-v_0}{\Delta t},[/itex] but why would you do that?

Also note that the formula [tex]f=\gamma^3ma[/tex] is only valid if the force is parallel to the velocity (linear motion).
 
  • #5
that would work for what I specified, but I forgot to mention I need to solve the equation with a smaller variable set
in the cases I need to solve I know the force the mass the starting velocity and the end velocity, the force the mass the starting velocity and the amount of time that passes, and the force the mass the starting velocity and the distance covered
 
  • #6
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An answer on the pixelpuffin's question depends on what we mean writing "acceleration" and "force". IMHO the best way to solve a problem is to read Landau's course. Vol.2 (see for example http://music.whu.edu.cn/Resources/eBooks/Physics/Field Theory/landau.pdf). "acceleration" see p.23, "force" see p.29. Anyone can find solution of a problem dealt with uniformely accelerated motion (p.24)
 
  • #7
PAllen
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that would work for what I specified, but I forgot to mention I need to solve the equation with a smaller variable set
in the cases I need to solve I know the force the mass the starting velocity and the end velocity, the force the mass the starting velocity and the amount of time that passes, and the force the mass the starting velocity and the distance covered
Each of these 3 cases is readily solved from the formulas dauto posted. You may need numerical methods for some of the cases, but they are still the right equations (assuming all measurements are made in one inertial coordinate system, and force is constant as measured by an observer stationary in said frame - as opposed to as measured by the accelerating object (that's the case Yuiop was referring to).

For example, if you take force, mass, starting velocity and end velocity as given, you are solving the integral equation for delta t. If you take the force, mass, starting velocity, and elapsed time as given, you just directly integrate for the ending velocity. Your third case is slightly more interesting - you need a double integral by dt to get distance, which you treat as given. Then you have to solve this for delta t.

[edit: actually, it is all easier than this: You want to re-arrange to f/m = gamma^3 a = gamma^3 v'.
Then, integrating by dt, you get (f/m)t = integral ( gamma^3 )dv
The integral can be solved in closed form, giving you the relation of v and t. From there, all you cases are solvable, with more integration in one case.]
 
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  • #8
PAllen
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Actually, this can be ridiculously simple. The m [itex]\gamma[/itex]^3v comes from:

f = dp/dt = m (d/dt)([itex]\gamma[/itex]v). Then,

f/m = (d/dt)(([itex]\gamma[/itex]v), so

ft/m = [itex]\gamma[/itex]v + k

This simply, you have v as f (t), and can solve all your cases.

Going a little further, to get v=v0 at t=0, [itex]\gamma[/itex]0 having the obvious meaning, then define:

T= ft/m + [itex]\gamma[/itex]0 v0

then v = c /√ (1 + c^2 / T^2)

and you can see that v -> c as t -> ∞.
 
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  • #9
that seems quite helpful
I'll look at this closer once i get some sleep
(screw late night algebra)
 

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