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So, as a personal project, I imagined a scenario where a projectile with an initial velocity $x$ is affected by air resistance (drag) but not gravity, and I wanted to find out the distance it would travel until it stops.

I know that drag force is determined by: $F_D= (1/2) \rho v^2 C_d A$, which I simplified to $F_D= bv^2$ by grouping all the constants.

And I know that $\text{work = force} \times \text{displacement}$. So in theory I should be able to find the displacement of the projectile when work done by drag is equal to the kinetic energy of the projectile. However, since in this case force is proportional to the velocity of the projectile, and the velocity would keep decreasing, I figured I should find the average force first and then multiply it by the displacement in order to get the work done by drag. Thus (with $x$ being the initial velocity and $F_{avg}$ being average force):

$$F_{avg}=\frac{1}{x} \int_0^xbv^2 dv$$

$$F_{avg}=\frac{1}{3} bx^2$$

$$Work=F_{avg} \times Displacement$$

$$Work=\frac{1}{3} bx^2 \times Displacement$$

And thus, because of conservation of energy, the kinetic energy of the projectile must be equal to the work done by the drag in order for the projectile to be stopped, thus:

$$\frac{1}{3} bx^2 \times Displacement= \frac{1}{2} mx^2$$

Which reduces to:

$$Displacement=\frac{3m}{2b}$$

At this point, I went "wait, wait, this can't be right!" This implies that the horizontal displacement of a projectile, assuming it is affected by drag and not by gravity, would not be affected by it's initial velocity. My gut instinct is that this can't possibly be true. So I guess my question is

  1. Where in my process above did I make a mistake and
  2. How do you find the displacement (or distance) where the projectile unaffected by gravity but affected by drag force finally stops, assuming you know the initial velocity ($x$)?
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  1. Where in my process above did I make a mistake and
  2. How do you find the displacement (or distance) where the projectile unaffected by gravity but affected by drag force finally stops, assuming you know the initial velocity (x)?

Firstly, your calculation of the work done by the drag force is wrong. The drag force is $F_D=bv^2$. The work done $dF$ over an infinitesimal displacement $dx$ is:

$dF=bv^2dx$, which unless we have a function $v(x)$ is really not a useful, integratable expression.

So let's try another approach. $F_D$ causes the object to decelerate and with no gravity acting we can write the equation of motion as:

$ma+F_D=0$, with $a$ the deceleration the object is subjected to.

$m\frac{dv}{dt}+bv^2=0$ is an ordinary differential equation, solved by separation of variables:

$-m\frac{dv}{v^2}=bdt$ and integrated between $(0,v_0)$ and $(t,v)$ we get:

$(\frac{1}{v}-\frac{1}{v_0})=\frac{b}{m}t$.

Slightly reworked we get:

$\Large{v(t)=\frac{m}{bt+\frac{m}{v_0}}}$.

Note that this a hyperbolic decay function where $v(t) \to 0$ only for $t \to +\infty$.

This means the object continues to slow down forever and there is no discrete displacement at which it truly comes to rest. The displacement at which it comes to rest is effectively $x=+\infty$.

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  • $\begingroup$ Thank you! Does the function at the end work in predicting the velocity of the object at any given point of time, assuming initial velocity v0 is known? Conversely, if we do know the final velocity v(t) and time t, can we use the equation to find v0? Just checking if my understanding here is correct. $\endgroup$ – curiousobserver Sep 29 '15 at 14:09
  • $\begingroup$ @curiousobserver: yes, if four parameters are known then the expression for $v(t)$ can always be used to find the fifth parameter $(v(t),v_0,m,b,t)$. One can also find $x(t)$ by integrating $dx(t)=v(t)dt$. $\endgroup$ – Gert Sep 29 '15 at 14:19

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