0
$\begingroup$

Consider initial motion of a particle with frictional damping that is proportional to the vector velocity with damping coefficient $(F=-rv)$ Assum an initial condition of $u = U$ and $v = V$ at $t = 0$. Derive and describe the solution for the trajectory of the particle in the absence of any forcing other than friction.

I tried looking this up and all I can find is harmonic oscillators but the question says friction only and does not specify to be on a spring.

Is there an ODE out there that can help me solve this?

  • 1
    What are U, V, r – 2017-02-08
  • 0
    Don't we just have $ma=mv'=-rv$? What is $U$? – 2017-02-08
  • 0
    $U$ is initial position, $V$ is initial velocity, and $r$ is the damping coefficient. The mass is also needed as noted in the comments. – 2017-02-08

1 Answers 1

1

We are only interested when only friction acts on the particle. Newton's second law gives,

$$F=ma=mv'=-rv$$

This means

$$\int m \frac{v'}{v} dt = \int -r dt$$

$$m\ln|v(t)|=-rt+c$$

Let $t=0$

$$m \ln|V|=c$$

So that,

$$m\ln |\frac{v(t)}{V}|=-rt$$

$$v(t)=Ve^{\frac{-rt}{m}}$$

Integrating both sides of this with respect to time then letting $t=0$,

$$u(t)=-\frac{m}{r}Ve^{\frac{-rt}{m}}+c_2$$

$$U=-\frac{m}{r}V+c_2$$

$$u(t)=(U+\frac{m}{r}V)-\frac{m}{r}Ve^{\frac{-rt}{m}}$$

This an exponential function with a horizontal asymptote,

$$U+\frac{m}{r}V$$

  • 0
    Would the particles path just decay exponentially? – 2017-02-08
  • 0
    Not exactly, it would increase @Abigail – 2017-02-08
  • 0
    I don't know what you would call something like $1-e^{-t}$. – 2017-02-08
  • 0
    Where $U$ used in solution. ? – 2017-02-08
  • 0
    Please see my edit @Abigail – 2017-02-08
  • 0
    I didn't write the full answer at first @MyGlasses – 2017-02-08
  • 0
    when you integrate both sides with respect to t, you come out with x(t) from v(t)...wouldn't it become y(t)? – 2017-02-08
  • 0
    What is $y(t)$, I think it just a matter of convention. Here I take $x(t)$ to mean position. – 2017-02-08
  • 0
    it should actually be u(t) since at t=0 , u(t) = U – 2017-02-08
  • 0
    Right ${}{}{}{}$ @Abigail – 2017-02-08