Complex substitution into the equation of motion.

  • #1
125
1

Homework Statement


The equation of motion of a mass m relative to a rotating coordinate system is
[tex]m\frac{d^{2}r}{dt^2} = \vec{F} - m\vec{\omega} \times (\vec{\omega} \times \vec{r}) - 2m(\vec{\omega} \times \frac{d\vec{r}}{dt}) - m(\frac{d\vec{\omega}}{dt} \times \vec{r})[/tex]

Consider the case F = 0, [tex]\vec{r} = \hat{x} x + \hat{y} y[/tex], and [tex]\vec{\omega} = \omega \hat{z}[/tex], with [tex]\omega[/tex] a constant.

Show that the replacement of [tex]\vec{r} = \hat{x} x + \hat{y} y[/tex] by z = x + iy leads to

[tex]\frac{d^{2}z}{dt^2} + i2\omega\frac{dz}{dt} - \omega^2z=0[/tex].

Note, This ODE may be solved by the substitution [tex]z=fe^{-i\omega t}[/tex]


Homework Equations


None.


The Attempt at a Solution



I've calculated that [tex]-\vec{\omega} \times (\vec{\omega} \times z) = \omega^2 z[/tex].

As far as figuring out how [tex]-2(\vec{\omega} \times \frac{d\vec{r}}{dt}) -(\frac{d\omega}{dt} \times \vec{r})[/tex] gives [tex]i2\omega\frac{dz}{dt}[/tex] I'm lost.
 

Answers and Replies

  • #2
125
1
I solved it. Turns out that [tex]-2(\vec{\omega} \times \frac{d\vec{r}}{dt}) = -2i\omega\frac{dz}{dt}[/tex] and since F = 0 then [tex] \frac{d\omega}{dt} = 0[/tex].
 

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