# Homework Help: Complex substitution into the equation of motion.

1. Mar 3, 2007

### ultimateguy

1. The problem statement, all variables and given/known data
The equation of motion of a mass m relative to a rotating coordinate system is
$$m\frac{d^{2}r}{dt^2} = \vec{F} - m\vec{\omega} \times (\vec{\omega} \times \vec{r}) - 2m(\vec{\omega} \times \frac{d\vec{r}}{dt}) - m(\frac{d\vec{\omega}}{dt} \times \vec{r})$$

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

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

$$\frac{d^{2}z}{dt^2} + i2\omega\frac{dz}{dt} - \omega^2z=0$$.

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

2. Relevant equations
None.

3. The attempt at a solution

I've calculated that $$-\vec{\omega} \times (\vec{\omega} \times z) = \omega^2 z$$.

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

2. Mar 3, 2007

### ultimateguy

I solved it. Turns out that $$-2(\vec{\omega} \times \frac{d\vec{r}}{dt}) = -2i\omega\frac{dz}{dt}$$ and since F = 0 then $$\frac{d\omega}{dt} = 0$$.