Question

Please help me to prove this

Answer 1

expand the cross products in terms of the regular unit vectors first.

Answer 2

and then ??

Answer 3

Replace x hat with 1 and y hat with i, then rewrite everything in terms of z.

Answer 4

i can't figure out how the next

Answer 5

Dividing everything by m gives and taking into account that F = 0 and omega is constant yields
\[\frac{d^2 \vec{r}}{dt^2} + 2(\vec{\omega}\times \frac{d\vec{r}}{dt}) +\vec{\omega}\times(\vec{\omega}\times \vec{r})= 0\]
since
\[ \vec{\omega} = \omega \hat{z} \]
this gives
\[ \frac{d^2 \vec{r}}{dt^2} +2 \omega(-\frac{dy}{dt} \hat{x} + \frac{dx}{dt}\hat{y}) +\omega\hat{z}(\omega z) -\vec{ r}(\omega^2)\]

We can lose the second to last term because z = 0 , so

\[\frac{d^2 \vec{r}}{dt^2} + 2\omega(- \frac{dy}{dt}\hat{x} + \frac{dx}{dt}\hat{y}) - \omega^2 \vec{r} = 0 \]
if i replace the vector r with z = x+iy, and set xhat = 1, yhat = i, this gives me
\[ \frac{d^2 z}{dt^2} + 2\omega (i \frac{dx}{dt} - \frac{dy}{dt}) - \omega^2 z = 0\]

last but not least, if i factor an i out of the middle term, that gives me
\[ \frac{d^2z}{dt^2} + 2i\omega z - \omega^2z = 0 \]

Answer 6

thank you so much @jemurray3

Answer 7

That should be z' in the middle, not z, sorry.

Answer 8

yes., i know that should be dz/dt, thank u., wanna help me again? please?

Answer 9

Sure