(Solved): Question \( \checkmark \) Answered step-by-step The equation of motion of a mass \( m \) relative ...

Question \( \checkmark \) Answered step-by-step The equation of motion of a mass \( m \) relative to a rotating coordinate system is \[ m \frac{d^{2} \mathbf{r}}{d t^{2}}=\mathbf{F}-m \omega \times(\omega \times \mathbf{r})-2 m\left(\omega \times \frac{d \mathbf{r}}{d t}\right)-m\left(\frac{d \omega}{d t} \times \mathbf{r}\right) . \] Consider the case \( \boldsymbol{F}=0, \mathbf{r}=\hat{\mathbf{x}} \boldsymbol{x}+\hat{\mathbf{y}} \), and \( \omega=\dot{w} \mathbf{x} \), with \( \omega \) constant. Show that the replacement of \( \mathbf{r}=\hat{\mathbf{x}} \boldsymbol{x}+\hat{\mathbf{y}} \boldsymbol{b} \mathrm{by}_{z=x+i y} \) leads to \[ \frac{d^{2} z}{d t^{2}}+i 2 w \frac{d z}{d t}-w^{2} z=0 . \] AI Recommended Answer: In the first step, we find the equatio View Answer \( \left.\mathrm{m} \backslash \operatorname{frac}\left\{\mathrm{d}^{\wedge}\{2\} \mathrm{x}\right\} \mathrm{dt} \mathrm{t}^{\wedge}\{2\}\right\}+\mathrm{m} \backslash \mathrm{frac}\left\{\mathrm{d}^{\wedge}\{2\}\right. \) mes \( \backslash \operatorname{mathbf}\{\mathrm{r}\})-2 \) The equation of motion is solved for \( x, y \), and \( z \) using standard linear equations. In the second step, we use the equation of \( n \) w. \( S S \backslash \) omega \( \backslash \backslash \operatorname{frac}\left\{\mathrm{d}^{\wedge}\{2\} \mathrm{x}\right\}\left\{\mathrm{d}^{\wedge}\{2\}\right\}+\backslash \operatorname{frac}\left\{\mathrm{d}^{\wedge}\{2\} \mathrm{y}\right\}\left\{\mathrm{dt} \mathrm{t}^{\wedge}\{2\}\right\}+\backslash \) frac \( \left\{\mathrm{d}^{\wedge}\{2\} \mathrm{z}\right\}\left\{\mathrm{dt} \mathrm{t}^{\wedge}\{2\}\right\} \).