From $z = \dot{\Theta}^2 \operatorname{sgn}(\dot{\Theta})$ to $\dot{\Theta} = \operatorname{sgn}(z) \sqrt{\left| z \right|}$

Question

I read on a scientific paper (*) the following equations:

$$ z = \dot{\Theta}^2 \operatorname{sgn}(\dot{\Theta}) $$

and then:

$$ \dot{\Theta} = \operatorname{sgn}(z) \sqrt{| z |} $$

Could you tell me how to pass from the first equation to the second one please?

(*) Dynamics and stability of a rimless spoked wheel: a simple 2D system with impacts by Michael J. Coleman

Answer 1

$\def\sgn{\mathrm{sgn}}$

Since $z = \dot{\Theta}^2 \sgn(\dot{\Theta})$ we have $|z| = \dot{\Theta}^2$ and then $\dot{\Theta} = \pm \sqrt{|z|}$

Which sign we choose? Well, from $z = \dot{\Theta}^2 \sgn(\dot{\Theta})$ we see that $\sgn(z) = \sgn(\dot{\Theta})$, so $\dot{\Theta} = \sgn(z) \sqrt{|z|}$.

Answer 2

Let's examine the two cases; $(1)\,\dot{\Theta}\ge 0$ and $(2)\,\dot{\Theta}< 0$.

CASE $1$:

For $\dot{\Theta}\ge 0$, we see that

$$z=\dot{\Theta}^2 \text{sgn}\left(\dot{\Theta}\right)\implies z\ge0$$

and hence

$$\dot{\Theta}=\sqrt z$$

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CASE $2$:

For $\dot{\Theta}<0$, we see that

$$z=\dot{\Theta}^2 \text{sgn}\left(\dot{\Theta}\right)\implies z<0$$

and hence

$$\dot{\Theta}=-\sqrt {-z}$$

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Putting it all together, we have

$$\dot{\Theta}=\text{sgn}(z)\sqrt{|z|}$$

And we are done!