Spinning spring

Let $r_0$ be the initial radius, $\omega_0$ be the initial angular speed, $r$ be the radius after expansion, $\omega$ be the angular speed after expansion, $m$ be the mass and $k$ be the spring constant.

At the beginning the angular momentum is $m r_0^2 \omega_0$ and after expansion $m r_0^2 \omega_0$ . By conservation of the momentum, $\omega = \frac{r_0^2}{r^2} \omega_0$ .

The inital energy is $\tfrac{1}{2} m r_0^2 \omega_0^2$ . On the other hand, the energy when the radius is maximal (or minimal) is $\tfrac{1}{2} m r^2 \omega^2 + 2 \pi^2 k (r-r_0)^2$ (the other part of the kinetic energy $\frac{1}{2} m \dot{r}^2$ is zero when the radius is max or min [huge thanks to Mark Hennings for pointing this out]). So conservation of energy states that (at the optimal radius): $\tfrac{1}{2} m r_0^2 \omega_0^2 = \tfrac{1}{2} m r^2 \omega^2 + 2 \pi^2 k (r-r_0)^2$ Pluging in the formula for $\omega$ and gathering some terms: $0 = \tfrac{1}{2} m \omega_0^2 r_0^2 ( \frac{r_0^2}{r^2} - 1)+ 2 \pi^2 k (r-r_0)^2$ Since $\tfrac{1}{2} m \omega_0^2 = 2 \pi^2 k$ one gets upon dividing $0 = ( \frac{r_0^2}{r^2} - 1)+ (\frac{r}{r_0}-1)^2$ Lastly using $r_0=1$ and multiplying by $r^2$ there remains a polynomial: $r^4-2r^3+ 1 =0$ . One can solve the polynomial numerically or factor the "obvisous" $r=1$ root out, to get $r^3-r^3-r-1 =0$ and then use the formula. Either way the solution is: ${\left(\frac{1}{9} \, \sqrt{33} + \frac{19}{27}\right)}^{\frac{1}{3}} + \frac{4}{9 \, {\left(\frac{1}{9} \, \sqrt{33} + \frac{19}{27}\right)}^{\frac{1}{3}}} + \frac{1}{3} \approx 1.83928675521416$