More power gives more velocity

Here's a solution in terms of time (starting from rest at $t = 0$ ):

$\large{E = P \, t = \beta \, \omega^2 \\ \omega = \sigma \, t^{1/2}}$

Determine an expression for $\theta$ :

$\large{\theta = \int_0^t \omega \, dt = \sigma \int_0^t t^{1/2} \, dt \\ \theta = \gamma \, t^{3/2}}$

The following is therefore apparent:

$\large{\omega \propto \theta^{1/3}}$

Here is my solution to this: