Constant power in rotational dynamics I am having trouble understanding and applying the concept of constant power (e.g. a motor) in rotational dynamics. We have that:
$$P=\tau\omega$$
Therefore if we imagine a physical system with a motor, supplied with a constant voltage and current such that $P=P_{0}$ is a constant, driving a disk with moment of inertia $I$. We surely have a differential equation of the following form:
$$I\dot{\omega}=\frac{P}{\omega}-\gamma \omega$$
Where $\gamma$ is some frictional drag coefficient. This is a separable ODE which yields (according to Wolfram Alpha):
$$\omega(t)=\pm\sqrt{\frac{P-e^{\frac{2\gamma(cI-t)}{I}}}{\gamma}}$$
This doesn't seem realistic as using initial conditions $\omega(0)=0$ we find that:
$$c=\left\{\frac{\ln(\pm i\sqrt{P})}{\gamma}\right\}$$
What am I misunderstanding here?
 A: Wolfram Alpha wont give you the coefficients of integration properly.
The solution is 
$$ t = \int \frac{1}{\dot \omega}\,{\rm \omega} = \int \limits_{\omega_0}^\omega \frac{I \omega}{P-\gamma \omega^2}\,{\rm d} \omega $$
$$ t = - \frac{I}{2 \gamma} \ln \left( \frac{P-\gamma \omega^2}{P-\gamma \omega_0^2} \right) $$
where $\omega_0$ is the initial speed. When $\omega > \omega_0$ then the argument in the logarithm is less than one, but positive. This yields a positive time. The top speed is $\omega_f = \sqrt{ \frac{P}{\gamma}}$ and when $\omega \rightarrow \omega_f$ then $t \rightarrow \infty$.
The speed as a function of time is 
$$ \omega(t) = {\rm e}^{-\left(\frac{\gamma}{I} t\right)} \sqrt{\omega_0^2 + \frac{P}{\gamma} ({\rm e}^{\left(\frac{2 \gamma}{I} t\right)}  -1)} $$
Remember if initially at rest, the power is infinite. In reality there is a part of linear power (constant torque) before the constant power part.
To solve for angle use $\theta = \int \frac{\omega}{\dot \omega}\,{\rm d} \omega$ with appropriate limits.
