I'm trying to derive the equation of motion of a snowball down a slope, given some known factors, in the form of either its angular acceleration, acceleration, velocity, or angular velocity expressed as a function of time. I've reached this point;
$$g\sin \theta=6w\frac{dr}{dt}+\frac 75 \frac{rdw}{dt}$$
Where $\theta$ is the angle of incline of the slope, $w$ is its angular velocity, $r$ is its radius, and $g$ is gravity. $dr/dt$ is rate of change of radius and $dw/dt$ is rate of change of angular velocity. I now need to integrate this in such a way that radius is expressed as some function of angular velocity. The original question I was following at this point assumed that theta was equal to zero (i.e the slope is flat) in order to simplify calculations. I'm not sure how to work with a differential equation when there seems to be 3 variables. Please help.
By definition of angular velocity we can argue that:$$r\omega=gt\sin \theta $$ or $$\frac{1}{\omega}=\frac{r}{gt\sin\theta}$$ also we can write the differential equation of question as following:$$g\sin \theta=4.6(\omega\frac{dr}{dt})+1.4(\omega\frac{dr}{dt}+r\frac{d\omega}{dt})=4.6(\omega\frac{dr}{dt})+1.4g\sin \theta$$finally we have:$${-0.4g\sin\theta}\frac{1}{\omega}=4.6\frac{dr}{dt}$$ this leads to $$-0.4\frac{r}{t}=4.6\frac{dr}{dt}\to \ln t=C-11.5\ln r\to r=Ct^{-\frac{2}{23}}\quad,\quad \omega=Cg\sin\theta t^{\frac{25}{23}}$$