Perhaps it should. As far as the rest of the problem goes, I was able to determine that ##\phi=\omega t## and thus ##\dot{\phi}=\omega## so there is only one degree of freedom. I even got something as a result that looks like SHM:
##\ddot{r}=\left(\omega^2-\frac{k}{m}\right)r - \frac{k}{m}R...