Question

You probably have noticed that, as you increase ω, there will be a value, ωcrit, for which r(ω) goes to infinity. find ωcrit. answer in terms of k and m.

Answer 1

For the mass m, attached to a spring, to move in a circle centripetal force and restoring force of the spring must be equal. Centripetal force is given with this formula:
$F_c=\frac{mv^2}{r}=\frac{mw^2r^2}{r}=mw^2r$
Restoring force only occurs if the spring is stretched. If L is the length of unstretched spring we have the following formula for restoring force:
$F=k(r-L)$
r is the length of a circle that mass m is traveling along. 
As said above, these two forces have to be equal:
$F=k(r-L)=mw^2r$
We solve for r:
$F=k(r-L)=mw^2r\\ kr-kL=mw^2r\\ mw^2r-kr=-kL\\ r(mw^2-k)=-kL\\ r=\frac{-kL}{mw^2-k}\\ r(w)=\frac{kL}{k-mw^2}$
r(w) will go to infinity when denominator is equal to zero:
$k-mw^2=0\\ k=mw^2\\ w^2=\frac{k}{m}\\ w_{crit}=\sqrt{\frac{k}{m}}$
Please keep in mind that Hooke's law has it's limitations, and before we reach our critial value of angular velocity spring will be strecthed to a point where Hooke's law does not aply anymore.