Answer
$\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$
Work Step by Step
When the disk is at the top of its path, the center of mass is shifted up by $\frac{1}{30}R$. The mass of the disk is $\frac{15}{16}M$, since $M-(\frac{1}{4})^2M =\frac{15}{16}M$. Thus, using conservation of energy, it follows:
$ I\omega_{max}^2=I\omega_{min}^2+\frac{15}{16}M(\frac{1}{30}R)g$
Thus, using the result from problem 65 and simplifying, we find:
$\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$
