Answer
The distance should be greater than $223.4m$
Work Step by Step
From the given information, since the rotational motion should not be faster than 2 revolutions per minute, the minimum period $T_{min}=0.5min=30s$. Furthermore, $a=g=9.8m/s^2$
We have $$T=\frac{2\pi r}{v} (1)$$
We also have $v^2=ar$ or $v=\sqrt{ar} (2)$
From (1) and (2), $$T=\frac{2\pi r}{\sqrt{ar}}=\frac{2\pi \sqrt r}{\sqrt a}$$ $$r=\Big(\frac{T\sqrt a}{2\pi}\Big)^2$$
Apply $T_{min}$ and $a$ here, we will have $$r_{min}=223.4m$$
So the distance should be greater than $223.4m$
