Answer
$v = \sqrt{\frac{m_2~g~r}{m_1}}$
Work Step by Step
For the hanging mass $m_2$ to remain at rest, the tension $T$ in the string must be equal to the weight of $m_2$
$T = m_2~g$
The tension $T$ provides the centripetal force to keep $m_1$ moving around in a circle. We can find an expression for the speed of $m_1$.
$T = \frac{m_1~v^2}{r}$
$m_2~g = \frac{m_1~v^2}{r}$
$v^2 = \frac{m_2~g~r}{m_1}$
$v = \sqrt{\frac{m_2~g~r}{m_1}}$
