Answer
$v = \sqrt{\frac{m_2~g~r}{m_1}}$
Work Step by Step
The weight of $m_2$ should be equal to the centripetal force required to keep $m_1$ moving in a circle. We can find the required speed of $m_1$ as;
$F_c = \frac{m_1~v^2}{r} = m_2~g$
$v^2 = \frac{m_2~g~r}{m_1}$
$v = \sqrt{\frac{m_2~g~r}{m_1}}$