Answer
$v=10.5m/s$
Work Step by Step
We can determine the required speed as follows:
$\Sigma F_y=0$
$\implies T-W_A=0$
$\implies T=W_A=m_A g~~~$eq(1)
But we know that $T=m_b\frac{v^2}{r}~~~$eq(2)
Comparing eq(1) and eq(2), we obtain:
$m_Ag=m_B\frac{v^2}{r}$
This can be rearranged as:
$v=\sqrt{(\frac{m_A}{m_B})gr}$
We plug in the known values to obtain:
$v=\sqrt{(\frac{15}{2})(9.81)(1.5)}$
This simplifies to:
$v=10.5m/s$
