## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v=10.5m/s$
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: $v=10.5m/s$