Answer
The tension in the string connecting the two blocks is $2.0~N$
The tension in the string connecting the inner block to the pole is $3.8~N$
Work Step by Step
The angular speed is $(1.5~rev/s)(2\pi~rad/rev) = 3\pi~rad/s$
We can find the tension $T_2$ in the string connecting the two blocks. Note that this tension provides the centripetal force to keep the outer block moving around in a circle:
$T_2 = m_2~\omega^2~r_2$
$T_2 = (0.030~kg)~(3\pi~rad/s)^2~(0.75~m)$
$T_2 = 2.0~N$
The tension in the string connecting the two blocks is $2.0~N$
We can find the tension $T_1$ in the string connecting the inner block to the pole:
$\sum F = m_1~\omega^2~r_1$
$T_1-T_2 = m_1~\omega^2~r_1$
$T_1 = T_2 + m_1~\omega^2~r_1$
$T_1 = 2.0~N + (0.050~kg)~(3\pi~rad/s)^2~(0.40~m)$
$T_1 = 3.8~N$
The tension in the string connecting the inner block to the pole is $3.8~N$.
