Answer
The tension in rod A is $4\pi^2f^2(m_Ar_A+m_Br_B)$.
The tension in rod B is $4\pi^2f^2m_Br_B$
Work Step by Step
The tension $T_A$ in rod A provides the centripetal force for both $m_A$ and $m_B$. The tension $T_B$ in rod B provides the centripetal force for $m_B$.
First we can find the tension $T_B$ in rod B:
$v_B = f\times(2\pi r_B)$.......(1)
$T_B = m_B\frac{v_B^2}{r_B}$.........(2)
Substituting (1) in (2):
$T_B = m_B\frac{(f\times(2\pi r_B))^2}{r_B}$
$T_B = 4\pi^2f^2m_Br_B$
The tension in rod B is $4\pi^2f^2m_Br_B$
Now we can find the tension in rod A:
$v_A = f\times(2\pi r_A)$.......(3)
$T_A = m_A\frac{v_A^2}{r_A} + T_B$.......(4)
Substituting (3) into (4) and solving:
$T_A = m_A\frac{(f\times(2\pi r_A))^2}{r_A}+ T_B$
$T_A = 4\pi^2f^2m_Ar_A + 4\pi^2f^2m_Br_B$
$T_A = 4\pi^2f^2(m_Ar_A+m_Br_B)$
The tension in rod A is $4\pi^2f^2(m_Ar_A+m_Br_B)$.
