Answer
The tension in the rod is $~~5mg$
Work Step by Step
In part (a), we found that $v_0= \sqrt{2~g~L}$
We can use conservation of energy to find the speed $v_B$ at point $B$:
$\frac{1}{2}mv_B^2 = \frac{1}{2}mv_0^2+mgL$
$v_B^2 = v_0^2+2gL$
$v_B^2 = (\sqrt{2gL})^2+2gL$
$v_B^2 = 4gL$
$v_B = \sqrt{4gL}$
We can find the tension in the rod:
$T-mg = \frac{mv_B^2}{L}$
$T = mg+\frac{m~(\sqrt{4gL})^2}{L}$
$T = mg+\frac{4mgL}{L}$
$T = 5mg$
The tension in the rod is $~~5mg$
