Answer
$v = 7.2~m/s$
Work Step by Step
We can use the parallel axis theorem to find the rotational inertia:
$I = \frac{1}{12}ML^2+Md^2$
$I = \frac{1}{12}(3.0~kg)(4.0~m)^2+(3.0~kg)(1.0~m)^2$
$I = 7.0~kg~m^2$
We can find the angular speed at the vertical position:
$K = \frac{1}{2}I~\omega^2$
$\omega^2 = \frac{2K}{I}$
$\omega = \sqrt{\frac{2K}{I}}$
$\omega = \sqrt{\frac{(2)(20~J)}{7.0~kg~m^2}}$
$\omega = 2.39~rad/s$
We can find the linear speed of the end B:
$v = \omega~r$
$v = (2.39~rad/s)(3.0~m)$
$v = 7.2~m/s$
