Answer
$\alpha = 1.4~rad/s^2$
Work Step by Step
We need to consider the torque on the person due to the gravitational force and due to the pole's weight.
We can find the magnitude of the initial angular acceleration:
$\tau = I~\alpha$
$\alpha = \frac{\tau}{I}$
$\alpha = \frac{m_1g~d_1-m_2g~d_2}{I}$
$\alpha = \frac{(70.0~kg)(9.8~m/s^2)(0.050~m)-(14.0~kg)(9.8~m/s^2)(0.10~m)}{15.0~kg~m^2}$
$\alpha = 1.4~rad/s^2$
