Answer
(a) $\alpha = \frac{3g}{2l}$
(b) $a = \frac{3g}{2}$
Work Step by Step
(a) We can use a torque equation to find the angular acceleration $\alpha$ of the rod. The torque is provided by gravity. Therefore,
$\tau = I \alpha$
$r\cdot Mg = (\frac{1}{3}Ml^2)~\alpha$
$\alpha = \frac{3~r~g}{l^2} = \frac{(3)(l/2)(g)}{l^2}$
$\alpha = \frac{3g}{2l}$
(b) Now, we find the linear acceleration through the formula:
$a = \alpha ~l = (\frac{3g}{2l}~rad/s^2)(l)$
$a = \frac{3g}{2}$
