Answer
$a = 6.9~m/s^2$
Work Step by Step
We can find an expression for the rotational inertia about the fulcrum:
$I = mL_1^2+mL_2^2$
$I = m~(L_1^2+L_2^2)$
$I = m~(0.20^2+0.80^2)$
$I = 0.68~m$
We can find the angular acceleration:
$\sum~\tau = I~\alpha$
$\alpha = \frac{\sum~\tau}{I}$
$\alpha = \frac{mg~L_2-mg~L_1}{0.68~m}$
$\alpha = \frac{g~(L_2-L_1)}{0.68}$
$\alpha = \frac{(9.8)~(0.80-0.20)}{0.68}$
$\alpha = 8.647~rad/s^2$
We can find the magnitude of the initial acceleration of particle 2:
$a = \alpha~L_1$
$a = (8.647~rad/s^2)(0.80~m)$
$a = 6.9~m/s^2$
