Answer
$a = 1.57~m/s^2$
Work Step by Step
We can find an expression for $T_1$:
$T_1-m_1~g = m_1~a$
$T_1 = m_1~g + m_1~a$
We can find an expression for $T_2$:
$m_2~g - T_2 = m_2~a$
$T_2 = m_2~g - m_2~a$
We can consider the torque on the pulley to find the magnitude of the acceleration:
$\sum~\tau = I~\alpha$
$R~T_2-R~T_1 = (\frac{1}{2}MR^2)~(\frac{a}{R})$
$T_2-T_1 = \frac{1}{2}M~a$
$(m_2~g - m_2~a)-(m_1~g + m_1~a) = \frac{1}{2}M~a$
$\frac{1}{2}M~a+m_1a+m_2a = m_2~g-m_1~g$
$a = \frac{g~(m_2-m_1)}{\frac{M}{2}+m_1+m_2}$
$a = \frac{(9.8~m/s^2)~(0.600~kg-0.400~kg)}{\frac{0.500~kg}{2}+0.400~kg+0.600~kg}$
$a = 1.57~m/s^2$
