## College Physics (4th Edition)

$a = \frac{m_2~g}{m_2 + m_1 + \frac{I}{R^2}}$
Let $T_1$ be the tension in the rope attached to block $m_1$. We can find an expression for $T_1$: $T_1 = m_1~a$ Let $T_2$ be the tension in the rope attached to block $m_2$. We can find an expression for $T_2$: $m_2~g - T_2 = m_2~a$ $T_2 = m_2~g - m_2~a$ Let $R$ be the radius of the pulley. We can consider the torque from the tension in each rope acting on the pulley: $\tau_2-\tau_1 = I~\alpha$ $R~T_2 - R~T_1 = I~(\frac{a}{R})$ $(m_2~g-m_2~a) - (m_1~a) = I~(\frac{a}{R^2})$ $m_2~g = m_2~a + m_1~a + \frac{Ia}{R^2}$ $a = \frac{m_2~g}{m_2 + m_1 + \frac{I}{R^2}}$