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An Atwood machine: masses $m_1 > m_2$ over a **massive** pulley (uniform disc, mass $M$, radius $R$, $I = MR^2/2$). The acceleration of the masses is:
A$(m_1 - m_2)g/(m_1 + m_2)$ (massless pulley result)
B$(m_1 - m_2)g/(m_1 + m_2 + M/2)$
C$(m_1 - m_2)g/(m_1 + m_2 + M)$
DZero
Answer & Solution
Correct answer: B. $(m_1 - m_2)g/(m_1 + m_2 + M/2)$
Let $T_1, T_2$ be tensions on each side. $m_1g - T_1 = m_1 a$; $T_2 - m_2g = m_2 a$. Torque: $(T_1 - T_2)R = I\alpha = (MR^2/2)(a/R)$ ⇒ $T_1 - T_2 = Ma/2$. Add the three: $(m_1 - m_2)g = (m_1 + m_2 + M/2)a$.
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