RBD CYU 12

Let $r$ be the radius of the inner drum and $R = \eta r$ be the radius of the outer drum.
Block $m_2$: hangs from the right side of the outer drum. $$ a_2 = R \alpha = \eta r \alpha $$ Block $m_1$: hangs from a movable pulley. The string wraps over the left side of the outer drum ($R$) and the right side of the inner drum ($r$). When the pulley rotates to lift $m_2$ (CCW), the outer drum pays out string (left side goes down), while the inner drum winds up string (right side goes up). The net velocity of the loop is determined by the difference in tangential velocities: $$ v_{loop} = v_{outer} – v_{inner} = R\omega – r\omega $$ The acceleration of the movable pulley $m_1$ is half of this: $$ a_1 = \frac{(R – r)\alpha}{2} = \frac{(\eta – 1)r \alpha}{2} $$ Relation: Substituting $\alpha = \frac{a_2}{\eta r}$: $$ a_1 = \frac{(\eta – 1)r}{2} \cdot \frac{a_2}{\eta r} = \frac{\eta – 1}{2\eta} a_2 $$ $$ a_2 = \frac{2\eta}{\eta – 1} a_1 $$

Let $T_1$ be the tension in the loop ($m_1$) and $T_2$ be the tension for $m_2$.
Equation for $m_1$ (downward accel): $$ m_1 g – 2 T_1 = m_1 a_1 \implies T_1 = \frac{m_1(g – a_1)}{2} $$ Equation for $m_2$ (upward accel): $$ T_2 – m_2 g = m_2 a_2 \implies T_2 = m_2(g + a_2) $$

The net torque on the massless pulley is zero. – $T_1$ acts on the left at radius $R$ (pulling down, creating CCW torque). – $T_1$ acts on the right at radius $r$ (pulling down, creating CW torque). – $T_2$ acts on the right at radius $R$ (pulling down, creating CW torque). $$ \tau_{net} = T_1 R – T_1 r – T_2 R = 0 $$ $$ T_1 (R – r) = T_2 R $$ $$ T_2 = T_1 \frac{R – r}{R} = T_1 \frac{\eta – 1}{\eta} $$

Substitute tensions into the torque equation: $$ m_2(g + a_2) = \frac{\eta – 1}{\eta} \cdot \frac{m_1(g – a_1)}{2} $$ Substitute kinematic relation $a_2 = \frac{2\eta}{\eta – 1} a_1$: $$ m_2\left(g + \frac{2\eta}{\eta – 1} a_1\right) = \frac{m_1(\eta – 1)}{2\eta} (g – a_1) $$ Solving for $a_1$: $$ a_1 = g \frac{(\eta – 1) [ m_1(\eta – 1) – 2\eta m_2 ]}{m_1(\eta – 1)^2 + 4\eta^2 m_2} $$

Values: $m_1 = 9.0, m_2 = 1.0, \eta = 2, g = 10$. $$ a_1 = 10 \cdot \frac{1 \cdot [9(1) – 4(1)]}{9(1) + 16(1)} = 10 \cdot \frac{5}{25} = 2.0 \text{ m/s}^2 $$