To solve this problem, let's approach it in steps. We have an Atwood's machine where a pulley is involved with rotational inertia, and we need to consider the forces and torques acting on each component of the system.

Given dataTo solve this problem, let's approach it in steps. We have an Atwood's machine where a pulley is involved with rotational inertia, and we need to consider the forces and torques acting on each component of the system.

Given

• $m_1 = 12.0 \, \text{kg}$
• $m_2 = 22.0 \, \text{kg}$
• Pulley mass $M = 7.10 \, \text{kg}$
• Pulley radius $r = 0.200 \, \text{m}$

Steps to Solution

1. Define the forces:

   • The gravitational force on each mass:
     • $F_{g1} = m_1 \cdot g = 12.0 \cdot 9.81 = 117.72 \, \text{N}$
     • $F_{g2} = m_2 \cdot g = 22.0 \cdot 9.81 = 215.82 \, \text{N}$
   • Tension forces:
     • $T_1$ (tension in the cord connected to $m_1$)
     • $T_2$ (tension in the cord connected to $m_2$)

2. Set up the equations of motion for $m_1$ and $m_2$:

   • For $m_1$: $T_1 - m_1 \cdot g = m_1 \cdot a$
   • For $m_2$: $m_2 \cdot g - T_2 = m_2 \cdot a$

3. Consider the rotational motion of the pulley:

   • The pulley’s rotational inertia $I = \frac{1}{2} M r^2 = \frac{1}{2} \cdot 7.10 \cdot (0.200)^2 = 0.142 \, \text{kg} \cdot \text{m}^2$
   • The torque on the pulley due to the tensions is $\tau = T_2 \cdot r - T_1 \cdot r$.
   • Using Newton's second law for rotation: $\tau = I \cdot \alpha$, where $\alpha$ is the angular acceleration.
   • Since $a = \alpha \cdot r$, we can substitute $\alpha = \frac{a}{r}$, leading to $(T_2 - T_1) \cdot r = I \cdot \frac{a}{r}$.

4. Combine equations:

   • Substitute $I = 0.142 \, \text{kg} \cdot \text{m}^2$: $(T_2 - T_1) \cdot 0.200 = 0.142 \cdot \frac{a}{0.200}$
   • Simplify to get: $T_2 - T_1 = 0.71 \, a$.

5. Solve for $a$:

   • From the equations for $m_1$ and $m_2$: $T_1 = m_1 \cdot (g + a) \quad \text{and} \quad T_2 = m_2 \cdot (g - a)$