To solve the problem of an Atwood's machine with the given parameters, we will apply Newton's second law and the rotational dynamics of the wheel.

Given Data:

• Mass of block $m_1 = 4.00 \, \text{kg}$
• Mass of block $m_2 = 2.00 \, \text{kg}$
• Moment of inertia of the wheel $I$ (we'll leave it as a variable for now)
• Radius of the wheel $r = 0.120 \, \text{m}$
• Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$

Step 1: Equations of Motion

For block $m_1$ (the heavier block): $m_1 g - T_1 = m_1 a$

For block $m_2$ (the lighter block): $T_2 - m_2 g = m_2 a$

Where:

• $T_1$ is the tension in the cord on the side of $m_1$
• $T_2$ is the tension in the cord on the side of $m_2$
• $a$ is the linear acceleration of the blocks

Step 2: Torque on the Wheel

The torque $\tau$ exerted by the tensions on the wheel is given by: $\tau = T_1 r - T_2 r = I \alpha$ Where $\alpha$ is the angular acceleration of the wheel. The relationship between linear acceleration $a$ and angular acceleration $\alpha$ is: $a = r \alpha \quad \Rightarrow \quad \alpha = \frac{a}{r}$

Step 3: Substituting $\alpha$

Substituting $\alpha$ into the torque equation: $T_1 r - T_2 r = I \left(\frac{a}{r}\right)$ Simplifying: $T_1 - T_2 = \frac{I a}{r^2} \quad \text{(1)}$

Step 4: Solve the System of Equations

Now, we have three equations:

1. $m_1 g - T_1 = m_1 a$ (2)
2. $T_2 - m_2 g = m_2 a$ (3)
3. $T_1 - T_2 = \frac{I a}{r^2}$ (1)

From Equations (2) and (3):

Rearranging for $T_1$ and $T_2$: $T_1 = m_1 g - m_1 a \quad \text{(4)}$ $T_2 = m_2 g + m_2 a \quad \text{(5)}$

Substitute (4) and (5) into (1):

$(m_1 g - m_1 a) - (m_2 g + m_2 a) = \frac{I a}{r^2}$ $m_1 g - m_2 g - m_1 a - m_2 a = \frac{I a}{r^2}$ $(m_1 - m_2)g = (m_1 + m_2 + \frac{I}{r^2}) a$

Step 5: Solving for $a$:

$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + \frac{I}{r^2}}$

Step 6: Substitute Values

Substituting the known values: $a = \frac{(4.00 - 2.00) \cdot 9.81}{4.00 + 2.00 + \frac{I}{(0.120)^2}}$ $= \frac{2.00 \cdot 9.81}{6.00 + \frac{I}{0.0144}} = \frac{19.62}{6.00 + \frac{I}{0.0144}} \quad \text{(6)}$

Step 7: Finding Tensions

Once $a$ is found, substitute back into (4) and (5) to find $T_1$ and $T_2$.

Step 8: Angular Acceleration

Finally, the angular acceleration $\alpha$ can be found from: $\alpha = \frac{a}{r}$

Summary

To find specific numerical values for $a$, $T_1$, $T_2$, and $\alpha$, we need the value of $I$. Please provide that if available, or I can proceed with calculations assuming a certain moment of inertia. Do you want to continue with specific assumptions or provide the moment of inertia?