To find the angular acceleration $\alpha$ of the two-disk system, we need to analyze the torques created by the two hanging masses and then apply Newton's second law for rotational motion:

$\sum \tau = I \alpha$

where $I$ is the total moment of inertia of the two-disk system, and $\tau$ is the net torque acting on the system. We’ll go step-by-step through the solution.

Step 1: Find the Moments of Inertia of the Disks

For a solid disk rotating about an axis through its center, the moment of inertia $I$ is given by:

$I = \frac{1}{2} M R^2$

where $M$ is the mass of the disk and $R$ is its radius.

1. Moment of inertia of the larger disk ($I_{\text{large}}$): $I_{\text{large}} = \frac{1}{2} M_{\text{large}} R_{\text{large}}^2 = \frac{1}{2} \cdot 1.3 \, \text{kg} \cdot (0.20 \, \text{m})^2 = 0.026 \, \text{kg} \cdot \text{m}^2$

2. Moment of inertia of the smaller disk ($I_{\text{small}}$): $I_{\text{small}} = \frac{1}{2} M_{\text{small}} R_{\text{small}}^2 = \frac{1}{2} \cdot 0.50 \, \text{kg} \cdot (0.10 \, \text{m})^2 = 0.0025 \, \text{kg} \cdot \text{m}^2$

So, the total moment of inertia of the system is:

$I_{\text{total}} = I_{\text{large}} + I_{\text{small}} = 0.026 + 0.0025 = 0.0285 \, \text{kg} \cdot \text{m}^2$

Step 2: Calculate the Torque Produced by Each Hanging Mass

The torque $\tau$ produced by a force $F$ at a radius $R$ is given by:

$\tau = R F$

Since the force on each hanging mass is due to gravity, we have $F = mg$ for each mass, where $g = 9.8 \, \text{m/s}^2$.

1. Torque due to $m_2 = 0.230 \, \text{kg}$ on the larger disk: $\tau_{\text{large}} = R_{\text{large}} \cdot m_2 \cdot g = 0.20 \, \text{m} \cdot 0.230 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 0.4508 \, \text{N} \cdot \text{m}$

2. Torque due to $m_1 = 0.060 \, \text{kg}$ on the smaller disk: $\tau_{\text{small}} = R_{\text{small}} \cdot m_1 \cdot g = 0.10 \, \text{m} \cdot 0.060 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 0.0588 \, \text{N} \cdot \text{m}$

Since the masses are trying to rotate the system in opposite directions, the net torque $\tau_{\text{net}}$ will be the difference between these torques:

$\tau_{\text{net}} = \tau_{\text{large}} - \tau_{\text{small}} = 0.4508 - 0.0588 = 0.392 \, \text{N} \cdot \text{m}$

Step 3: Solve for the Angular Acceleration

Using Newton’s second law for rotational motion, we have:

$\alpha = \frac{\tau_{\text{net}}}{I_{\text{total}}}$

Substitute $\tau_{\text{net}} = 0.392 \, \text{N} \cdot \text{m}$ and $I_{\text{total}} = 0.0285 \, \text{kg} \cdot \text{m}^2$:

$\alpha = \frac{0.392}{0.0285} \approx 13.75 \, \text{rad/s}^2$

Final Answer

The magnitude of the angular acceleration of the disks is approximately:

$\alpha \approx 13.75 \, \text{rad/s}^2$

Would you like further details on any specific step, or help with related questions? Here are some follow-up questions to consider:

1. How would the angular acceleration change if the masses of the hanging weights were increased?
2. What would happen to the system if the radii of the disks were doubled?
3. How does the rotational inertia of each disk affect the overall angular acceleration?
4. If there was friction at the axis, how would that affect the net torque and angular acceleration?
5. How can we experimentally measure angular acceleration in such a system?

Tip: When dealing with rotating systems, always analyze each force’s direction to determine if they add or subtract in producing the net torque.