An Atwood's machine consists of blocks of masses m1 = 13.0 kg and m2 = 19.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 7.30 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

To find the acceleration of the system and the tension in the cord, we can use Newton's second law for each mass and the torque equation for the pulley.

Let's start by finding the acceleration of the system. We'll assume that the positive direction is upwards.

1. Calculate the net force on each mass:
- For mass m1: F_net1 = m1 * g - T, where g is the acceleration due to gravity (9.8 m/s^2) and T is the tension in the cord pulling upwards.
- For mass m2: F_net2 = m2 * g - T, where T is the tension in the cord pulling downwards.

2. Applying Newton's second law to each mass:
- For mass m1: F_net1 = m1 * a, where a is the acceleration of the system.
- For mass m2: F_net2 = m2 * a, where a is the acceleration of the system.

Since the magnitudes of the forces are the same, we can equate the two equations for F_net and solve for the acceleration:

m1 * g - T = m2 * g + T
m1 * g = m2 * g + 2T
m1 * g - m2 * g = 2T
g * (m1 - m2) = 2T
T = (m1 - m2) * g / 2

Now, we can find the tension in the cord.

3. Calculate the moment of inertia of the pulley:
- The moment of inertia of a solid cylinder is given by I = (1/2) * M * r^2, where M is the mass of the pulley and r is the radius of the pulley.

4. Apply the torque equation to the pulley:
- The net torque on the pulley is equal to the product of its moment of inertia and its angular acceleration.
- The torque generated by the tension in the cord is equal to T * r (since the radius of the pulley is r).
- The torque generated by the weight of the pulley, which acts at its center, is equal to (1/2) * M * g * r.

Setting up the torque equation:

T * r - (1/2) * M * g * r = I * alpha

where alpha is the angular acceleration.

5. Calculate the angular acceleration:
- The angular acceleration can be obtained by dividing the linear acceleration by the radius of the pulley: alpha = a / r.

6. Substitute the moments of inertia and solve for the angular acceleration:
T * r - (1/2) * M * g * r = [(1/2) * M * r^2] * (a / r)
T - (1/2) * M * g = (1/2) * M * a
T = (1/2) * M * g + (1/2) * M * a
T = (1/2) * M * (g + a)

Now, we have the tension in the cord in terms of the acceleration. We can substitute the expression for the tension in the previous equation we obtained for T to solve for the acceleration:

(m1 - m2) * g / 2 = (1/2) * M * (g + a)

Now, we can solve for the acceleration a:

(m1 - m2) * g = M * (g + a)
(m1 - m2) * g = M * g + M * a
a = (m1 - m2) * g / M

By substituting the given values for m1, m2, M, and g, you can find the acceleration a.