Analysis of the problem:

In this problem, we have a string wrapped around a cylinder of mass m and radius r. The string is also connected to a block of the same mass m with the help of another pulley. The system is initially at rest, and we need to find the angular acceleration of the cylinder.

Free body diagram:

To analyze the system, let's draw the free body diagram of the cylinder and the block.

- The cylinder experiences a tension force T from the string, directed tangentially to the cylinder's surface.
- The block experiences a tension force T from the string, directed vertically upwards.
- Both the cylinder and the block experience the force of gravity, directed downwards.

Equations of motion:

Next, let's write the equations of motion for the cylinder and the block.

- For the cylinder, we can write the torque equation:
τ = Iα,
where τ is the net torque acting on the cylinder, I is the moment of inertia of the cylinder, and α is the angular acceleration of the cylinder.

The net torque acting on the cylinder is given by:
τ = TR - f,
where TR is the torque due to the tension force and f is the frictional torque.

The torque due to the tension force is given by:
TR = rT,
where r is the radius of the cylinder.

The frictional torque is given by:
f = μmgR,
where μ is the coefficient of friction between the cylinder and the surface it is rolling on, m is the mass of the cylinder, and R is the radius of the cylinder.

Substituting the expressions for TR and f into the torque equation, we get:
rT - μmgR = Iα.

- For the block, we can write the equation of motion:
T - mg = ma,
where T is the tension force, m is the mass of the block, g is the acceleration due to gravity, and a is the acceleration of the block.

Since the cylinder and the block are connected by the string, their accelerations are the same, i.e., a = αr.

Substituting the expression for a into the equation of motion for the block, we get:
T - mg = mαr.

Solving the equations:

We have two equations:
rT - μmgR = Iα, (1)
T - mg = mαr. (2)

To solve these equations, we need to eliminate T. From equation (2), we can express T in terms of α:
T = mg + mαr.

Substituting this expression for T into equation (1), we get:
r(mg + mαr) - μmgR = Iα.

Simplifying the equation, we have:
mgr + mαr^2 - μmgR = Iα.

Rearranging the terms, we get:
α(r^2 + R) = (mgr - μmgR)/I.

Finally, we can express the angular acceleration α as:
α = (mgr - μmgR)/(I(r^2 + R)).

Conclusion:

The angular acceleration of the cylinder is given by:
α =