Two masses are suspended from a pulley as shown in the figure below (the Atwood machine). The pulley itself has a mass of 0.20 kg, a radius of 0.15 m, and a constant torque of 0.35 m·N due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1 = 0.30 kg and m2 = 0.70 kg? (Neglect the mass of the string.)

workings pls

To find the magnitude of the acceleration of the suspended masses, we can use Newton's second law of motion. According to this law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, we have two masses, m1 and m2, which are connected by a string and suspended from a pulley. The pulley itself has a mass, radius, and a constant torque.

To start, we need to draw a free-body diagram for each of the masses. Let's consider m1 first.

Free-body diagram for m1:
- There is tension in the string pulling m1 upward.
- There is the weight force (mg1) pulling m1 downward.

Now, let's consider m2.

Free-body diagram for m2:
- There is tension in the string pulling m2 downward.
- There is the weight force (mg2) pulling m2 downward.

From the free-body diagram of m1, we can write the equation for its net force:

T - mg1 = m1 * a1

From the free-body diagram of m2, we can write the equation for its net force:

mg2 - T = m2 * a2

Next, we need to consider the pulley. The friction between the rotating pulley and its axle exerts a constant torque. The torque can be calculated as the product of the pulley's radius and the torque force:

τ = r * F_torque

Since torque (τ) is the moment of inertia (I) times the angular acceleration (α), and the pulley's mass is given, we can calculate its moment of inertia as:

I = 0.5 * m_pulley * r^2

where m_pulley is the mass of the pulley and r is its radius.

We can equate the torque to the torque force as:

τ = F_torque

Since τ = I * α and F_torque = r * F_n, where F_n is the normal force acting on the pulley, we can rewrite the equation as:

I * α = r * F_n

Plugging in the values:

(0.5 * m_pulley * r^2) * α = r * F_n

Simplifying the equation:

m_pulley * α = 2 * F_n

Now, we can calculate the net force acting on the pulley:

F_net = F_n - F_torque

And equating it to the mass of the pulley times its acceleration, we get:

m_pulley * a_pulley = F_n - F_torque

We can substitute F_n from the previous equation:

m_pulley * a_pulley = 2 * F_net - F_torque

Now, we have equations for m1, m2, and the pulley. We can solve these simultaneous equations to find the acceleration of the system.

The last step is to sum the accelerations of m1, m2, and the pulley (since they are all connected) and equate it to zero:

a1 + a_pulley - a2 = 0

Substituting the equations we derived earlier, we get:

(T - mg1) / m1 + (2 * F_net - F_torque) / m_pulley - (mg2 - T) / m2 = 0

Now, we can solve this equation to find the acceleration of the system.