Answer:
[tex]a = 2.77~{\rm m/s^2}[/tex]
Explanation:
Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.
The moment of inertia of a ring is
[tex]I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127[/tex]
The mass on the left is heavier, that is the pulley is rotating counterclockwise.
By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.
[tex]\tau = I \alpha[/tex]
Here, the net torque is the sum of the weight on the left and the weight on the right.
[tex]\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}[/tex]
Applying Newton's Second Law gives the angular acceleration
[tex]\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}[/tex]
The relation between angular acceleration and linear acceleration is
[tex]a = \alpha R[/tex]
Then, the linear acceleration of the masses is
[tex]a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}[/tex]
