An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass md = 1.8 kg and radius rd = 0.08 m. the other end of the string is attached to a massless axle through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 3.6 kg and radius rs = 0.2 m. the system is released from rest. 1) what is magnitude of the linear acceleration of the hoop

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N 
the slow masses that must be quicker are the pulley, ring, and the rolling sphere. 
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m 
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m 
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

answer and explanation;

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