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An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.1 kg and radius rd = 0.1 m. the other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 4.3 kg and radius rs = 0.18 m. the system is released from rest. 1) what is magnitude of the linear acceleration of the hoop?

Respuesta :

shinmin
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table. 
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center 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 
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 

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