Pulley, Motor and Weight Torque Problem

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SUMMARY

The discussion centers on calculating the acceleration of a 1140 kg mass being lowered by a motor, which exerts a tension of 10,400 N in the cable. The pulley has a moment of inertia of 79.7 kgm² and a radius of 0.741 m. The correct acceleration, derived from the equations of motion and torque, is 0.609 m/s². The user expresses confusion regarding the correctness of their solution when submitted to LON-CAPA.

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Pulley and Motor - Torque Problem

Homework Statement


When the motor in the figure below lowers the m = 1140kg mass, it produces a tension of 1.04E+4N in the cable on the right side of the pulley. The pulley has a moment of inertia of 79.7kgm^2 and a radius of 0.741m. The cable rides over the pulley without slipping. Determine the acceleration of the m = 1140kg mass. Use g=9.81m/s^2.
Draw free-body diagrams of the mass and the pulley. Do not assume that the tension in the cable is the same on both sides of the pulley.



Homework Equations


T=rxf

a=α*r where α = angular acceleration
----> a/r=α

The Attempt at a Solution


mg-[itex]T_1{}[/itex] = ma
[itex]T_2{}[/itex] = 10 400 N

r*([itex]T_1{}[/itex]-[itex]T_2{}[/itex])=I*α
r*([itex]T_1{}[/itex]-[itex]T_2{}[/itex])=I*(a/r)
[itex]T_1{}[/itex]-[itex]T_2{}[/itex]=I*(a/r^2)
(11 183 - 11 40a) - (10 400) = (79.7/0.741^2)a
783-1140a = 145.15a
783 = 1285.15a
a= 0.609 m/s/s

I'm not sure what I'm doing wrong. The steps all make sense to me, but when I've tried submitting the answer on LON-CAPA it says it's incorrect. Anyone have any ideas?
 
Last edited:
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Can you add a picture, sketch, or scan of the set-up. One picture is worth a lot of words.
 

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