What is the solution to the Atwood machine problem with given variables?

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The discussion centers on solving an Atwood machine problem involving two masses and a pulley with specific variables. The user is struggling to derive the correct acceleration of the masses, which is stated to be 1.2 m/s². They initially attempted to use the net torque equation but are unable to reconcile their calculations with the provided answer. A suggestion is made to utilize free-body diagrams for each mass to accurately determine the tensions in the strings, emphasizing that tension equations only hold true when acceleration is zero. The conversation highlights the complexity of the Atwood machine problem and encourages revisiting fundamental principles to find the solution.
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I have been working on this problem off-and-on for a couple of days now and cannot seem to get the correct answer. The problem is an Atwood machine problem. The following are the variables:

Mass of pulley = .20 kg
Radius of pulley = .15 m
Clockwise frictional torque = .35 m*N
Mass 1 on right side = .40 kg
Mass 2 on left side = .80 kg

The problem wants to know the acceleration of the masses. The correct answer is 1.2 m/s^2.

My thought was to set the net torque of the system equal to moment of inertia times angular acceleration. Take that number and multiply by the radius to get the tangential acceleration. I have tried the following to get the angular acceleration:

T2 - T1 - Tf = Iα

Which equates to:
m2gR - m1gR - Tf = Iα (solve for α)

With the above formula, which must obviously be the wrong approach, I cannot seem to match the answer in the back of the book. What am I missing here?
 
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Ah, the dreaded Atwood machine! In our General Physics lab, we call it the "toe-cruncher." :eek:

T_1 = m_1 g and T_2 = m_2 g only if the acceleration is zero. Do a free-body diagram for each mass to figure out the tension in each string in terms of the acceleration.
 
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If you search the threads, I'm very sure you will find that someone else has already dealt with this kind of problem (I know because I've posted such problems before).
 
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