Acceleration of a mass lowered by a motor (Help with Non-Ideal Pulley)

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Homework Help Overview

The discussion revolves around a non-ideal pulley system involving a mass being lowered by a motor. The problem incorporates concepts of torque and Newton's second law, with specific parameters provided for the mass, tension, moment of inertia, and radius of the pulley.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the formulation of equations based on free body diagrams for both the mass and the pulley. There is an exploration of the implications of sign conventions in the context of the problem, particularly regarding the direction of acceleration.

Discussion Status

Some participants have provided feedback on the original poster's approach, noting that while the equations were well-developed, there may be confusion regarding the sign of the acceleration. The discussion reflects an ongoing exploration of the problem rather than a resolution.

Contextual Notes

Participants note the importance of sign conventions in physics problems, suggesting that the choice of positive direction can significantly affect the interpretation of the results. There is also mention of the specificity of the problem's requirements, which may lead to confusion.

Kermit_the_Phrog
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Summary: Non - ideal pulley question, should be easy but has got me good

Hey guys, looking for some help on this pulley question. It involves torque, and works with Newton's 2nd law in conjunction with a non-ideal pulley.

Text of question:
" When the motor in the figure below lowers the m = 1160kg mass, it produces a tension of 1.06E+4N in the cable on the right side of the pulley. The pulley has a moment of inertia of 75.7kgm^2 and a radius of 0.747m. The cable rides over the pulley without slipping. Determine the acceleration of the m = 1160kg mass. Use g=9.81m/s^2. "

Diagram:
1570661347930.png


Attempt

I found free body diagrams for both the mass and the pulley, and boiled them down to two equations, two unknowns as follows -

T2 = mg-ma

and

(r^2)T2 - (r^2)T1 = Ia

But when I added the equations together (subbed the first into the second), I got a final answer a = 0.602 m/s^2 , which was wrong.*Note - my final formula was:

a = (r^2)(mg - T1) / (I + (r^2)m)can anyone help me out?

[Moderator's note: Moved from a technical forum and thus no template.]
 
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Kermit_the_Phrog said:
But when I added the equations together (subbed the first into the second), I got a final answer a = 0.602 m/s^2 , which was wrong.

So as it Turns out, my answer was correct, BUT my sign was wrong: the correct answer was -0.602 m/s^2.

This makes me more confused honestly.

can anyone help me out explain this?
 
In your equations, you chose downward and ccw as the positive direction , which is fine , and that’s the way I would have done it. But apparently the solution likes downward to be in the negative direction, per typical convention. But since it was given that the mass is lowered, its picky in my mind to not consider your answer as correct.
 
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PhanthomJay said:
In your equations, you chose downward and ccw as the positive direction , which is fine , and that’s the way I would have done it. But apparently the solution likes downward to be in the negative direction, per typical convention. But since it was given that the mass is lowered, its picky in my mind to not consider your answer as correct.

Thank you very much, it's good to know that I didn't make any sort of fatal error and that the body of my problem solving was okay. I guess the problem was just very picky on those minute details. Have a great rest of your day!
 
Thank you. You did a very good job in developing your equations. Ribbet, ribbet...
ribbet_ribbet.jpg
ribbet_ribbet.jpg
 
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