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Angular acceleration of a falling arm

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Assume your arm is a solid rod, 0.740 m in length, free to rotate about your shoulder (so its
    moment of inertia is (1/3)ML^2). If you hold your arm out sideways, and then completely relax
    your muscles, what is the initial downward angular acceleration of your arm?

    2. Relevant equations
    τ=LF=Iα
    F=mg
    I=(1/3) ML^2

    3. The attempt at a solution
    I'm not given a specific formula to use, but I think the above formula for Torque applies. I'm not using conservation of energy because I dont have a value for the height to use for potential energy.

    If I plug the formulas for inertia and Fg into the torque formula:

    Lmg=(1/3)ML^2 α

    α=3g/L = (3*9.81)/0.740 = 39.77 but the answer key says the answer is 19.9 (half).


    What am I doing wrong here?
     
    Last edited: Mar 24, 2014
  2. jcsd
  3. Mar 24, 2014 #2

    SammyS

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    What is the location of the center of mass of the arm ?
     
  4. Mar 24, 2014 #3
    L/2, the very center of the length?
     
  5. Mar 24, 2014 #4

    SammyS

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    Yes.

    That's where gravity effectively acts.

    So, what is the torque produced by gravity about the shoulder ?
     
  6. Mar 24, 2014 #5
    τ=(L/2)F=(L/2)mg
    But the value of the length in the formula for inertia will remain the full length right??
     
  7. Mar 24, 2014 #6

    SammyS

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    Right !
     
  8. Mar 24, 2014 #7
    awesome!! Thank you so much for the help!!
     
  9. Mar 24, 2014 #8

    SammyS

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    You're welcome.

    By The Way, welcome to PF !
     
  10. Mar 24, 2014 #9
    Thanks!
     
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