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Unsure where g term comes from? in this equation

  1. Mar 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a uniform rod of mass 12 kg and length 1.0 m. At its end the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall from rest. Determine
    a) the angular acceleration of the rod as it passes through the horizontal.

    2. Relevant equations
    I started with torque = Ia = ML/2


    I got my equation down to

    a= ML / [2(1/3 M L ^2)]


    3. The attempt at a solution
    however I am unsure what to do after this step, the answer key has it further reduced to 3g/2L, but I am confused where the g comes from?

    Thanks
     
  2. jcsd
  3. Mar 7, 2015 #2

    SammyS

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    Hello trojan15. Welcome to PF !

    In your equation, Iα = ML/2 , what does M represent?

    How is it that the right hand side represents torque?
     
  4. Mar 7, 2015 #3

    gneill

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    Hello trojan15, Welcome to Physics Forums.

    How did you arrive at your torque equation, Ia = ML/2? The units aren't correct.

    I think you'll first want to investigate the moment of inertia ##I## of a slender rod pivoted about an end. Then consider the torque on the rod at the instant it's horizontal.

    [edit: Ah! Beaten to the punch by SammyS!]
     
  5. Mar 8, 2015 #4
    So I went back and examined where I went wrong, however, I hit another snag.



    So I did a= torque/Inertia, which got me to:


    a= rF/ [1/3 ML ^2]

    a= maF/ 1/3ML^2

    and then I get stuck because I have an acceleration value on both sides, and I feel like i'm going in the wrong direction
     
  6. Mar 8, 2015 #5

    gneill

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    What force is causing the torque on the rod?
     
  7. Mar 10, 2015 #6
    So I went back and examined where I went wrong, however, I hit another snag.



    So I did a= torque/Inertia, which got me to:


    a= rF/ [1/3 ML ^2]

    a= maF/ 1/3ML^2

    and then I get stuck because I have an acceleration value on both sides, and I feel like i'm going in the wrong direction
     
  8. Mar 10, 2015 #7

    SammyS

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    Are you saying that r = ma ? Surely not.

    Let's go back to your original post:
    That has an error in it.

    What is the force which pulls down on the rod (thus providing the torque) ?
     
  9. Mar 11, 2015 #8
    I realized the ML/2 part was wrong, and instead substituted
    a= ML / [2(1/3 M L ^2)]
    is this even the right step?
     
  10. Mar 11, 2015 #9

    SammyS

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    Simply answer this question.

    What force produces the torque? Don't even give a formula, just name it.
     
  11. Mar 11, 2015 #10
    The rod rotating?
     
  12. Mar 11, 2015 #11

    SammyS

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    That's not a force.

    Let's change the starting conditions.

    Suppose the rod were to start in a horizontal position. What force would cause the rod to move downward?
     
  13. Mar 12, 2015 #12
    gravity!
     
  14. Mar 12, 2015 #13

    SammyS

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    Correct.

    What force does gravity exert on the rod? (Now write an expression which includes the mass, M, of the rod.)
     
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