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Pivoting Stick

  1. Nov 5, 2007 #1
    [SOLVED]Pivoting Stick

    1. The problem statement, all variables and given/known data
    A uniform stick of mass M = 1.3 kg and length L = 2.3 m is pivoted at one end. It is held horizontally and released. Assume the pivot is frictionless. Find the angular acceleration (in rad/s) of the stick immediately after it is released.

    Continuation: Find the magnitude in newtons of the force Fo exerted on the stick by the pivot immediately after it is released.

    http://img134.imageshack.us/img134/1377/prob06azg9.gif

    2. Relevant equations
    torque=Ia



    3. The attempt at a solution

    I set T=Ia
    t=1.15(1.3)(9.81)=14.67
    I=mr^2=1.3(1.15)^2=1.72

    14.67=1.72a
    a=8.53rad/s(s)

    It tells me that is the wrong answer

    For the second part, shouldn't the force be 0?(0is not the right answer)
     
    Last edited: Nov 5, 2007
  2. jcsd
  3. Nov 5, 2007 #2

    Doc Al

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    Staff: Mentor

    OK.
    Not OK. What's the rotational inertia of a thin rod about one end? (When analyzing rotational motion, you can't treat an extended body as if its mass were concentrated at its center of mass.)
     
  4. Nov 5, 2007 #3
    Ah, thanks, using I =1/3mL^2 I get the correct answer.

    So for the second part is the force=9.81(1.3)-6.39(1.15)(1.3)

    or in other words the force on the rod it it was not connecting to a pivot point, minus the acceleration of the rod while it is connected to the pivot point?

    EDIT: i just put in my answer as caclulated above and it was correct.

    thanks Doc Al for the Help
     
  5. Nov 5, 2007 #4

    Doc Al

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    Staff: Mentor

    Cool. The way to think of the second part is just to apply Newton's 2nd law to the vertical direction:
    F - mg = ma
    (where a is the acceleration of the center of mass, which you can figure out from the first answer)
     
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