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Homework Help: Moment of Inertia and Angular Momentum

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data
    This was a TMSCA (Texas Math and Science Competition) from the physics section.

    A thin meter stick of mass 200g is positioned vertically on a frictionless table. It is released, slips, and falls. Find the speed of its center of mass just before it hits the table.


    2. Relevant equations
    To the right of this question there is the equation for the Moment of Inertia of the stick
    I=(1/12)mL^2


    3. The attempt at a solution

    Answer seems to be 2.71 according to the Key
    Answer that I got was 3.13 m/s
     
  2. jcsd
  3. Feb 26, 2010 #2

    rock.freak667

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    The stick is rotating about one end, not its center, so use the parallel axis theorem to get the moment of inertia about one end. That should give you the correct answer.
     
  4. Feb 26, 2010 #3
    Could you show me step by step how to do it
     
  5. Feb 26, 2010 #4

    rock.freak667

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    Post what you did. I am guessing your problem other than the moment of inertia?
     
  6. Feb 27, 2010 #5
    I found the gravitational potential energy at the center of the meter stick (.5 meters)

    PEg= (.2)(9.81).5) = .981

    Then I made .981 equal to the Kinetic Energy of an object in rotation

    .981= (1/2)Iw^2

    Where I is Moment of Inertia and w is angular velocity
    I=MR^2
    1.962=(.2)(.5^2)w^2
    w= 6.2641

    Then V= WR
    V= 6.2641 * (1/2)
    V=3.13 m/s
     
  7. Feb 27, 2010 #6

    rock.freak667

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    your error is there, for a rod, I=(1/12)ML2 about its center. But it is not rotating about its center, it is rotating about its end. Thus you need to use the parallel axis theorem. Do you know the parallel axis theorem?

    I=Ic+mr2
     
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