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Homework Help: Angular velocity of a falling rod?

  1. Mar 25, 2010 #1
    A uniform rod of mass 0.62 kg is 6 m long.
    The rod is pivoted about a horizontal, fric-
    tionless pin at the end of a thin extension (of
    negligible mass) a distance 6 m from the cen-
    ter of mass of the rod. Initially the rod makes
    an angle of 48◦ with the horizontal. The rod
    is released from rest at an angle of 48◦ with
    the horizontal, as shown in the figure.
    What is the angular speed of the rod at
    the instant the rod is in a horizontal position?
    The acceleration of gravity is 9.8m/s2 and the
    moment of inertia of the rod about its center
    of mass is I = (1/12)mL2
    Answer in units of rad/s.


    2. Relevant equations

    U = mgh
    KE = .5Iw2


    3. The attempt at a solution

    So, I set U = KE

    mgh = .5Iw2

    Then I plugged in the moment of inertia

    mgh = .5(1/12)(mass of rod)(length of rod)2w2

    Cancelled out the m and solved for the height of the center of mass

    g(6sin(48degrees)) = .5(1/12)mL2w2

    and finally, I plugged in values and solved.

    (9.8)(6sin(48 degrees)) = .5(1/12)(6)2w2

    And solved for w, which I found to equal 5.397339822.

    This is wrong, but I have no idea where my error is. Can you guys help me spot it??
     

    Attached Files:

  2. jcsd
  3. Mar 25, 2010 #2

    thrill3rnit3

    User Avatar
    Gold Member

    parallel axis theorem, since the system is not rotating about its center of mass
     
  4. Mar 25, 2010 #3
    Yeah, you have to use Steiner's theorem to work out the inertia at that point.
     
  5. Mar 25, 2010 #4
    So if I read my stuff right, instead of inserting (1/12)mL2 for momentum, I would insert (1/12)mL2 + md2?
     
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