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Rotational moments of Inertia

  1. Dec 5, 2004 #1
    Pivot point on a Uniform Rod and Acceleration

    I'm stuck on this question, which seems like it should be fairly simple:

    A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.
    theres a graphic of it here:
    [​IMG]

    ok so what I've done so far is:
    Rmg(sinθ) = Iα
    Rmg(sinθ) = (1/3)mL^2α

    getting the expression I = (1/3)mL^2 from my text book for a uniform rod with a pivot at its end.

    working that out I get:
    (0.575)(9.8)(sin21) = (1/3)(1.15)^2 α
    α = 4.58 rad/s

    then converting α to a with:
    a = Rα
    a = (.575)(4.58)
    a = 2.63 m/s^2

    but apparently this isn't right.
    can anyone help me out?
     
    Last edited: Dec 5, 2004
  2. jcsd
  3. Dec 5, 2004 #2
    nevermind... my stupid mistake should have used cos instead of sin :redface:
     
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