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Falling rod

  1. Feb 6, 2006 #1
    A thin 1.0-m rod pivoted at one end falls (rotates) frictionlessly from a vertical position, starting from rest. What is the angular speed of the rod when it is horizontal? [Hint: Consider the center of mass and use the conservation of mechanical energy.]

    So far I'm thinking that intial potential energy = final kinetic energy, so
    mgh = K, where h = 0.5 m (the center of mass)

    I can't seem to use K = 1/2 * Icm * w2 + 1/2 * M * vcm2
    because I don't know vcm... or can I find it?
  2. jcsd
  3. Feb 6, 2006 #2


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    Staff: Mentor

    Can you express w(cm) in terms of v(cm) and then solve the equation?
  4. Feb 6, 2006 #3
    hmm.. i'll express v in terms of w instead...

    mgh=.5Iw2 + .5M(rw)2
    2mgh=Iw2 + Mr2w2
    w2=(2mgh)/(I + Mr2)

    h = 0.5m, I = 1/2 ML2, where L = 0.5m, r = 0.5m

    so cancelling out M,
    w2 = (2gh)/(1/2 * 0.52 + 0.52)
    w = 5.1 rad/s

    but the answer is supposed to be 5.4 rad/s...
  5. Feb 7, 2006 #4

    Doc Al

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    Staff: Mentor

    Recheck your expression for the moment of inertia of a rod about its center of mass.

    Realize that you can also treat the rod as being in pure rotation about the pivot, so:
    [tex]m g h = 1/2 I \omega^2[/tex]
    (where I is the moment of inertia about the pivot)
  6. Feb 7, 2006 #5
    I should have used I = 1/3 ML2

    thanks! I got the right answer now.
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