1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Falling rod angular speed

  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


    User Avatar

    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

    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook