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Homework Help: Rotational Motion

  1. Feb 17, 2010 #1
    I ran across this problem studying for my exam, and I'm not sure how to solve it:

    1. The problem statement, all variables and given/known data

    A thin rod of Mass M and length L is supported by a pivot a distance of L/4 from its left end. A second support a distance 3L/4 from the left end prevents it from falling. A clay ball of mass M/2 drops from a height H above the beam, strikes the left end and completely sticks to the rod. The beam swings upward to make an angle [tex]\Theta[/tex] with the horizontal.

    a. Calculate the moment of Inertia, Iz, of the beam and clay ball together around the pivot point, after the clay sticks to the beam.

    b. Calculate the angle [tex]\Theta[/tex] to which the beam rises. Express your answer in terms of [tex]\omega[/tex], V, and/or Iz.

    3. The attempt at a solution

    For a, I was thinking of finding the moment of Inertia at the new center of mass after the ball sticks, then using the parallel axis theorem to get Iz, but that seems too messy..

    For b, I think there must be some way to do it with energy conservation, but I really don't know where to start.

    Thanks for the help.
  2. jcsd
  3. Feb 17, 2010 #2


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    Hi Poisonous! :smile:

    (have a theta: θ and an omega: ω :wink:)
    That's far too complicated …

    just calculate the moment of inertia of the whole (about the pivot) as the sum of the moments of inertia of its parts. :wink:
    How can energy be conserved? It's a perfectly inelastic collision (the two final velocities are the same).

    You can use conservation of energy after the collision, but for the collision itself, you need … ? :smile:
  4. Feb 17, 2010 #3
    For a: Iz = 1/12ML2 + (M/2)(L/4)2


    For b: For the collision itself you would need conservation of angular momentum? So,

    Iclay * ωclay + 0 = Iz * ωz

    so, ωz = MVL/8Iz

    So you know the rotational kinetic energy at the beginning, and that it should be all potential energy at the point of [tex]\theta[/tex], but how do you relate the energies with [tex]\theta[/tex]?

    Thanks for the help.
    Last edited: Feb 17, 2010
  5. Feb 17, 2010 #4


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    No, 1/12 would be at the end, you want L/4 from the end.
    (what happened to that θ i gave you? :wink:)

    You relate θ to h. :smile:
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