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Homework Help: Angular Momentum of a uniform rod

  1. Dec 16, 2009 #1
    A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod.

    What is the final angular speed of the rod?

    Ok, so the answer is 6v/19L, but I don't know how to get there. I have been using angular momentum concepts, but I don't get that answer it wants.

    Been on this question for many hours, thanks in advance :smile:
     
  2. jcsd
  3. Dec 16, 2009 #2

    ideasrule

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    Since there's no external torque, the system's angular momentum is conserved. (The system consists of the bullet and the rod.) So what's the angular momentum of the system before the collision? What is it after, in terms of L and v?
     
  4. Dec 16, 2009 #3
    Quick question--what did you get for the combined I after collision? Should be something like mL^2(1/3 +1/16) where m = mass os rod.
     
  5. Dec 16, 2009 #4
    Before:
    [tex]m_{B}=m_{r}/4[/tex]
    [tex]v=L\omega/2[/tex]
    [tex]\omega=2v/L[/tex]

    [tex]P_{net}=m_{r}L^{2}(0)/3+m_{B}\frac{L}{2}\omega=m_{r}L^{2}v/4[/tex]

    After:
    [tex]I_p=I_{rod}+m_BL^2/4=19/48*m_rL^2[/tex]

    [tex]P_{net}=19/48*m_rL^2\omega[/tex]

    Solution:
    [tex]m_rL^2v/4=19/48*m_rL^2\omega\Rightarrow\omega=12v/19[/tex]

    I got close but I mest up somewhere :frown:
     
  6. Dec 16, 2009 #5
    Yeah I did get that Moment of Inertia
     
  7. Dec 16, 2009 #6

    ideasrule

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    Check this step. Your L's should cancel.
     
  8. Dec 16, 2009 #7

    ideasrule

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    Try calculating the angular moment of the rod and of the bullet separately, then adding them together. Once you correct the previous mistake I pointed out, you'll see why.
     
  9. Dec 16, 2009 #8
    [tex]P_{net}=m_rv/4=m_rL^2\omega/3+m_b(L/2)^2\omega=m_rL^2/3+m_rL^2\omega/16[/tex]

    That gives me:
    [tex]P_{net}=19/48*m_rL^2\omega\Rightarrow\omega=12v/19L^2[/tex]

    Not quite there yet.
     
  10. Dec 16, 2009 #9
    Got it:
    [tex]P_{initial}=m_rvL/8[/tex]

    Then it works. Thanks, I realized that the angular momentum of the bullets should have been the same, your suggestion worked :).
     
    Last edited: Dec 16, 2009
  11. Mar 24, 2010 #10
    Sorry to bump this thread, but could someone explain to me why the last term is divided by four and not 3? In other words, why is the formula for angular inertia changing from (m*l^2)/3 to (m*l^2)/4?
     
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