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Angular momentum vs energy

  1. Mar 22, 2008 #1
    the problem is the following:

    we have a vertical wooden bar pivoted from the top end, length [tex] 2 l [/tex], mass [tex] M [/tex]

    a bullet with mass [tex] m [/tex] hits it in the middle at velocity [tex] v [/tex] and gets stuck

    i am asked to find the angular velocity [tex] \omega [/tex] of the system bar+bullet immediately after the hit

    i do know this calls for applying the conservation of energy or angular momentum, for some reason however i get different results

    Both if them involve the moment of inertia of the combined system, [tex] I_{\Sigma}=\frac{1}{3} M (2l)^2 + m l^2 = \frac{4}{3} M l^2 + m l^2[/tex]

    Conservation of angular momentum gives me [tex]m v l = \omega I_{\Sigma} [/tex], from which [tex] \omega = \frac{m v l}{I_{\Sigma}} = \frac{m v l}{\frac{4}{3}M l^2+m l^2} = \frac{m}{\frac{4}{3}M+m} \cdot \frac{v}{l} [/tex]

    Whereas conservation of energy says [tex] \frac{m v^2}{2} = \frac{\omega^2 I_{\Sigma}}{2} [/tex], which gives [tex] \omega = \sqrt{\frac{m}{I_{\Sigma}}} v = \sqrt{\frac{m}{\frac{4}{3}M + m}}\cdot \frac{v}{l} [/tex]

    So the big question is where did I mess up this time. I know it's something really basic because I can't see it. Usually i ask a deskmate or someone to have a look if they spot something really simple but since nobody's around I had to come here.

    P.S. while you're at it, why do my [tex] m [/tex], [tex] v [/tex] and [tex] \omega[/tex] look superscripted but [tex] M [/tex] and [tex] 2 l [/tex] don't ?
     
  2. jcsd
  3. Mar 22, 2008 #2

    Doc Al

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    Staff: Mentor

    Your only mistake is in thinking that mechanical energy is conserved--it's not. This is an example of a perfectly inelastic collision.

    To use Latex in the middle of a line of text and have it appear even, use "itex" as your delimiter, not "tex". It gives you [itex]m[/itex] instead of [tex]m[/tex].
     
    Last edited: Mar 22, 2008
  4. Mar 22, 2008 #3
    so the angular momentum one is correct ? (just clarifying...)

    and the loss in energy is the good ol' [tex] \int F ds[/tex] over the distance the bullet travels into the rod !
     
  5. Mar 22, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes.

    Good luck calculating that! To find the loss in energy, just calculate the final KE and compare it to the initial.
     
  6. Mar 22, 2008 #5
    o no wasn't going to do that, good luck indeed
    just realising where the energy went.
    case closed anyway
     
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