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Inelastic Collision and Rotational Inertia

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A mass M slides down a smooth surface from height h and collides inelastically with the lower end of a rod that is free to rotate about a fixed axis at P as shown below. The mass of the rod is also M, the length is L, and the moment of inertia about P is ML2/3.

    The angular velocity of the rod about the axis P just after the mass sticks to it will be:
    (A) √(gh/2)
    (B) √(2gh)/L
    (C) (3/4L)√(2gh)
    (D) (3L)/√(2gh)
    (E) (9√(2gh))/L

    2. Relevant equations

    There's a bunch.

    3. The attempt at a solution

    I started with conservation of energy, so mgh=(1/2)mv2 and thus v=√(2gh). I don't really know where to go next. I know inelastic means momentum is conserved and kinetic energy is not.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

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    Hi Victorzaroni! :smile:
    angular momentum is also conserved :wink:
     
  4. Feb 21, 2012 #3
    oh okay. So The angular momentum of the block when it gets to the bottom equals to angular momentum of the rod when it hits it? How do I express that in an equation? Is the r in w=v/r the height of the circle?
     
  5. Feb 21, 2012 #4

    tiny-tim

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    of the rod-plus-block, yes
    (you mean half the height?) :smile:

    what does r have to do with it? this is a rod, you'll have to use moment of inertia
     
  6. Feb 21, 2012 #5
    Okay. So: rmv=Iw=>rmv=(M+M)r2w?
     
  7. Feb 21, 2012 #6

    tiny-tim

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    (have an omega: ω :wink:)
    uhh? :confused:

    what's the moment of inertia of a rod?
     
  8. Feb 22, 2012 #7
    It's (1/3)ML^2 (about the end) I meant to write that.
     
  9. Feb 22, 2012 #8

    tiny-tim

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    then do so! :smile:
     
  10. Feb 22, 2012 #9
    Do what?
     
  11. Feb 22, 2012 #10

    tiny-tim

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    combine …
     
  12. Feb 22, 2012 #11
    okay. So: rmv=Iw => rm√(2gh)=(1/3)(m+m)L2
    I'm just confused about what the r is. Is that L?
     
  13. Feb 22, 2012 #12

    tiny-tim

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  14. Feb 22, 2012 #13
    Unless I made a math error, I end up with (3√(2gh))/(2L), and that's not a choice. :(
     
  15. Feb 22, 2012 #14

    tiny-tim

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    oh wait, it's not …
    … the "1/3" is only for the rod, not for the other mass
     
  16. Feb 22, 2012 #15
    Oh okay yea I didn't think that that made much sense. So Im just equating: Lmsqrt(2gh)=(1/3)mL^2
     
  17. Feb 23, 2012 #16

    tiny-tim

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    but you still need to add on the angular momentum for the other mass
     
  18. Feb 23, 2012 #17
    But the other mass is sliding? So how does it have angular momentum?
     
  19. Feb 23, 2012 #18

    tiny-tim

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    ah

    everything has angular momentum (about any point that isn't on its line of motion)

    conservation of angular momentum always means conservation of angular momentum about a particular point

    you've chosen, as that point, the pivot at the top of the rod

    you chose it because there's an unknown force at the pivot, and the only way you can avoid it is to make its moment zero

    so now you need to find the angular momentum of the sliding mass about the pivot

    i thought you already knew that, and had done it? :confused:
     
  20. Feb 23, 2012 #19
    Well that's what I did and it was wrong. You said the right side of the equation was wrong because of the (m+m). I get now why we can equate things because of their angular momentum, but im missing a piece and I don't what it is
     
  21. Feb 23, 2012 #20

    tiny-tim

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    yes, because the 2nd m was wrong, it should have been m/3

    (the 1st m was right)
     
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