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Give me back my bullets

  1. Nov 18, 2005 #1

    kp

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    A bullet of mass m and an initial speed v strikes, and is embedded in, the end of an uniform rod of mass 2m and length L originally at rest. The pivots about a fixed axis at it center

    what is the angular speed of the rod after the collision? (this, i dont know)

    what is the angular momentum of the bullet/rod system with respect to the axis through the rods's pivot? (this, i think i know,but probably dont)

    (what i know) This is an conservation of angular momentum/inelastic collision problem? I done some of these before with kids jumping on merry-go-rounds, an such, but it is the symbolic
    answers i think throws me off. The speed of the system i terms of what. I know the policy here is to show your work, however work is a vector quantity, and i have been going in around in circles with this problem. So my net work as of this moment is

    mrv = (1/12(2m)l^2 + mr^2) wf
    solve for then wf, and wf equals all this mess --> mrw/( 1/12(2m)l^2 +mr^2) which i know isn't right. So i'm counting on
    someone much smarter than I, to help me out of the woods on this easy problem
     
  2. jcsd
  3. Nov 19, 2005 #2

    andrevdh

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    Consider the conservation of angular momentum. Bullet before collision
    [tex]L_b=I_b\omega_b=\frac{mLv}{2}[/tex]
    for [itex]r=L/2[/itex]. After collision
    [tex]L_b=I_c\omega_c[/tex]
    where the subscript refer to the combined (bullet and rod) quantities.
     
  4. Nov 19, 2005 #3

    cepheid

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    Can you clarify something andrevdh? If we're saying that the angular momentum of the bullet-rod system is conserved, and the rod is clearly rotating afterwards, then there must be some initial angular momentum. You seem to have defined this initial angular momentum as, "the angular momentum of the bullet around the pivot point of the rod,"

    Lb = mLv/2

    But I find it counter-intuitive to speak of this angular momentum, "of the bullet," because the bullet is not travelling in a circular path of radius L/2 about the pivot point, it is travelling in a straight line. I understand that this is nevertheless the equivalent angular momentum that the bullet transfers to the bullet-rod, but it's almost as if this angular momentum doesn't really "exist" until the instant that the bullet strikes the rod, and translational motion becomes rotational motion.
     
  5. Nov 20, 2005 #4

    kp

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    :confused: hmmm...if mL/2v = momentum of the bullet before the collision..?

    and L/2 =r, then mL/2v = mvr = back to where i started...:cry: :cry:
     
  6. Nov 20, 2005 #5

    andrevdh

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    In kinematics we define instantaneous quantities by investigating the motion in the limit [itex]\Delta t \rightarrow \ 0[/itex]. Surely, in the limit we can say that the bullet is rotating around the pivot point for such an instantaneous interaction. We can work the angular momentum out for any moving object, independant of it's path. In most case the radius would change, but in this case it is virtually constant for the interaction which is of short duration. If you draw the radius vector from the pivot point to the bullet it will rotate about the pivot as it follows the bullet.
     
    Last edited: Nov 20, 2005
  7. Nov 20, 2005 #6

    andrevdh

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    Why do you say that this is'nt right (with r = L/2)? It seems fine to me (with v in stead of w), just develop it further.
     
    Last edited: Nov 20, 2005
  8. Nov 20, 2005 #7

    kp

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    instantaneous angular momention..I get that know
    ..so that's why you can define the linear momentun of the bullet as mvr.

    I am assuming my answer is not right. This is an problem on an extra-credit worksheet, so I dont have the answer for reference. So, your help is greatly appreciated.
     
  9. Nov 20, 2005 #8

    andrevdh

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    What do you get for [itex]\omega_c[/itex] if you develop the equation?
    Do you need any more help?
     
  10. Nov 20, 2005 #9

    kp

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    well after further developement this mess--> w = (1/12(2m)L^2 +mr^2)

    cleans up too w = (3/5)v/r

    I dont know if its right, but it looks nicer :approve:
     
    Last edited: Nov 20, 2005
  11. Nov 21, 2005 #10

    andrevdh

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    Yes, I get the same (with r = L/2). Which means you could have done it all by yourself! You were only a bit unsure of your result. This can be avoided if you apply your physics and mathematics at each step of the way. The magnitude of the angular momentum, [itex]L[/itex], of a moving object about a pivot is is given by
    [tex]L=rp\sin(\theta)[/tex]
    , where p is it`s linear momentum and theta is the angle between the radius vector and p, which can be evaluated for any moving object, even if it is not moving in a circle. Good luck with the physics!
     
    Last edited: Nov 21, 2005
  12. Nov 21, 2005 #11

    kp

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    "big-ups" to you andre for keeping headed in the right direction :cool:

    2.) what is the angular momentum of the bullet/rod system with respect the the rods pivot? hmm..
    not sure about this..
     
    Last edited: Nov 21, 2005
  13. Nov 21, 2005 #12

    andrevdh

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    Well you have answered this one also yourself. In your equation above you have the angular momentum of the bullet on the lefthand side of the equation. On the righthand side you have the final angular momentum of the combined system. So what you are saying is that the angular momentum of the system is conserved. What question 2 is asking is for the right hand side of the equation, but wait, it is equal to the lefthand side which is all known quantities!
     
  14. Nov 21, 2005 #13

    kp

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    yeah...see, I am making this problem much harder
    than it really is, Thanks again!
     
  15. Nov 22, 2005 #14

    andrevdh

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    It has been a pleasure.
     
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