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Thin rod at rest gets shot with bullet and rotates? Ang momentum problem? help!

  1. Nov 17, 2012 #1
    A uniform thin rod of length 0.500 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact?

    Solution with diagram on page 4m problem #11.55:
    http://www.ifm.liu.se/edu/coursescms/TFYA16/lessons/Le-8-extra.pdf [Broken]

    Moment inertia of system: I = Irod + mr^2

    Solution:
    conservation of ang momentum:

    rmvsinθ = (1/12 ML^2 + mr^2)ω


    So conservation of ang momentum is

    Linitial = Lfinal

    but this problem uses: ang momentum of a particle = ang momentum of a rigid body?

    What i dont understand:

    Why is the Moment inertia of system I = Irod + mr^2? what i did was before looking up the solution was:

    ML^(2)/12 ω = rmvsinθ + MR^(2)ω

    because ang momentum inital was just the rod rotating by itself and then comes the bullet so ang momentum of a bullet plus ang momentum of the rod. rmvsinθ = (Irod + mr^2)ω doesnt make sense to me. Also can someone explain to me what a rigid body is? because i know rigid body is L = Iω. Are rigid bodies like a rod, ring, disk or shell?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 17, 2012 #2

    tiny-tim

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    hi nchin! :smile:

    (try using the X2 button just above the Reply box :wink:)
    a rigid body is anything, of any shape, that doesn't change shape

    (ie it has no separately-moving parts …

    eg a rod with a ring free to move along it is not a rigid body :wink:)
    the angular momentum of the bullet after the collision is rmvf

    but vf = rω, doesn't it?

    so rmvf = mr2ω :smile:

    (but since the bullet has become part of the rod, it's easier to use the moment of inertia of the bullet-plus-rod about the centre of the rod, which is the sum of the two moments of inertia, I + mr2

    mr2 of course comes from the parallel axis theorem, 0 + mr2 :wink:)
     
  4. Nov 17, 2012 #3
    so the moment of inertia can only be used when two moments of inertia come together as one?

    so what about the ang momentum before? since the rod is at rest it would just be I(0) + rmvsinθ = rmvsinθ, right?
     
  5. Nov 17, 2012 #4
    also when we calculate the r, that would be half of the length of the rod? i thought radius only applied to circles.
     
  6. Nov 17, 2012 #5

    tiny-tim

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    hi nchin! :smile:
    the moment of inertia of a large body can only be used when the large body moves as one

    if its parts move separately, they will have separate angular velocities, and maybe separate centres of rotation, and so we must use the separate moments of inertia, and calculate each part's angular momentum separately :wink:
    right :smile:
    which r are you talking about?

    the r in r x mv is the displacement from the point about which you're taking moments

    the r in the parallel axis theorem is usually written d (so as not to confuse it with radius!), and is the distance from the centre of mass of the part to the centre of rotation
     
  7. Nov 17, 2012 #6
    Got it. Thanks!
     
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