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Homework Help: Angular speed of rod after projectile collides into it

  1. Jul 21, 2012 #1
    [SOVLED] Angular speed of rod after projectile collides into it

    1. The problem statement, all variables and given/known data

    2. Relevant equations
    L = r x p
    Iparallel = ICM + md2

    3. The attempt at a solution
    L = r x p = mvir = mvid/2
    mvi(d/2) = (Iparallel + md2
    mvi(d/2) = (1/12M(d/2)^2 + m(d/2)^2)ω

    Plugging and solving eventually leads me to...
    ω = (mvi)/(1/24Md+1/2md)
    ...Which is apparently incorrect.
    Last edited: Jul 21, 2012
  2. jcsd
  3. Jul 21, 2012 #2


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    What's the moment of inertia of a thin rod about its center?
  4. Jul 21, 2012 #3
  5. Jul 21, 2012 #4


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    And what should r be in this problem? Hint: I don't think it's d/2.

    As a matter of interest, how good is the submission system at algebra? Will it understand that a/(1/24b+1/2c)=24a/(b+12c), or will it only recognise expressions that your prof entered? The only one I've ever used could only do exact string comparison, so if the answer was 0.3 and you typed .3, that was marked wrong...
  6. Jul 21, 2012 #5
    Why wouldn't it be d/2? If the diameter of the circular motion is d, then wouldn't the radius be d/2?
  7. Jul 21, 2012 #6


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    You have the correct functional form for the second moment of inertia of the rod, but are misinterpreting the thing you called r. It's the length of the rod (d, in this case). You can easily look it up, or derive it from first principles by integrating [itex]\mu r^2dr[/itex] (where [itex]\mu[/itex] is the mass per unit length of the rod) over the range -d/2 to d/2.
  8. Jul 21, 2012 #7
    Oh okay, I thought r was supposed to be the distance between the rotation point and the end of the rod.

    I got the right answer of ω = (mvi)/(1/6Md+1/2md)

    Thanks for your help!
  9. Jul 21, 2012 #8
    Okay, I've tried part B of the question and this is what I've done so far.

    Kfinal = 1/2Iω2
    Kinitial = 1/2mv2

    I divided both of those to get (Iω2)/(mvinitial)

    Calculating that further, and I get this:

    I'm guessing that it's not Kfinal/Kinitial, but (Kinitial - Kfinal)/Kfinal?

    EDIT: Okay, I got the answer. Thank you all!
    Last edited: Jul 21, 2012
  10. Jul 21, 2012 #9


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    Almost. After subtracting, divide by Kinitial rather than Kfinal. (It's like one of those percent decrease problems in algebra.)
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