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Angular Momentum: mud hitting door

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A solid wood door 1.00m wide and 2.00m high is hinged along one side and has a total mass of 42.0kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500kg, traveling perpendicular to the door at 14.0m/s just before impact.
    Find the angular speed of the door.


    2. Relevant equations
    [itex]L_{initial}=L_{final}[/itex]

    [itex]L=Iw=rp[/itex]


    3. The attempt at a solution
    [itex]L_{initial}=L_{door}+L_{mud}=0+mvr=(0.5)(14)(0.5)=3.5[/itex]

    [itex]L_{final}=I_{total}+w_{f}[/itex]

    [itex]I_{total}=I_{door}+I_{mud}[/itex]

    I'm not sure how to calculate [itex]I_{door}[/itex], which gets me stuck here.
    I had read online that it is:

    [itex]I_{door}=\frac{1}{3}(42)(0.5)^2=3.5[/itex]

    Mud I know:

    [itex]I_{mud}=Iw=(0.5)(0.5)^2=1.125[/itex]

    Assuming I calculated the door correctly,

    [itex]L_{initial}=L_{final}[/itex]

    [itex]3.5=(3.5+0.125)w_{f}[/itex]

    [itex]w_{f}=0.966[/itex]

    But this is wrong, and I'm not sure where I messed up. Thanks in advance!
     
  2. jcsd
  3. Nov 19, 2011 #2
    I figured out the correct answer, but in order to get it, I needed to say that

    [itex]I_{door}=\frac{1}{3}(42)(1)^{2}[/itex]

    I don't understand why I should use 1 for the radius of the door, when the force was applied to the centre of the door.
     
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