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Homework Help: Moment of Inertia for a door

  1. May 2, 2010 #1
    Moment of Inertia - Unsolved

    1. The problem statement, all variables and given/known data
    A 22 kg solid door is 220 cm tall, 92 cm wide.
    a) What is the door's moment of inertia for rotation on its hinges?
    b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?


    2. Relevant equations
    [tex]I = \Sigma m_i*r^{2}_i[/tex]
    [tex]I_{parallel axis} = I_{cm} + md^{2}[/tex]

    3. The attempt at a solution
    I was able to get part a by using a given formula for a plane with an axis of rotation around it's edge: [tex]\frac{1}{3}Ma^2 = \frac{1}{3}(22)(.93)^2 = 6.2 kg * m^2[/tex].

    For part b, I cannot seem to get the correct answer. I applied the parallel axis theorem, which I understand to be the second equation I listed above. From a given list, I used the equation for a plane with the axis of rotation around it's center: [tex]\frac{1}{12}Ma^2[/tex]. My resulting formula ended up as this: [tex]I_{parallel axis} = (\frac{1}{12}(22)(.92)^2) + ((22)(.17^2))[/tex], which results in [tex]2.2 kg * m^2[/tex], but it was determined to be incorrect.

    What am I doing wrong here? Any help would be greatly appreciated.
     
    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2

    tiny-tim

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    Hi tangibleLime! :smile:
    You've used 17 cm from the centre of mass. :wink:
     
  4. May 2, 2010 #3
    Hm, I guess I'm a little bit confused as to how these coefficients are calculated (the 1/3, 1/12, etc..).

    I tried to use [tex]I_{parallel axis} = (\frac{1}{3}(22)(.92)^2) + ((22)(.17^2))[/tex], which I think would be taking the moment of inertia from the edge, as in part a, and moving it 17cm to the right. This results in [tex]6.8[/tex], which is apparently incorrect.

    I assume that my final answer is going to be less than 6.2 (the answer from part a) since the axis of rotation is being moved towards the center. That being said, the answer of 6.8 doesn't make sense anyways.
     
  5. May 3, 2010 #4

    tiny-tim

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    Hi tangibleLime! :smile:

    (just got up :zzz: …)
    The "d" in Icm + md2 must be from the centre of mass.

    That's where the 1/3 for the end of a rod or plate comes from …

    Icm for a rod or plate has 1/12, you add d2 = 1/4,

    and 1/12 + 1/4 = 4/12 = 1/3. :wink:
     
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