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Distance between 2 axis in Parallel Axis Theorem

  1. Nov 19, 2016 #1
    1. The problem statement, all variables and given/known data
    The moment of inertia for a perpendicular axis through the center of a uniform, thin, rectangular metal sheet with sides a and b is (1/12)M(a2 + b2). What is the moment of inertia if the axis is through a corner?

    The answer is given as this was a powerpoint lecture and it is: I = 1/3 M(a^2 + b^2)
    I'm not looking for how to solve the whole thing however as the moment of Inertia is given in the context of the problem. I'm trying to understand how D was found to be: (a^2/2^2 + b^2 /2^2).

    2. Relevant equations
    I = I (cm) + Md^2

    3. The attempt at a solution
    D is meant to be the distance between the the new axis and the center of mass, if I personally were to solve this I'd state that the center of mass is half the distance between the axis that contains a and the axis that contains b. If that's the case, the distance from the center of mass to a diagonal would be a^2 + b^2 because you're increasing both a and b values by the same distance it originally was from the center of mass.
     
  2. jcsd
  3. Nov 19, 2016 #2
    For clarification: You're trying to understand how the distance between two different axis on a sheet depends on the lengths of the sheet?

    I'm asking cause I see questions, but then answers to those questions in later sentences.
     
  4. Nov 20, 2016 #3
    I trying to understand why D = (a^2/2^2 + b^2 /2^2). To me it seems as if it should simply be D = (a^2 + b^2)
     
  5. Nov 20, 2016 #4

    haruspex

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    a and b are the dimensions of the plate. In what sense does an axis contain them?
    The distance from the centre to a diagonal is zero. You want the distance to a corner.
    It's a/2 parallel to one axis, then b/2 parallel to the other. What does Pythagoras have to say on the matter?
     
  6. Nov 20, 2016 #5
    Thank you so much! The second part of your comment really helped me. I've noticed how you're consistently on this forums helping out people whether it comes to high level physics or highschool stuff and it's really well appreciated.
     
  7. Nov 20, 2016 #6

    haruspex

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    You are welcome.
     
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