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Moment of inertia computation

  1. Dec 5, 2015 #1
    This seems to be a crucial detail that I just glossed over, but when finding the inertia tensor of an object, is the origin always situated at the object centre of mass?

    For example: In the link (http://hepweb.ucsd.edu/ph110b/110b_notes/node26.html ), is it necessary to do the integral from -s/2 to s/2 in each dimension as opposed to 0 to s if finding the inertia through the CoM?

    Also, how exactly can one find the moment of inertia of an object on an arbitrary axis? Referring back to the link, if one wanted to find the moment of inertia on an axis making 30 degrees with the horizontal (but still running through the CoM), how exactly could the inertia tensor be transformed to do this?
  2. jcsd
  3. Dec 5, 2015 #2

    Vanadium 50

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    The moment of inertia that you need is the one about the pivot point - the point about which the object is rotated. It is usually convenient to calculate this by taking the momen of inertia about the center of mass and use the parallel axis theorem to move to the point you want.
  4. Dec 5, 2015 #3
    Your moment of inertia tensor is only true for where you center your origin. In most cases centering your coordinate system at the center of mass and calculating the tensor from there is complicated. The usual method for these problems is first calculating your inertia tensor in a "convenient" coordinate system (usually at the base of your solid) and then performing a simple matrix operation on your tensor to "move it" into the center of mass coordinate system. If you'd like me to show you how to do so I can, although most textbooks on the subject will have an example on it as well.
  5. Dec 5, 2015 #4
    I think I got it. Although I would be happy to read any material you might have.
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