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Moment of inertia

  1. Nov 6, 2005 #1
    A uniform thin rod has a length 3.14 m and mass 2.11 kg. Find the moment of inertia in kg-m2 about an axis that is perpendicular to the rod and passes through the rod at a distance of 0.67 m from the end of the rod.

    I know that if the axis was in the center of the rod the equation would be 1/12 * M *L^2 . M =mass L = length.

    I don't know what to do with the axis being in a different place. is it the same eqation(doubt it) or the left half minus the right half or what?
  2. jcsd
  3. Nov 6, 2005 #2


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    Use the Parallel axis or Steiner's Theorem.
  4. Nov 6, 2005 #3
    Have you heard of the Parallel Axis Theorem?
  5. Nov 6, 2005 #4


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    Did you learn about the parallel axis theorem?
  6. Nov 6, 2005 #5
  7. Nov 6, 2005 #6


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    ECHO, Echo, echo!! ;)
  8. Nov 6, 2005 #7
    well thanks for the quick relpies, ummm yeah here in my book it is I = I(cm) +Mh^2

    is I(cm) = 1/12 * M *L^2?
    M is total mass
    and h would be..... distance from the center of mass?
  9. Nov 6, 2005 #8
    This is true.

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