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Moment of Inertia - Need Guidance!

  1. Nov 7, 2005 #1
    A pendulum is made of a rod whose moment
    of inertia about its center of mass is
    1/12ML^2 where its mass is 4.9 kg and and its length
    is 3.3 m, and a thin cylindrical disk, whose
    moment of inertia about its center of mass
    is 1/2MR^2, where its mass is 1.7 kg and its
    radius is 1.6 m.

    What is the moment of inertia of the pen-
    dulum about the pivot point? Answer in units
    of kg*m^2.

    I tried adding the moment of Inertias --> ([1/12 (4.9) (3.3)^2 ]+[1/2 (1.7)(1.6)^2] = 6.62275.

    This is obviously wrong. I have know clue how to approach this. I think there's a theorem (parallel theorem?) that will help but I don't know how to apply it. Help is much appreciated.
  2. jcsd
  3. Nov 7, 2005 #2
    The disc is at the END of the rod. Whats the distance of the disk from the pivot point then?
  4. Nov 8, 2005 #3


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