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Moment of inertia of a cylinder in unusual position

  1. Sep 5, 2015 #1
    True or false.jpg
    See image. It would be better if you can show me the calculation process.

    EDIT: the axis of rotation goes right through the centre of the cylinder.

    NEVER MIND! GOT IT!
     
    Last edited: Sep 5, 2015
  2. jcsd
  3. Sep 5, 2015 #2
    My opinion: for a thin disc whose radius is ##r## and mass is ##m,## with the axis perpendicular to the disc, the moment of inertia is ##I_0=\frac{1}{2}mr^2,## which may be easy to find. Then with the perpendicular axis theorem, if the axis is parallel to the disc and passes the center, the moment of inertia should be ##I_i=\frac{1}{2}I_0=\frac{1}{4}mr^2.##
    With this result and the parallel axis theorem, we may be able to solve the problem. For the disc away from the center ##x##(based on your diagram), the disc's moment of inertia is ##\frac{1}{4}mr^2+mx^2##(of course ##m## should be adjusted. Then, just integrate them we may get the answer.
     
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