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Mass Moment of Inertia

  1. Jan 8, 2008 #1
    hi.i've been asked to post this question here for a concrete answer..

    the above question has been solved which the answer is 0.1706

    and here's the another question that i need a help

    and here's my another approach

     
  2. jcsd
  3. Jan 12, 2008 #2
    For the first part you need to use the parallel axis theorem. It states the following:

    [tex]I_{p}=I_{cm} + mk^{2}[/tex]

    Where [tex]I_{p}[/tex] is the mass moment of inertia about any point, [tex]I_{cm}[/tex] is the mass moment of inertia about the object's center of mass, and k is the distance between the object's center of mass and the axis of revolution.

    Clean up some of the units before you post.
    I'm going to assume that you meant that the mass of the rod is 4kg and the speed of rotation is 800rpm? I am not really sure about that last unit.

    Anyways, there will be no moment on the rod (and thus torque on the bar) unless it is accelerating. If it is simply rotating at a constant velocity then the only thing you need to take into consideration is the weight of the rod. Maybe I could help you more if you explained your reasoning verbally.
     
  4. Jan 12, 2008 #3
    mass of the rod is 4m.. so the total mass would be 0.4x4 = 1.6 kg..
    yup..revolution per minute is 800 rpm.. sry about dat..
    and the rod is homogeneous..

    for the 2nd question..
    it's only the rod which is parallel to the z-axis is to be considered rite?
    so to calculate the moment about x axis, i should consider the center of mass of the rod parallel to z-axis..

    is it correct?
     
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