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Moment of inertia (compound object)

  1. Nov 26, 2006 #1
    I have a disk parallel to the floor rotating about an axis perpindicular to the floor, going through the center of the disk.
    Its moment of inertia is [tex]I=\frac{1}{2}mR^2[/tex]
    I also have a ring (which I will forbear to describe. A picture is worth a thousand words [or 30-ish, in this case], so please see attatchment) with intertia [tex]I=\frac{1}{2}m(R_1^2+R_2^2)[/tex]
    All fine and dandy. But when I place one on top of the other (still rotating about the same axis), to find the moment of inertia, do I just add their individual moments? Or something else?

    Attached Files:

  2. jcsd
  3. Nov 26, 2006 #2


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    It can easily be seen from the definition of moment of inertia that if you know the moment of inertia of two objects rotating about the same axis the combined moment of inertia is the sum of the two individual moments of inertia.
    Last edited: Nov 26, 2006
  4. Nov 26, 2006 #3
    Thank you!
    I confess--I am deplorably shaky on the definition of the moment of inertia, and all of torque in general.
  5. Nov 26, 2006 #4
    The reason moments of inertia are just given in formulas is because the real definition involves nasty summations and integrals. Take every particle and multiply by the distance from axis squared. For a hoop, its obviously MR^2, but theres a bunch of calculus behind the other formulas, so be thankful for them.

    Torque is the rotational equivalent of force. It also depends on the distance from the axis. The longer the arm, the more leverage and the more torque. It is defined as the perpendicular distance, so you need right angles. If a force acts at an angle, the torque is given as rFsintheta, because the sin of the angle yields the perpendicular force. This is also known as a cross product. I hope this helps
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