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Parallel axis theorem + rotational inertia

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Two particles, each with mass m = 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.6 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed w = 0.30 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?
    http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2010.%20Rotation/Problems/c10x10_12.xform_files/nw0536-n.gif

    2. Relevant equations
    parallel-axis theorem: I = Icom + Mh^2
    rotational inertia for thin rod: I = 1/12ML^2
    K = 1/2Iomega^2

    3. The attempt at a solution
    Inet = Irod1 + Iparticle1 + Irod2 +Iparticle2
    Inet = 1/12Md^2 + Md^2 + 1/12M(2d)^2 + M(2d)^2

    help!
     
  2. jcsd
  3. May 5, 2009 #2
    Think more carefully about applying the parallel axis theorem to the rods.
     
  4. May 5, 2009 #3
    Inet = 1/12Md^2 + Md^2 + 1/12(2M)(2d)^2 + 2M(2d)^2 ?
     
  5. May 5, 2009 #4
    In your formula for the parallel axis threorem, h means distance
    from O to the center of mass.

    This is d/2 for the inner rod and 3d/2 for the outer.
     
    Last edited: May 6, 2009
  6. May 5, 2009 #5
    oh. h means distance from O to com. not just d.

    Inet = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2 ?
     
    Last edited: May 5, 2009
  7. May 5, 2009 #6
    Nope.

    You can take each rod separately (as you seem to be trying to do),
    or you could treat them as one single rod.
    You are muddling these two approaches.
     
  8. May 5, 2009 #7
    sigh. i guess i would like to treat them as one rod. so this is my answer (i noticed that i wrote a wrong number, so i'm fixing it here):

    Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = 1/12M(1/2)d^2 + md^2 + 1/12(2M)(3d/2)^2 + m(2d)^2

    i think the particles are okay, but not the rods. is that right?
     
  9. May 5, 2009 #8
    Yes the particles are fine.
    But look at the parallel axis theorem again. you don't seem to be applying
    it.
    And the mass of the outer rod isn't 2M is it?
     
  10. May 5, 2009 #9
    sorry, i had class. okay, how about this?

    Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12(1/2)d^2 + Md^2) + md^2 + (M/12(3d/2)^2 + Md^2) + m(2d)^2
     
  11. May 6, 2009 #10
    actually, why do you have to use the parallel axis theorem here? why can't you just add up the rotational inertias for rods and particles? someone please help me out soon! thanks!
     
  12. May 6, 2009 #11
    David
     
    Last edited: May 6, 2009
  13. May 6, 2009 #12
    Because the rotation is about O, not the c of m.
     
  14. May 6, 2009 #13
    Icom = 1/12 ML^2
    where L is the length of the rod
     
  15. May 6, 2009 #14
    so... d/2 and 3d/2 are for the Mh^2 not Icom? that seems weird to me. wait let me try.
     
  16. May 6, 2009 #15
    Inet = Irod1 + Iparticle1 + Irod2 + Iparticle2 = (M/12 d^2 + M(1/2 d)^2) + md^2 + (M/12 d^2 + M(3d/2)^2) + m(2d)^2
     
  17. May 6, 2009 #16
  18. May 6, 2009 #17
    BINGO!

    Nearly. But the mass of the rods is M not m.
     
  19. May 6, 2009 #18

    Doc Al

    User Avatar

    Staff: Mentor

    In that problem the rotational inertia of the rod about point A is given. You don't have to calculate it at all.
     
  20. May 6, 2009 #19
    In that problem, the moment of inertia of the rod about A was given
    in the question.

    Try this one:
    Using the parallel axis theorem, show that the moment of inertia
    of a rod about one end is 1/3 M L^2.
     
  21. May 6, 2009 #20
    You beat me to it!

    BTW I think "rotational" inertia is a much better term than
    "moment of" inertia. I shall try to adopt it.

    David
     
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