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Moment of inertia problem

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A rigid body is made of three identical thin rods, each with length L = 0.530 m, fastened together in the form of a letter H, as suggested by the figure here. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

    2. Relevant equations
    This is really a conservation of energy question but the hard bit is to find out the moment of inertia of this object.
    K+P=K+P
    K=0.5I*w^2
    3. The attempt at a solution
    ==> 2mgL=(1/2)I*w^2 +(1/2)mgL
    and from this equation w can be easily found. However, i canot seem to get the correct Rotational Inertia. Can anyone show me how to find the moment of inertia for this system??
     

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  2. jcsd
  3. Apr 14, 2016 #2

    gneill

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    What have you tried? What equations are relevant for dealing with the moment of inertia of thin rods?
     
  4. Apr 14, 2016 #3

    haruspex

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    Which rods contribute to the MoI about that axis? Find the MoI of each.
     
  5. Apr 14, 2016 #4
    That is where i got stuck. For some reason i believe the Moment of inertia for this system is 1/3mL^2+mL^2, but it doesnt lead to the correct answer
     
  6. Apr 14, 2016 #5
    I know that only the two rods on the RHS of the 'H' contribute to the moment of inertia. But I have no clue how to find the MoI. Is it the parallel axis theorem?
     
  7. Apr 14, 2016 #6

    gneill

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    The parallel axis theorem is involved for one of the rods (which one?).

    What is the MoI of a thin rod about its center of mass?
     
  8. Apr 14, 2016 #7

    haruspex

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    Depending on your starting formulae, the parallel axis theorem may apply to both rods.
     
  9. Apr 14, 2016 #8

    gneill

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    True, but I've found that the formula most often given for a thin rod is the one for it rotating about a line perpendicular to the rod and passing though its center of mass.

    upload_2016-4-14_21-8-40.png

    That would be the y-axis or z-axis in the figure.
     
  10. Apr 14, 2016 #9
    The one parallel to rotation axis i suppose?
     
  11. Apr 14, 2016 #10

    haruspex

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    gneill meant the other one (see post #8), but arguably it can be used for both.
     
  12. Apr 14, 2016 #11
    I know the outer rod has I=(1/3)mL^2 +mL^2, but how do i find the inner perpendicular rod's?
     
  13. Apr 14, 2016 #12

    haruspex

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    Ah, the dangers of quoting formulae without understanding when they apply.
    The 1/3 formula you quote is for a rod of what length rotating about what axis?
     
  14. Apr 14, 2016 #13
    TBH i have no idea. Only these cases are included in the text book and i got that one from the internet. So I=1/2mL^2 ?
     

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  15. Apr 14, 2016 #14
    so after a quick research, 1/3mL^2 is rotation about an axis thru one end of the rod and 1/2mL^2 + mL^2 is the MoI for the parallel rod(by parallel axis theorem). Does that look right?
     
  16. Apr 14, 2016 #15

    haruspex

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    None in that image show a factor 1/3, so where did that come from?
     
  17. Apr 14, 2016 #16

    haruspex

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    An axis perpendicular to the rod and through one end, yes.
    No. Where did the 1/2 term come from?
     
  18. Apr 14, 2016 #17
    thats the MoI of the rod thru a parallel axis(ie Icom) and I=0.5mL^2+mL^2 by parallel axis theorem
     
  19. Apr 14, 2016 #18

    haruspex

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    The "i.e." is not appropriate. The direction of the axis and the point it passes through are independent.
    To apply the parallel axis theorem you first need to identify an axis parallel to the given axis and passing through the mass centre of the body.
    What does that parallel axis look like in this case, from the viewpoint of the rod?
     
  20. Apr 14, 2016 #19
    Is the axis pointing thru the rod? Thats parallel tocthe original axis
     
  21. Apr 14, 2016 #20

    haruspex

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    Along the rod, yes. What do you think its moment of inertia would be about such an axis? (You won't find this one in any table, it's too simple!)
     
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