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Homework Help: Physical pendulum: frequency of rotated square

  1. Jul 5, 2010 #1
    1. A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner. If it is rotated slightly and then released, at what frequency will it swing back and forth? hooksquare.gif

    2. Relevant equations

    frequency of a physical pendulum:

    Untitled1.png

    parallel-axis theorem:

    I=Icm+MD2

    3. The attempt at a solution

    Each thin stick has an I of (1/12)ML2. There are four of them, and the distance from the center of mass of each rod to the center of the square is L/2 so the I for the square (using parallel-axis theorem) is I=(4)((1/12)mL2)+m(L/2)2

    Then plug it into the frequency equation...d here is from the axis of rotation to the center of mass, so L/sqrt(2)

    Untitled2.png

    Simplified a little:

    Untitled3.png


    Is this right at all? Thanks.
     
  2. jcsd
  3. Jul 5, 2010 #2

    kuruman

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    In the expression

    [tex]f=\frac{1}{2 \pi}\sqrt{\frac{mgd}{I}}[/tex]

    about what axis is I, the center of mass or the point of support?
     
  4. Jul 5, 2010 #3
    I guess it should be the point of support...
     
  5. Jul 5, 2010 #4

    kuruman

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    That's what is. You need to recalculate. Since you already have the moment of inertia about the CM, all you need to do is apply the parallel axis theorem once more.
     
  6. Jul 5, 2010 #5
    so Icm=(4)((1/12)mL2)+m(L/2)2

    which simplifies to (7/12)mL2.

    Then the "new" I would be

    I=(7/12)mL2 + m(L/sqrt(2))2 ?
     
  7. Jul 6, 2010 #6

    kuruman

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    Almost. Read the problem. If the square object has mass m what is the mass of one of the arms?
     
  8. Jul 6, 2010 #7
    m/4...which m is that though? My first thought is that it's the first one in the "old" inertia equation...is that the only one I would need to switch? Because the others reflect the mass of the entire object, and I'm pretty sure they're correct.
     
  9. Jul 6, 2010 #8

    kuruman

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    Pretend that each arm has mass m,i.e. symbol "m" stands for the "mass of one side", then find the moment of inertia of the square about its CM correctly. (It is not (7/12)mL2)
    Use the parallel axis theorem consistently (i.e. pretend that the square's mass is 4m) to find Isupport. Having done that, divide all occurrences of m by 4 to take into account the fact that each side does not have mass m but m/4.
     
  10. Jul 6, 2010 #9

    ehild

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    It would be a bit simpler to derive the moment of inertia directly from its definition.

    [tex]I = \int{\rho s^2 dl}[/tex]

    where q is the density of the rods, and s is the distance of a small piece of length dl from the axis and the integration goes around the rectengular frame.

    ehild
     
  11. Jul 6, 2010 #10
    I am really confused now. I thought in the parallel axis theorem, m was supposed to be the entire mass of the object...I understand that each stick is m/4, but after that, you've lost me.

    And my professor isn't big on calculus, so I don't think I'm supposed to integrate anything...
     
  12. Jul 6, 2010 #11

    kuruman

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    Then follow my suggestion. It completes what you started, but correctly, so it should be easy for you to finish.
     
  13. Jul 6, 2010 #12
    Okay, trying again:

    Icm=4(1/12)(m/4)(L2)+(m/4)(L/2)2
    Icm=(7/48)mL2

    then

    Isupport=(7/48)mL2+4m(L/sqrt(2))2
    Isupport=(103/48)mL2

    then divide m by 4 so

    Isupport=(103/192)mL2

    But that seems weird...so another way:

    Icm=4(1/12)(m/4)(L2)+(4m)(L/2)2
    Icm=(13/12)mL2

    then

    Isupport=(13/12)mL2+4m(L/sqrt(2))2
    Isupport=(37/12)mL2

    then divide m by 4 so

    Isupport=(37/48)mL2

    Are either of those correct?
     
  14. Jul 6, 2010 #13

    kuruman

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    Where do you get the 7? It seems you think that (L/2)2=L2/2 instead of L2/4.
     
  15. Jul 6, 2010 #14
    Icm=(4)(1/12)(m/4)(L2)+(m/4)(L/2)2
    =(1/3)(m/4)(L2)+(m/4)(L2)(1/4)
    =(1/12)mL2+(1/16)mL2
    =(16/192)mL2+(12/192)mL2
    =(28/192)mL2
    =(7/48)mL2
     
  16. Jul 6, 2010 #15

    kuruman

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    The factor of 4 must multiply both terms in the sum and it should be
    ICM=(4)*[(1/12)(m/4)(L2)+(m/4)(L/2)2]
    because there are 4 sides each of moment of inertia
    (1/12)(m/4)(L2)+(m/4)(L/2)2
     
  17. Jul 6, 2010 #16
    So I=(1/3)mL^2

    Then Isupport is (partially simplified)

    (1/3)mL^2 + 2mL^2
    =(7/3)mL^2

    Then if I divide that by 4, it's (7/12)mL^2...
     
  18. Jul 6, 2010 #17

    kuruman

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    That is correct.

    I don't think so. Please show me exactly how you got this result.
    Why would you do that? One more time, when you write

    ICM=4*[(1/12)(m/4)(L2)+(m/4)(L/2)2],

    the symbol "m" stands for the mass of the four sides combined. Each side has mass m/4, that's why m/4 appears in the above expression. The overall factor of 4 is there because you have four sides to the square. In other words,

    ICM = 4*(Iof one side about the center of the square).
     
    Last edited: Jul 6, 2010
  19. Jul 6, 2010 #18
    I don't know why I divided by four...thanks for being so patient with me.

    The Isupport I did before is wrong (found a mistake in my work) so

    Trying again:
    Isupport= (1/3mL^2) + m(L/sqrt(2))^2
    = 5/6mL^2
     
  20. Jul 6, 2010 #19

    kuruman

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    That looks about right.
     
  21. Jul 6, 2010 #20
    Okay. and then I just plug that into the frequency equation:

    Untitled8.png
     
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