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What is the period of a compound pendulum?

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A compound pendulum consists of a thin rod of length 1.4 m and a disc of radius 0.2 m. The centre of the disc is attached to the end of the rod and the pendulum pivots about the opposite end of the rod. Both the mass of the rod and the mass of the disc are the same, each being 3.9 kg. What is the period of the pendulum?

    [g=9.81 ms/2, MI rod about its end 1/3mL2, MI disc through centre, perpendicular to the plane 1/2mr2]

    2. Relevant equations
    Compound pendulum: T = 2π√(I/mgL)
    1/3mL2
    1/2mr2


    3. The attempt at a solution
    I tried adding the two inertias together for I = 1/3mL2 + 1/2mr2] but that didn't really work out. Since r<<L, I considered this to be a simple pendulum and used 1/2mr2] to find I in the compound pendulum equation, but this method was also incorrect. Am I supposed to use the parallel axis theorem? Should I have used the simple pendulum equation instead? Please help.
     
  2. jcsd
  3. Mar 15, 2015 #2

    ehild

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    Use parallel axis theorem to get the moment of inertia of the disk with respect to the pivot.
     
  4. Mar 15, 2015 #3
    Would the moment of inertia be: 1/3ml2 + 1/4mr2 then? I'm not 100% sure how to apply the parallel axis theorem correctly.
     
    Last edited: Mar 15, 2015
  5. Mar 15, 2015 #4

    ehild

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    The total moment of inertia is the sum of that of the rod and that of the disk with respect to the pivot. Apply parallel axis theorem for the disk. http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
     
  6. Mar 15, 2015 #5
    Okay, so I'll say that d2=(r/2)2 and apply it to the disk. I should get: 1/4mr2 (disk) + 1/3ml2. Would you say that this is correct?
     
  7. Mar 15, 2015 #6

    ehild

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    ????
    The moment of inertia of the disk is its moment of inertia with respect to its centre + md2, with d the distance between the centre of the disk from the pivot.
     
  8. Mar 15, 2015 #7
    Hmmmm okay. So the moment inertia of the disk with respect to the pivot is: 1/2+mr2+m(r/2)2? Adding this to the inertia of the rod for the total inertia.
     
  9. Mar 16, 2015 #8

    ehild

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    NO.Look at parallel axis theorem in your notes or book.What does it say?

    compoundpendulum.JPG
     
  10. Mar 16, 2015 #9
    There's nothing in my notes about parallel axis theorem, that's why I'm so confused! I don't know, is it: 1/3mL2+1/2mr2+m(1.4)2???
     
  11. Mar 16, 2015 #10

    BvU

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    No grounds for confusion. You did know about this parallel thingy from post #1 and can always google to find out what and why. It's not rocket science (which itself is also overrated :smile:). Simply the moment of inertia of a point mass at distance d - and then if it isn't a point mass you have to add the moment of inertia around its center of mass.

    What you have looks good, but your notation is asking for trouble. (L in post #9 is not the L in post #1 equation 1).
     
  12. Mar 16, 2015 #11
    Woooo, finally on the right track! Hmmmm I'm looking at L in post #9 and #1 and I see no difference, how are they different?
     
  13. Mar 16, 2015 #12

    BvU

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    One is the length of the rod
    The other is the distance from the pivot point to the center of mass of the pendulum
     
  14. Mar 16, 2015 #13
    Are they not the same value of 1.4?
     
  15. Mar 16, 2015 #14

    BvU

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  16. Mar 16, 2015 #15
    So I should use L = I/mr to find the L in 1/3mL2+1/2mr2+m(1.4)2?
    Which length would I use in T = 2π√(I/mgL)? The rod length of 1.4?
     
  17. Mar 16, 2015 #16

    BvU

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    Oh boy, no! Did you check the link I gave ?

    Ehild made a clear drawing, let's use that notation: 1.4 m is d, the length of the rod.

    If you remember, for the moment of inertia of the whole thing we now have I = 1/3 m d2+ ( 1/2 mr2 + md2 )

    That leaves the symbol L for the distance from the pivot point to the center of mass of the pendulum, to be calculated.
    By you :smile: .
     
    Last edited: Mar 16, 2015
  18. Mar 16, 2015 #17
    Awesome, so the L that I calculate on my own will be used in T = 2π√(I/mgL). Right?
     
  19. Mar 16, 2015 #18

    BvU

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    Good plan !
     
  20. Mar 16, 2015 #19
    Putting everything in:
    m=3.9kg
    I=10.27
    L=13.1666
    g=9.81
    I get 1.11 which is apparently the incorrect answer. :cry:
     
  21. Mar 16, 2015 #20

    BvU

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    Are you saying the center of mass is 13.17 m below the pivot point ?

    And how come I has no units ? (the value is OK)

    General rule for calculations:
    1. Strike all cancelling quantities
    2. Check dimensions
    3. Estimate order of magnitude -- or approx value if possible
    4. Employ calculator
    Caveat:
    m is the mass of the rod and the mass of the disc. In your T formula it's also the mass of the compound pendulum. Recipe for disaster.​
     
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