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Pendulum lab

  1. Oct 19, 2006 #1
    im suppose to do some calculations to prove whether or not the pendulum demonstrates the law of conservation of energy.

    mass of pendulum = 240.3g = 0.2403 kg
    diamter of pendulum bob = 3.50
    initial height of pendulum bob = 48 cm = 0.48 m
    length of pendulum string = 2.14 m
    time interval of photogate light interruption = 11.8 m/s

    so there's position 'A' and position 'B'
    for 'A"
    Emechanical = Ek + Eg
    = (0.5)mv^2 + mgh
    = 0 + (0.2403kg)(9.8)(0.48m)
    = 1.13J

    for 'B'
    Emechanical = Ek + Eg
    = (0.5)mv^2 + mgh
    = (.5)(0.2403kg)(11.8^2) + 0
    = 16.73J

    i must be missing something (my calculations aint proving anything).. any help will be appreciated!

    ~Amy
     
  2. jcsd
  3. Oct 19, 2006 #2

    marcusl

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    Did you calculate the speed correctly? You list it as time but give units of speed, suggesting an error there. Furthermore, 12 m/s is pretty fast for a 2m long pendulum.
     
  4. Oct 19, 2006 #3
    k, i think i see the mistake! it says "11.8 ms" and i thought they meant meters per second.

    so 11ms = 0.011s?

    so for B i'd go:
    Emechanical = Ek + Eg
    = (0.5)mv^2 + mgh
    = (.5)(0.2403kg)((d/0.011)^2) + 0 = 1.13

    im short on time but will figure this out tomorrow and post my results.


    ~Amy
     
  5. Oct 19, 2006 #4
    just thought of something else. to find the velocity i just take the diameter of the bob divided by 11ms?

    ~Amy
     
  6. Oct 20, 2006 #5

    OlderDan

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    Yes, but use the 11.8 ms as it was in your data. And yes, ms is milliseconds so 11.8 ms = .0118 s
     
  7. Oct 21, 2006 #6
    i changed all the units to regular units (meters, kg, s, etc.) just to play thing safe.

    for Ek (A)
    i got: mgh
    = (0.2403kg)(9.8 m/s^2)(0.48m)
    = 1.13J

    for Eg (B)
    i got: 0.5mv^2
    =(0.5)(0.2403kg)(2.966^2)
    =1.057J

    (to get the 2.966 velocity i took 0.035/0.01185s).

    and then i just explain that mechanical energy becomes heat energy so the total mechanical energy gradually decreases. ?

    ~Amy
     
  8. Oct 21, 2006 #7

    marcusl

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    Yes, a certain amount of potential energy is lost to friction and air resistance, although 6% may be a little high. All measurements have uncertainty or inaccuracy, and that will also be a factor. One final question: when you measured the height of the pendulum, did you measure at the center of the weight at both positions A and B? It's important to measure the heights at the center of gravity.
     
  9. Oct 21, 2006 #8
    i didnt do any of the measurements myself. im taking an independent learning course, and the measurements are listed in the book, im just suppose to do some qualculations and write a lab report explaining why and whether or the pendulum lab demonstrates the law of conservation of energy.

    slight ot: does the weight of the pendulum string make a difference? like a light weight vs. a heavy one?

    ~Amy
     
  10. Oct 21, 2006 #9

    marcusl

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    Got it. Thanks!

    Good question. What do think?
     
  11. Oct 21, 2006 #10
    shouldnt the string weight be added to the bob weight?

    ~Amy
     
  12. Oct 21, 2006 #11

    marcusl

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    Not exactly, because it isn't in the same place. Think about where its center of gravity (COG) is and where the bob's is. Any ideas on how you would add the string's mass to your problem?
     
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