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Swinging Pendulum and Conservation of Energy

  1. Apr 4, 2006 #1
    Hello all, I've been having major difficulty with a question that deals with a pendulum swing, and to find the maximum speed after the release

    The length of the pendulum is 85.5cm and the amplitude is 24.5cm

    I was thinking to find the vf= you have to do square root of 2gh and solve, but I'm always left with the wrong answer,

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 4, 2006 #2
    That seems like it should give the right answer. What are you using for g and h? Are they in the proper units?
     
  4. Apr 4, 2006 #3

    dav2008

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    Gold Member

    I don't know if that formula is correct or what it applies to but I would recommend not memorizing a bunch of formulas for specific cases (only memorize the ones that are hard to derive), but instead apply a general method that can be used to solve any equation.

    In this case, like you said in your title, you can use conservation of energy to equate gravitational potential energy at the top to kinetic and gravitational potential at the bottom.

    Like Euclid said make sure you are using proper units (meters not centimeters) and that you determined the height properly.
     
  5. Apr 4, 2006 #4
    yes i converted 85.5 to 0.855 m, and my final answer is 4.09 m/s but the text says 0.838 m/s
     
  6. Apr 4, 2006 #5
    You see, I did derive the formula, from equating 1/2mv^2 = mgh and i managed to get v= square root 2gh, and i cannot seem to get the answer, and I am using appropriate units. I just don't understand what the use of 24.5 cm, it says it is the amplitude.
     
  7. Apr 4, 2006 #6
    The term "amplitude" seems to be unclear here. I took it to mean the maximum height, but this gives 2.19 m/s for the max velocity.
    Usually the amplitude is given in terms of degrees.
     
  8. Apr 4, 2006 #7

    dav2008

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    I took the amplitude to mean the arc-length of half a pendulum swing and obtained an answer close to yours (I guess you used small angle approximation)
     
    Last edited: Apr 4, 2006
  9. Apr 4, 2006 #8
    The question says:

    Simple pendulum, 85.5 cm long, is held at rest so that its amplitude will be 24.5 cm as illustrated. (figure shown with a hanging pendulum and its length is 84.5 and the space between it after it has moved is 24.5). Use energy concepts to determine maximum speed of the pendulum bob after release.
     
  10. Apr 5, 2006 #9
    Euclid how did you get 2.09 m/s.....? Question is boggling me =(
     
  11. Apr 5, 2006 #10
    24.5cm must be the horizontal amplitude. This means [tex]l \sin \theta = 24.5[/tex]cm. For h, use [tex] h = l(1-cos\theta)[/tex]. This should give the correct answer.
     
  12. Apr 5, 2006 #11
    How is such applicable if no angle is given?
     
  13. Apr 5, 2006 #12
    I am using theta to denote the angle between the pendulum at its maximum height and at its minimum height.
    You don't need to know theta, since [tex] \cos \theta = \sqrt{1-\sin^2 \theta}=\sqrt{1-(24.5cm)^2/l^2}[/tex]
     
  14. Apr 5, 2006 #13
    I still cant get the answer of 0.838, or even close to.
     
  15. Apr 5, 2006 #14
    Show me your work, and I will try to figure out where you went wrong.
     
  16. Apr 5, 2006 #15
    ok, I did as you suggest square root of ==>

    1-(0.245)^2/(0.855)^2
    cos theta = 1.1334 and then i applied what you gave me when you said

    h= L (1-cos theta)
    = 0.855 (1-0.99)
    =1.71 x 10 ^-4

    then i put that back to v= square root 2gh
    = 2(9.8)(1.71 x 10 ^-4)
    =0.05789 m/s...

    Seems very wrong to me...?
     
  17. Apr 5, 2006 #16
    ok, I did as you suggest square root of ==>

    1-(0.245)^2/(0.855)^2
    cos theta = 1.1334 and then i applied what you gave me when you said

    h= L (1-cos theta)
    = 0.855 (1-0.99)
    =1.71 x 10 ^-4

    then i put that back to v= square root 2gh
    = 2(9.8)(1.71 x 10 ^-4)
    =0.05789 m/s...

    Seems very wrong to me...?
     
  18. Apr 5, 2006 #17
    sorry for the double post, but also may I point out that this technique has not even been remotely explained at all, i was wondering is there any other technique to solve this problem?
     
  19. Apr 5, 2006 #18
    cos can never be > 1
    try again
     
  20. Apr 5, 2006 #19
    To my knowledge, this is the best method to solve the problem.
    And it's a fundamental method anyway, so it would be beneficial to learn it.
     
  21. Apr 5, 2006 #20
    I don't understand I did what you asked me to, and I get 1.13 from the information you've given me
     
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