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Velocity of a Pendulum

  1. Jun 13, 2009 #1
    Okay, so I am a student in my first year of physics and I'm doing an EEI on collision; however I am stuck up to a part where I need to figure out the velocity of my pendulum without doing the experiments again (Long story). At the moment, the length of the chain/string is 55cm and the angle is at 90 degrees.

    I have also done some googling, and some have ended leading me to here. Is this formula suitable to find the velocity?

    v = √{2gL[1-cos(a)]}


    I applied my known data into the equation which ended up giving me an answer of about 3.8m/s. I personally suspect that it is incorrect, so I need some experts to help me. (Sorry if i sound like a dag lol.)
     
  2. jcsd
  3. Jun 13, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Dazz4C! Welcome to PF! :smile:

    Is your pendulum a string with a point mass on the end, or does it have a large mass and/or something heavy instead of a string?

    Your equation (with a being the angle from the vertical) comes from conservation of energy: KE + PE = constant.

    For a point mass, KE/m = 1/2 v2, and PE/m = gL(1 - cosa), and your equation is correct.

    But for a large mass, or for a bar instead of a string, KE/m and PE/m will be different. :smile:
     
  4. Jun 13, 2009 #3
    Re: Welcome to PF!

    It has a rectangular piece of wood (10cm x 4cm) @ 70g with a cushion, that weighs anywhere from 2g-10g. (It's simulating the crumple zone :P)

    *EDIT: Sorry, so which formula would I use?
     
    Last edited: Jun 13, 2009
  5. Jun 13, 2009 #4

    tiny-tim

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    is the wood instead of a string (and if so, where is the pivot?), or is it on the end of a string?

    in any case, you'll need to use the moment of inertia of a rectangle.

    Before I go any further, do you know what moment of inertia is, and how to use it to calculate energy? :smile:
     
  6. Jun 13, 2009 #5
    No idea :(

    I don't really understand what you mean by where is the wood.

    So here an illustration.
    vzvd3t.jpg
     
  7. Jun 13, 2009 #6

    tiny-tim

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    oh i see …

    ok, if that long straight line is string, then your original equation should work,

    with L being the length of the string plus half the height of the block.

    (you originally mentioned a chain)
     
  8. Jun 13, 2009 #7
    Ah, yes thankyou. When I meant chain; I kinda meant the weightless support for the bob. Didn't know how to express it.

    Thankyou again
     
  9. Jun 13, 2009 #8
    There is also another way of doing it.

    Since

    [tex]
    T=2 \pi \sqrt{\frac{l}{g}}
    [/tex]

    where, T-the period, l-length of the string, we can write a displacement equation (I'm not sure that's the correct term in English, but I hope you'll understand what I mean from the math).

    [tex]
    x=A \cos ( \omega t) (1)
    [/tex]

    where

    [tex]\omega = \frac{2\pi}{T}[/tex]

    and A - amplitude. Now differentiate (1) and you'll have a velocity equation.
     
    Last edited: Jun 14, 2009
  10. Jun 13, 2009 #9
    I'll stick to the original equation; it's probably a bit easier to understand. But thankyou for helping aswell.
     
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