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The Pumpkin Swing

  1. Feb 24, 2004 #1
    How fast is the pumpkin moving at the bottom if theta = 30 degrees, L = 5 m, g = 9.8 m/s^2, m = 10 kg.

    L is the length of the string holding the pumpkin. M is the mass of the pumpkin. It's like a pendulum, the pumpkin is positioned at 30 degrees from the vertical and let go. Assume no air resistance/friction.
     
    Last edited: Feb 24, 2004
  2. jcsd
  3. Feb 24, 2004 #2
    Why don't you try an energy argument?

    potential + kinetic energy = constant

    cookiemonster
     
  4. Feb 24, 2004 #3
    could you please explain what you mean a little? I am lost with that problem, too.

    This is what I'm thinking, but i dont know what to do with the angle and all and if i'm doing this correctly or not:

    TE = PE + KE
    PE = KE
    mgh = mv^2
    sqrt ((mgh)/m) = v
    sqrt (gh)= v
    sqrt (9.8m/s^2 * 5.0m) = v
    v = 7 m/s

    Please help me if you can, thanks in advance
     
    Last edited by a moderator: Feb 25, 2004
  5. Feb 25, 2004 #4
    Your approach is good, but your expression for potential energy is not correct.

    Let the position at theta = 0 (hanging straight down) have a potential of 0. So now the question is, "If the pendulum is at an angle of 30 degrees, how high is the bob?" Why don't you draw a triangle?

    cookiemonster
     
  6. Feb 25, 2004 #5
    can't or can?
     
  7. Feb 25, 2004 #6
    I hadn't read your post carefully enough to realize you needed help with the angle. I edited my post to include some helpful information.

    And just for a fun little exercise in English, the sentences with "can" and "can't" mean exactly the same thing!

    cookiemonster
     
    Last edited: Feb 25, 2004
  8. Feb 25, 2004 #7
    if i did it right, i got 1.25m high from the bottom point, is that correct?
     
  9. Feb 25, 2004 #8
    What triangle did you use?

    cookiemonster
     
  10. Feb 25, 2004 #9
    Hold on, i am about to post up the picture i drew

    http://mysite.verizon.net/vze3ss2y/sitebuildercontent/sitebuilderpictures/pumpkin.jpg

    i used the big triangle to find the hypotenuse of the small triangle (circled in green) then i used the laws of sine to determine the height of da pumpkin from the bottom point where PE = 0
     
    Last edited by a moderator: Feb 25, 2004
  11. Feb 25, 2004 #10
    Well, I'm not quite sure I understand your triangle. I'll describe the one you're going to want to use:

    We're going to let the string and bob form the hypotenuse. So the hypotenuse will always have length 5. Additionally, we want to get a value for y, so the other two sides will be in the up-down direction and left-right direction. Since we want the up-down and left-right dimensions to form a triangle with the string, put the up-down side directly below the place where the string is hanging and the left-right side right to the side of the bob. I assume the angle is of the hypotenuse off of the up-down side, so the angle on the top of this triangle will be 30. Now you can calculate the y height.

    Just remember that this is the height from the top, whereas you're looking for the height from the bottom.

    Edit: In response to your post and diagram (I understand it now), you're assigning the value of 5 to the wrong side. Your approach can work, but it's more difficult than it needs to be.

    cookiemonster
     
  12. Feb 25, 2004 #11
    could you please post a picture of what you're trying to say?


    edit: wait i think i know what you're talking about now
     
    Last edited by a moderator: Feb 25, 2004
  13. Feb 25, 2004 #12
    Hope it works... I don't have a webspace of my own (that I know how to use).

    cookiemonster
     

    Attached Files:

  14. Feb 25, 2004 #13
    thanks for the picture, it is very helpful.

    Ok, so when i solve for the adjacent side (up-down), i get 4.33m from the top, meaning the pendulum is .66m from the bottom. So when i solve the problem to find the velocity, for height in PE = mgh, i use .66m to get:

    PE = KE
    mgh = .5mv^2
    gh = .5v^2
    sqrt (gh/.5) = v
    sqrt ((9.8m/s^2 * .66m)/.5) = v
    v = 3.6 m/s

    Is that correct?
     
  15. Feb 25, 2004 #14
    Yup. Good job.

    cookiemonster
     
  16. Feb 25, 2004 #15
    thanks for all the help :smile:
     
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