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Pendulum without oscillation

  1. Oct 20, 2005 #1
    If a mass (m) at the end of a length (L) on a pendulum starts at an angle of θ from the vertical, what is the minimum inital velocity v0 it must have to just barely make it over the top and not oscillate?

    This is what I did:

    [tex]\Delta K=mgh\implies v_0=\sqrt{v^2-2gh}[/tex]

    but at the top, v is zero so it can be written as:

    [tex]v_0=\sqrt{-2gh}[/tex]

    and h in this case is L+Lcosθ, so using g=-10m/s2 I get:

    [tex]v_0=\sqrt{20L\left(1+\cos{\theta}\right)}[/tex]

    but whenever I plug in values for θ and L, I get the wrong answer. I can't see what I did wrong.

    Any ideas?

    Thanks a lot.
     
  2. jcsd
  3. Oct 20, 2005 #2
    What School do you go to?
     
  4. Oct 20, 2005 #3
    I'm in high school.
     
  5. Oct 21, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    At the top the pendulum has 0 velocity (actually, it should have just a tiny non-zero velocity but you can take 0 as "boundary" between going over the top and not making it to the top) and so has no kinetic energy but has potential energy, -mgL (relative to the height of the pivot- the center of the circle the pendulum moves in). Initially, it has velocity v0 and so kinetic energy (1/2)mv0[/sub]2 AND some potential energy. Have you taken into account the fact that the potential energy has to be measured with respect to the center of the circle?
     
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