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Straight up standing pendulum

  1. Jul 6, 2009 #1

    I'm working on a thing that could be simplified as a pendulum standing straight up. I've found some formulas on how a pendel moves when it's hanging on "the lower 180°" but can't figure out if it's possible to apply them on the other upper half.

    So what I'm trying to calculate is how many degrees a straight up standing (small initial angle/not standing stable) is turning/falling in 0.1 seconds. My measurement says it has m=1.9kg and r= 0.42m.

    Reguards // Viktor
  2. jcsd
  3. Jul 6, 2009 #2
    I guess it depends on which equations you're referring to. The equations of motion for an inverted pendulum are very similar to the equations of motion of a hanging pendulum, with the biggest difference being that the inverted pendulum tends to fall away from its equilibrium point, while the hanging pendulum falls towards it.

    It's not difficult to derive these equations. Have you tried yet?

  4. Jul 6, 2009 #3
    Theoretically, without vibrations or air motion, there should be an inverted pendulum position that would allow it to stand on end indefinitely. However, if you apply the uncertainty principle, specifically delta-x, delta-p <= h-bar to the pendulum mass, this uncertainty will require the pendulum to fall in a few seconds. I recall a similar problem in Quantum Mechanics class of a pencil standing on end.
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