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Statics: Potential Energy

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Includes picture, download question here.

    2. Relevant equations
    The potential energy equation ends up being
    [tex]V = 1/2 k x^2 + W y[/tex]

    3. The attempt at a solution
    xspring = 2 sin([tex]\Theta[/tex])
    ym = 2 cos([tex]\Theta[/tex])
    [tex]V = 1/2*40 (2 * sin(\Theta))^2 + 7 * 2 * g cos \Theta[/tex]
    [tex]dV/d\Theta = 160sin(\Theta)cos(\Theta)-14g*sin(\Theta) = 0[/tex]

    Solving returns theta = 0, 0.5404057 rad (31 degrees)

    That sounds about right, so now I look for stability

    [tex]d^2V/d\Theta^2 = 160cos(\Theta)^2-160sin(\Theta)^2-14g*cos(\Theta)[/tex]

    This is where I'm stuck. Using intuition, I feel the system should be UNstable at theta = 0 and find stability at theta = 31 degrees.

    Solving the second derivative, I get [tex]d^2V/d\Theta^2(0) > 0[/tex] and [tex]d^2V/d\Theta^2(.54) < 0[/tex]

    Is the math wrong or is my understanding of stability wrong?
    Last edited: Dec 5, 2008
  2. jcsd
  3. Dec 6, 2008 #2


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    The math looks OK.

    Did you sketch the potential? Try drawing it for [tex]-\pi/2 \le \theta \le \pi/2[/tex] (radians).
    How does the graph indicate whether the extremum is stable or unstable?
  4. Dec 6, 2008 #3
    I found out I did it exactly right. When the second derivative is negative, the system is unstable and when it's positive, the system is stable.

    What's going on is at the top position the spring is pulling on the rod and keeping it in place, so without excess force pushing on the rod, it will tend to stay in place, even if you push a fair amount, the spring will bounce back into place and the rod will be vertical again.

    If you find the equilibrium position at 31 degrees, it will stay but be quite fragile. if you push down, the entire system will give and the rod will lay flat. If you push the rod to the right, the spring will snap it back to the vertical position.

    So there we have it, after all my math was correct and my understanding of stability was wrong.

    My teacher was as blown away as I was.
  5. Dec 6, 2008 #4


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    As I said, nothing was wrong with your math, just with your intuition :smile:

    Attached is a graph of V(theta) zoomed in a bit (from -pi/4 to pi/4). You can clearly see here that the minimum at 0 is stable (imagine putting a ball close to it, it will roll down to 0). The maximum is clearly unstable, if you put a ball exactly there it will stay, but if you move it a little it will fall down.

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