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Elastic & gravitational potential energy

  1. Mar 22, 2009 #1
    Hi there.

    I have a bead of mass m, which slides down a frictionless parabolic wire in the form y=x^2, and is attached by elastic to the point (0,h), and I want to write down total energy f(x) (= elastic energy + gravitational potential energy) (no mention of kinetic energy...) of the bead at (x,x^2).

    I know that g.p.e is mgx^2, and that elastic energy is (k/2)(h^2 + x^2 - 2hx^2 + x^4), but I'm not sure whether I'm adding the g.p.e. to the elastic energy, or subtracting it, despite the fact that I want f(x) = elastic energy + g.p.e.

    Any clarification would be great :)

    Thanks.
     
  2. jcsd
  3. Mar 22, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Set Abominae! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Just ask yourself: will it make the bead go slower?

    ie will increasing the height make the bead go slower? will increasing the length of the elastic make the bead go slower? :wink:
    how did you get that? :confused:

    and what is the unstretched length of the elastic?
     
  4. Mar 22, 2009 #3
    The question states that:
    When the elastic has length d its elastic energy is [tex]\frac{k}{2}[/tex] [tex]d^{2}[/tex], where d> 0.....

    So when the bead is at [tex](x,x^{2})[/tex], we have that:
    [tex]d = \sqrt{x^{2}+(h-x^{2})^{2}}[/tex]

    (It'a a question from a math assignment, so its probably not totally accurate...)

    So that will be add, then?:shy:
     
  5. Mar 22, 2009 #4

    tiny-tim

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    hmm … that's unusual … but if the question says so, i suppose it's ok :smile:

    (perhaps the elastic starts further back, and goes round a peg at (0,h), and the unstretched length is up to the peg)
    i'm not sure what you're adding to what …

    anyway, give a reason, so we can see you're not guessing! :wink:
     
  6. Mar 22, 2009 #5
    Actually, thinking about it, I would imagine that I would subtract the g.p.e from the elastic energy to get f(x) (though I'm cautious about doing so, as the question says f(x)=elastic energy + g.p.e.).

    Another reason for doing this is that if I sum them to get f(x), when I start drawing my bifurcation diagram later in the question, I get:

    A single fixed point [tex]x=0[/tex] is stable for [tex]h<\frac{mg}{k}+\frac{1}{2} [/tex],

    and 3 fixed points

    [tex]x=0, \pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}[/tex] are all unstable for [tex]h>\frac{mg}{k}+\frac{1}{2}[/tex], and I've never seen a system where all 3 fixed points are unstable before...
     
  7. Mar 22, 2009 #6
    But I also get the same result of 3 unstable fixed points when I subtract the g.p.e. from the elastic energy.... :confused:
     
  8. Mar 23, 2009 #7

    tiny-tim

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    Are you taking into account the fact that gpe is minus mgh?
    urggh :yuck: … dunno anythng about bifurcation diagrams :redface:
     
  9. Mar 23, 2009 #8
    Here is a related problem that I. Newton solved in one day back in about 1697. Suppose we have a frictionless bead sliding on a wire from point A at (0,h) to point B at (x,0) by the fastest route. What is the shape of the wire from point A to point B? A straight line is the shortest distance, but a parabola is longer and faster. But a parabola is not the fastest. So what is?
     
  10. Mar 23, 2009 #9
    I forgot about that! Upon further thought, I would imagine that I would subtract the negative gpe, so be adding it to the elastic energy to give f(x). I say this by considering energy conservation - assuming we don't have any stupidly big oscillations, the higher the bead gets, the larger mgh gets, and the smaller the elastic energy.

    Likewise, when the bead is at x=0, it has it least gpe (0 if we take the x-axis as our 'zero height', but the elastic is at maximum extension (again, assuming well-behaved oscillations).

    So by energy conservation, I'm inclined to write that
    [tex]f(x) = (elastic energy) + mgx^{2}[/tex]

    Thoughts?
     
  11. Mar 26, 2009 #10
    I have the same problem for my maths assignment :) check ur calculations carefully. I have a single unstable fixed point at x=0 for h<mg/k+0.05, second derivative is >0 and 3 stable fixed points for h>mg/k+0.05, second derivative<0. Why do u have the opposite?
     
  12. Mar 26, 2009 #11
    [tex]f(x)=\frac{k}{2}(x^{2}+(h-x^{2})^{2}+mgx^{2}
    \Rightarrow f'(x)=2kx^{3}+kx-2hkx+2mgx[/tex]
    Solve [tex]f'(x)=0[/tex] to give [tex]x=0[/tex] for [tex]h\leq\frac{mg}{k}+\frac{1}{2}[/tex],
    and [tex]x=0, x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}[/tex] for [tex]h>\frac{mg}{k}+\frac{1}{2} [/tex]

    Now, [tex]f''(x)=6kx^{2}+k+2mg-2hk
    h<\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)=2mg+k-2hk<0 \Rightarrow[/tex] stable.
    (Note, [tex]h=\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)=0 \Rightarrow[/tex] system is structurally unstable.)

    [tex]h>\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)>0 \Rightarrow[/tex] unstable, and
    [tex]f''(\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}})=4hk-4mg-2k>4k(\frac{mg}{k}+\frac{1}{2})-4mg-2k=0 \Rightarrow[/tex] unstable.

    (Sorry for the mess!)

    I don't think there's any mistakes here... How did you go about solving it?
     
  13. Mar 27, 2009 #12
    [QUOTE f''(0)=2mg+k-2hk
    [tex]h>\frac{mg}{k}+\frac{1}{2} \Rightarrow f''(0)>0 \Rightarrow[/tex] unstable QUOTE]

    multiply h>mg/k+0.05 by 0.05k ( sign stays the same since k>0). U get 2kh>2mg+k i.e
    2mg+k-2kh<0 so f''(0)<0 stable ?? Same for the other two.
    I know it's a cusp catastrophe but how do u justify that?:confused:
    Do u have a symmetry question (with a see-saw) later on in your assignment? :smile:
     
  14. Mar 27, 2009 #13
    Actually this is ok. I had a wrong sign there, but u still have a mistake for x=0 so check it. We have ether one unstable at x=0 or one stable at x=0 and two unstable at +-sqrt(h-mg-1/2). Bifurcation looks much better now :) Thoughts?
     
  15. Mar 27, 2009 #14
    Yes, I now have:

    1 unstable fixed point [tex]x=0[/tex] for [tex]h<\frac{mg}{k}+\frac{1}{2}[/tex]

    1 fixed point [tex]x=0[/tex] for [tex]h=\frac{mg}{k}+\frac{1}{2}[/tex], and system is structurally unstable.

    1 stable fixed point [tex]x=0[/tex] and 2 unstable fixed points at
    [tex]x=\pm\sqrt{h-\frac{mg}{k}-\frac{1}{2}}[/tex] for [tex]h>\frac{mg}{k}+\frac{1}{2}[/tex]

    I do indeed have a symmetry question (check your private messages). I wasn't really too sure on whether this was a cusp catastrophe or not. I couldn't use any of the equations for the canonical cusp catastrophe, but was inclined to believe that it's a cusp catastrophe, as we just from a maximum to a mininum and two maximums (basically, the opposite of the canonical cusp catastrophe: just the same situation with all equations made negative). But, that's my only real argument....

    :confused:
     
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