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Elastic rod problem (having some math issue)

  1. Mar 10, 2016 #1
    Capture.PNG 1. The problem statement, all variables and given/known data
    I figured out the first part of the question, proving why |t| equals 1, but I have trouble solving the next part of the problem. I expressed F(r(s)) in terms of theta, but I cannot solve for a, b, and c using the equation I derived.

    2. Relevant equations

    Free energy minimization.
    Change of variable for a 2D geometry

    3. The attempt at a solution
    I first attemtped to convert the given equation F(r(s))= ∫(ds 1/2*k(d2r/ds2)^2) using polar coordinate.
    In order to replace (d2r/ds2)^2, I differentiated dr/ds=(cosθ(s),sinθ(s)) respect to s.
    (d2r/ds2)^2=(-dθ/ds *sinθ(s), dθ/ds*cosθ(s))^2 = (dθ/ds)^2*sin^2(θ(s))+(dθ/ds)^2*cos^2(θ(s))=(dθ/ds)^2.

    Then to replace ds, ds=dθ*ds/dθ.
    Eventually I turned F(r(s))= ∫(ds 1/2*k(d2r/ds2)^2,s=0 to L) into F=∫(dθ 1/2*k(dθ/ds), θ=θ(0) to θ(l))
    Is this correct?? Well, I thought it was a very simple and beautiful answer, but I could not solve the next problem using this equation.

    I do not know how I can minimize F=∫(dθ 1/2*k(dθ/ds), θ=θ(0) to θ(l)) when θ is given as the polynomial equation. By plugging in the polynomial equation,
    θ=b*s+c*s^2, a=0 due to initial condition θ(0)=0
    dθ/ds==s+2cs

    F=∫(dθ 1/2*k(dθ/ds)=F=∫(dθ 1/2*k*(b+2cs)). Then I expressed 'b' in terms of c using the initial condition θ(l)=0;

    This is where I am stuck... Could you please help me .. I have been struggling with it all day long.
     
  2. jcsd
  3. Mar 10, 2016 #2

    TSny

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    Welcome to PF!

    OK

    Did you overlook that ##\frac{\partial{\mathbf{r}}}{\partial{s}}## is squared in the expression for F?


    This is not the correct expression for dθ/ds. Typo?

    Overall, your approach looks correct. What do you get for F after making the corrections mentioned above?

    [Another approach that gets to the answer faster would be to use the Euler-Lagrange equation from calculus of variations. But it is not necessary for this problem.]
     
  4. Mar 10, 2016 #3
    That's a typo, but I plugged in the correct expression to F.



    [Overall, your approach looks correct. What do you get for F after making the corrections mentioned above?][/QUOTE]

    I still get this same equation, F=∫(dθ 1/2*k(dθ/ds)=F=∫(dθ 1/2*k*(b+2cs)), where θ(s)=a+bs+cs^2. I don't know how I can move further from this point..
     
  5. Mar 10, 2016 #4

    TSny

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    Note the power of 2 shown below.
     

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  6. Mar 10, 2016 #5
    (d2r/ds2)^2=(-dθ/ds *sinθ(s), dθ/ds*cosθ(s))^2 = (dθ/ds)^2*sin^2(θ(s))+(dθ/ds)^2*cos^2(θ(s))=(dθ/ds)^2

    Yes, from the above equatoin, I concluded (d2r/ds2)^2=(dθ/ds)^2 .
    But I also replaced ds with dθ*ds/dθ.

    ds (d2r/ds2)^2 --> dθ*ds/dθ * (dθ/ds)^2 --> dθ*dθ/ds.

    This is how I arrived at F=∫(dθ 1/2*k(dθ/ds).

    By differentiating θ(s)=a+bs+cs^2 with respect to s, dθ/s=b+2cs,

    F=∫(dθ 1/2*k(dθ/ds)=∫(dθ 1/2*k*(b+2cs))
     
  7. Mar 10, 2016 #6

    TSny

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    ##s## is the independent variable and ##\theta## is the dependent variable. So, the integration should be with respect to ##s##, not ##\theta##

    [Sorry I overlooked your change of variable in the integration. But you need to keep ##s## as the integration variable.]
     
    Last edited: Mar 10, 2016
  8. Mar 10, 2016 #7

    wow... Thank you so much! I feel so stupid haha. I eventually found 'a' and 'c' to be 0, so θ(s)=b*s for the minimum energy configuration, under the constraint that θ(s) is a quadratic equation of s. Do you think this is a reasonable answer? It appears to be a circle to me.
     
  9. Mar 10, 2016 #8

    TSny

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    Yes, I believe a circular arc is right. At least that's what I got when I worked it. The answer seems reasonable to me. I think the circular arc is the general answer (for these boundary conditions) even if you don't assume a quadratic dependence of ##\theta## on ##s##.
     
    Last edited: Mar 10, 2016
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