Elastic rod problem (having some math issue)

[kev931210]

Homework Statement

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

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.

 

[TSny]

Welcome to PF!

View attachment 97130

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.

OK

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))

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

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

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

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;

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.]

 

[kev931210]

Welcome to PF!

OKDid you overlook that ##\frac{\partial{\mathbf{r}}}{\partial{s}}## is squared in the expression for F?
I did not overlook. I differentiated t (which equals dr/ds) with respect to s, and I squared that expression.[This is not the correct expression for dθ/ds. Typo?]

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..

 

[TSny]

Note the power of 2 shown below.

 

[kev931210]

Note the power of 2 shown below.

(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))

 

[TSny]

##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.]

 

[kev931210]

##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.]

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.

 

[TSny]

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.

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##.