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Consider a thin plate with uniform density.

The plate is bounded above by the $x$-axis on the interval $[-a,a]$ and below by unknown, continous curve $y=f(x)$, so that $f(-a)=f(a)=0$

Then the vertical coordinate of the center of mass of the plate may be presented as

$$I[y]=\frac{\int_{-a}^{a}y^2dx}{2\int_{-a}^{a}y\,dx}$$

where $y=f(x)$ is still unknown function.

What is known is that the length of the curve $y=f(x)$ on the interval $[-a,a]$ must be $L>2a$.

Or formally:

$$\int_{-a}^{a}\sqrt{1+\left (\frac{dy}{dx}\right )^2}dx=L>2a$$

Now i want to find such a function $y=f(x)$ that gives minimum value of $I[y]$.

Note that $I[y]<0$ because the plate is located below $x$-axis.

So we have a constrained variational problem.

But the real problem here is the structure of the functional $I[y]$. It seems that standard methods do not work here.

In any case, based on the physical considerations there must exist unique solution.

Suggestions or even solutions ?

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  • $\begingroup$ In view of @Andrei's comment, could you confirm that $L$ is a constant and also that $y$ is differentiable? $\endgroup$ – mucciolo Dec 31 '17 at 3:12
  • $\begingroup$ @mucciolo Yes L is constant. $\endgroup$ – Martin Gales Jan 2 '18 at 16:19
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I will just give a hint, since it looks like a lot of work. You need to think of this as a minimization problem depending on multiple variables: what you need is to minimize $$I[y,I_1,I_2]=\frac{I_1}{I_2}$$ subject to the following constraints: $$I_1=\int_{-a}^ay^2dx\\I_2=\int_{-a}^a2ydx\\L=\int_{-1}^a\sqrt{1+y'^2}dx$$ Disclaimer: I've got the idea from this paper (chapter 4, page 17)

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  • $\begingroup$ Hi, thank you for the answer. I need to think about this paper. $\endgroup$ – Martin Gales Jan 2 '18 at 16:28

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