Bend a bow strongly

Let us place the bow in a reference frame as shown in the Figure. The shape of the bow is described by the $y(x)$ function, the dashed line is at $x=d$ . The torque acting at a cross section of the bow is $\tau=F x$ , where $F$ is the tension force in the string. The inverse of the radius of curvature $\rho$ is proportional to the torque. The inverse radius of curvature can be expressed with the derivatives of $y(x)$ in several different ways; here we list two:

$\frac{1}{\rho}=\frac{y''}{(1+y'^2)^{3/2}} \qquad$ Eq.1

$\frac{1}{\rho}= \frac {d^2y}{dx ds} \qquad$ Eq. 2

The first equation is the standard formula from differential geometry. The second equation has an unusual "mixed" derivative, where $ds$ is a line segment along the bow. We can derive it from Eq. 1 by using $ds^2=dx^2+dy^2$ .

If we use the first expression we get a second order nonlinear differential equation for the shape of the bow in the form of

$\frac{y''}{(1+y'^2)^{3/2}}=2c x$ ,

where $2c$ is a constant that depends on the force, the elastic parameters and the cross section of the bow. The solution of this differential equation is by no means simple. The result is the so called "elastica" curve, first derived in a special case by James Bernoulli and in the most general case by Euler. Actually, the special case Bernoulli derived corresponds to the boundary conditions imposed in the problem here. Our solution is based on his ideas. (For a great historic perspective see this link ).

Instead of Eq. 1, we use Eq. 2 for the radius of curvature:

$\frac {d^2y}{dx ds}=2cx$ .

We integrate with respect to $x$ to get

$\frac {dy}{ ds}=cx^2 \qquad$ Eq. 3

that satisfies the boundary condition of $dy/ds=0$ at $x=0$ . We can use the identity $ds^2=dx^2+dy^2$ again to express $\frac{dy}{dx}$ with $\frac{dy}{ds}$ :

$\frac{dy}{dx}=\frac{dy/ds}{\sqrt{1-(dy/ds)^2}}$

and use Eq. 3 to get

$\frac {dy}{dx}= \frac{cx^2}{\sqrt{1-c^2x^4}}$ , or

$\frac {dy'}{dx'}= \frac{x'^2}{\sqrt{1-x'^4}}$ ,

if we introduce the scaled variables $x'=\sqrt{c} x$ and $y'=\sqrt{c} y$ . The solution is

$y'=\int_0^{x'} \frac{\eta^2 d\eta}{\sqrt{1-\eta^4}}$

The result can be expressed in terms of elliptic integrals. The point $x'=1$ corresponds to the middle point of the bow, $d=1$ . The corresponding $y'$ value is $\ell/2$

$\frac{\ell}{2}=\int_0^{1} \frac{\eta^2 d\eta}{\sqrt{1-\eta^4}}=E(-1)-K(-1)\approx 0.59907$

where $K$ and $E$ are elliptic integrals of the first and second kind, respectively. Therefore $\ell=1.19814$ and the ratio of the two measurements is $d/\ell= 0.8346$

Read

"Physics of Continuous Matter. Exotic and everyday phenomena in the macroscopic world" by Benny Lautrup. The Niels Bohr Institute.

Slender Rods, Chapter-10, Section-10.3 and Figure 10.4 in the following link:

http://www.cns.gatech.edu/~predrag/GTcourses/PHYS-4421-04/lautrup/2.8/rods.pdf.

Take the Bow Angle=(π/2). Find the "Integral dx √cos(x) from 0 to π/2" = 1.19814, using WolframAlpha.

Use Equation-10.25 to calculate d and L.

If wave number is taken as unity, d=√2 and L=(√2)(1.19814). Therefore,

d/L=(1/1.19814).

Answer~0.834