A rubber band has mass m and force constant k. When relaxed, the band forms a ring of radius r. The band is placed horizontally on a vertical frictionless cone as shown (angle $2\theta$ is given). The radius R of the ring formed by the band is  $R=r+\frac{mg}{\alpha {\pi }^{2}k}\cot \theta$. Find the value of  $\alpha$

Consider the potential energy of the band as a function of R. To be in equilibrium, the band must be at a position of minimum potential energy. Let the elastic potential energy be $U_i=\frac{1}{2}k{(2\pi R-2\pi R)}^2$ and the gravitational potential energy measured with respect to the vertex of the code  $U_i=-mg(R-r)\cot \theta$
The total potential energy is  $U=2k{\pi }^2{(R-r)}^2-mg(R-r)\cot \theta$
At the minimum position, 
$\frac{dU}{dR}=4{\pi }^{2}k(R-r)-mg\cot \theta =0$
Which has solution
$R=r+\frac{mg}{4{\pi }^{2}k}\cot \theta$