Problem source

A rubber band band of length $2\pi$ and modulus of elasticity $\lambda$ encircles a smooth cylinder of unit radius, whose axis is horizontal.
A particle of mass $m$ is attached to the lowest point of the band, and hangs in equilibrium at a distance $x$ below the axis of the cylinder. Obtain an expression in terms of $x$ for the stretched length of the band in equilibrium. 
What is the value of $\lambda$ if $x=2$?

Solution source

\begin{center}


\begin{tikzpicture}[scale=1]
    \coordinate (O) at (0,0);
    \def\a{40};
    \def\r{3};
    \coordinate (X) at (0, {-\r/cos(\a)});
    \coordinate (X1) at ({\r*cos(360-90+\a)},{\r*sin(360-90+\a)});
    \coordinate (X2) at ({\r*cos(360-90-\a)},{\r*sin(360-90-\a)});
    
    \draw (O) circle (\r);
    \draw (X1) -- (X) -- (X2);
    \draw[dashed] (X) -- (O) -- (X1);
    \draw[dashed] (O) -- (X2);
    \pic [draw, angle radius=.7cm, "$\alpha$"] {angle = X--O--X1};

    \draw[-latex, blue, ultra thick] (X) -- ++($0.5*(X1)-0.5*(X)$) node[right] {$T$};
    \draw[-latex, blue, ultra thick] (X) -- ++($0.5*(X2)-0.5*(X)$) node[right] {$T$};

    \draw[-latex, blue, ultra thick] (X) -- ++(0,{-\r/2}) node[right] {$mg$};
    
\end{tikzpicture}

\end{center}

If $\alpha$ is as labelled then $\cos \alpha = \frac{1}{x}, \sin \alpha = \frac{\sqrt{x^2-1}}{x}, \tan \alpha = \sqrt{x^2-1}$. We also have the full length of the rubber band is $2\pi - 2\alpha +2\tan \alpha$ so the extension is $2 \l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r$

Therefore $T = \frac{\l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\lambda}{\pi}$.

If $x = 2$, $T = \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda, \sin \alpha = \frac{\sqrt{3}}{2}$
\begin{align*}
\text{N2}(\uparrow): && 2T\sin \alpha - mg &= 0 \\
\Rightarrow && \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda \sqrt{3} &= mg \\
\Rightarrow && \lambda &= \frac{\sqrt{3}\pi}{(3\sqrt{3}-\pi)}mg
\end{align*}